Answer:
The distance traveled by the sperm whale is 12.46 meters.
Explanation:
Given that,
Acceleration of a sperm whale, [tex]a=0.12\ m/s^2[/tex]
Initial speed of the whale, u = 1.1 m/s
Final speed of the whale, v = 2.05 m/s
We need to find the distance traveled by the sperm whale. It can be calculated using third equation of motion as :
[tex]v^2-u^2=2ad[/tex]
d is the distance covered
[tex]d=\dfrac{v^2-u^2}{2a}[/tex]
[tex]d=\dfrac{(2.05)^2-(1.1)^2}{2\times 0.12}[/tex]
d = 12.46 meters
So, the distance traveled by the sperm whale is 12.46 meters. Hence, this is the required solution.
A small glass bead charged to 5.0 nCnC is in the plane that bisects a thin, uniformly charged, 10-cmcm-long glass rod and is 4.0 cmcm from the rod's center. The bead is repelled from the rod with a force of 840 μNμN.What is the total charge on the rod?
Answer:
The total charge on the rod is 47.8 nC.
Explanation:
Given that,
Charge = 5.0 nC
Length of glass rod= 10 cm
Force = 840 μN
Distance = 4.0 cm
The electric field intensity due to a uniformly charged rod of length L at a distance x on its perpendicular bisector
We need to calculate the electric field
Using formula of electric field intensity
[tex]E=\dfrac{kQ}{x\sqrt{(\dfrac{L}{2})^2+x^2}}[/tex]
Where, Q = charge on the rod
The force is on the charged bead of charge q placed in the electric field of field strength E
Using formula of force
[tex]F=qE[/tex]
Put the value into the formula
[tex]F=q\times\dfrac{kQ}{x\sqrt{(\dfrac{L}{2})^2+x^2}}[/tex]
We need to calculate the total charge on the rod
[tex]Q=\dfrac{Fx\sqrt{(\dfrac{L}{2})^2+x^2}}{kq}[/tex]
Put the value into the formula
[tex]Q=\dfrac{840\times10^{-6}\times4.0\times10^{-2}\sqrt{(\dfrac{10.0\times10^{-2}}{2})^2+(4.0\times10^{-2})^2}}{9\times10^{9}\times5.0\times10^{-9}}[/tex]
[tex]Q=47.8\times10^{-9}\ C[/tex]
[tex]Q=47.8\ nC[/tex]
Hence, The total charge on the rod is 47.8 nC.
Final answer:
Using Coulomb's Law, the charge on the rod is calculated to be 7.5 *10⁻⁶ C by rearranging the formula and solving for the rod's charge with the provided force and the charge on the bead.
Explanation:
To determine the total charge on the rod, we can use Coulomb's Law, which states that the electric force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. In this case, the glass bead is assumed to act like a point charge, and we can treat the charge distribution on the thin glass rod as uniform.
The formula for Coulomb's Law is:
F = k * |q₁ * q₂| / r²
where:
F is the magnitude of the force between the charges,
k is Coulomb's constant (8.988 * 10⁹ N m²/C²),
q₁ and q₂ are the amounts of the charges, and
r is the distance between the centers of the two charges.
Given that the force F is 840 * 10⁻⁶ N and charge on bead q₁ is 5.0 * 10⁻⁹ C, and the distance r is 4.0 * 10⁻² m, we can rearrange the formula to solve for the charge on the rod (q₂):
q₂ = F * r² / (k * q₁)
Plugging in the values, we can calculate as follows:
q₂ = (840 *10⁻⁶ N) * (4.0 *10^⁻² m)² / (8.988 *10⁹ N m²/C² * 5.0 *10⁻⁹ C)
Calculating the charge on the rod:
q₂ = 0.0075 C
or
q₂ = 7.5 * 10⁻⁶ C
This value represents the total charge on the rod.
Apply the junction rule to the junction labeled with the number 1 (at the bottom of the resistor of resistance R2). Answer in terms of given quantities, together with the meter readings I1 and I2 and the current I3.
Answer:
This is the correct question.
Apply the junction rule to the junction labeled with the number1 (at the bottom of the resistor of resistance R_2).
Answer in terms of given quantities,together with the meter readings I_1 and I_2 and the current I_3.
b) Apply the loop rule to loop 2 (the smaller loop on the right).Sum the voltage changes across each circuit element around this loop going in the direction of the arrow. Remember that the current meter is ideal.
Express the voltage drops in terms ofV_b, I_2, I_3, the given resistances, and any other givenquantities.
c) Now apply the loop rule to loop 1(the larger loop spanning the entire circuit). Sum the voltagechanges across each circuit element around this loop going in thedirection of the arrow.
Express the voltage drops in terms ofV_b, I_1, I_3, the given resistances, and any other givenquantities.
Explanation:
a. Junction rule: Kirchhoff current law (KCL)
Then, total current entering a junction equals to current leaving the junction.
Therefore,
i1=i2+i3
b. Apply Kirchoff voltage law ( KVL) to loop 2
Then, sum of voltage in the loop equals to zero.
-i3R3+i2R2=0.
Then,
i2R2=i3R3
c. Apply KVL to the loop 1
-Vb+i1R1+i3R3
Therefore,
Vb=i1R1+i3R3.
The circuit diagram is in the attachment
Kirchhoff's first rule, or the junction rule, states that all currents entering a junction must equal all currents leaving that junction. For the network with a junction labeled 1 and currents I1, I2, and I3, the application of the rule yields the equation: I1 = I2 + I3.
Explanation:To apply Kirchhoff's first rule, also known as the junction rule, we need to consider that all currents entering a junction must equal all currents leaving that junction. This principle is based on the conservation of charge where the total amount of electricity is maintained. Given the question, we are working with a junction labeled 1 at the bottom of the resistor of resistance R2, with meter readings I1 and I2, and current I3.
According to the junction rule for this scenario, we would express this as: I1 = I2 + I3, where I1 is the current flowing into the junction, and I2 and I3 are currents flowing out. This equation states that the current entering the junction (I1) must be equal to the sum of the currents leaving the junction ( I2 and I3).
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A baseball of mass m1 = 0.45 kg is thrown at a concrete block m2 = 7.25 kg. The block has a coefficient of static friction of μs = 0.74 between it and the floor. The ball is in contact with the block for t = 0.185 s while it collides elastically.
a. Write an expression for the minimum velocity the ball must have, vmin, to make the block move.
b. What is the velocity in m/s?
Answer:
a. [tex]v = \frac{mu_sm_2g\Delta t}{m_1}[/tex]
b. 21.64 m/s
Explanation:
Let g = 9.81m/s2
a. The weight of the block is product of its mass and gravitational acceleration
[tex]W = m_2g = 7.25*9.81 = 71.1225N[/tex]
which is also the normal force acting on the block from the floor so it stays balanced.
N = 71.225N
The static friction of the block is product of its normal force from the floor and the friction coefficient
[tex]F_s = \mu_sN = \mu_sW = mu_sm_2g[/tex]
For the block to move, the force generated by the impact must be at least equal to the static friction.
[tex]F = F_s = mu_sm_2g[/tex]
The impulse is product of this force and time duration of impact.
[tex]I = F\Delta t = mu_sm_2g\Delta t[/tex]
As impulse is generated by change in momentum of the ball, which is product of its mass and velocity v
[tex]I = \Delta p = m_1\Delta v[/tex]
[tex]mu_sm_2g\Delta t = m_1 v[/tex]
[tex]v = \frac{mu_sm_2g\Delta t}{m_1}[/tex]
b. [tex]v = \frac{mu_sm_2g\Delta t}{m_1} = \frac{0.74*7.25*9.81*0.185}{0.45} = 21.64 m/s[/tex]
From Newton's second law, the minimum velocity the ball must have, to make the block move is 21 m/s
COLLISION
There are two types of collision
Elastic collisionInelastic collisionIn elastic collision, both momentum and energy are conserved.
Given that a baseball of mass m1 = 0.45 kg is thrown at a concrete block m2 = 7.25 kg. The block has a coefficient of static friction of μs = 0.74 between it and the floor. The ball is in contact with the block for t = 0.185 s while it collides elastically.
For the concreate block to move, the force applied must be greater than the friction between the block and the floor.
The frictional force = μN
where N = mg
Friction = 7.25 x 9.8 x 0.72
Friction = 51.156
Let assume that the force applied will be equal to the friction. From Newton's second law,
F = Change in momentum / time taken.
That is
F = [tex]m_{1}[/tex]V / t
since the ball is starting from rest, the initial velocity u = 0
a. The expression for the minimum velocity the ball must have, to make the block move will be
F = [tex]m_{1}[/tex]V / t
Make V the subject of formula
V = Ft / [tex]m_{1}[/tex]
substitute F into the formula
V = μ[tex]m_{2}[/tex]g t / [tex]m_{1}[/tex]
b. The velocity in m/s will be calculated by substituting all the parameters into the formula above.
V = (51.156 x 0.185) / 0.45
V = 9.46386 / 0.45
V = 21 m/s
Therefore, the minimum velocity the ball must have, to make the block move is 21 m/s
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For the following statements, choose the word or words inside the parentheses that serve to make a correct statement. Each statement has at least one and may have more than one correct answer. a. For a sample of an ideal gas, the product pV remains constant as long as the (temperature, pressure, volume, internal energy) is held constant. b. The internal energy of an ideal gas is a function of only the (volume, pressure, temperature). c. The Second Law of Thermodynamics states that the entropy of an isolated system always (increases, remains constant, decreases) during a spontaneous process. d. When a sample of liquid is converted reversibly to its vapor at its normal boiling point, ( q, w, p, V, T, U, H, S, G, none of these) is equal to zero for the system. e. If the liquid is permitted to vaporize isothermally and completely into a previously evacuated chamber that is just large enough to hold the vapor at 1 bar pressure, then ( q, w, U, H, S, G ) will be smaller in magnitude than for the reversible vaporizatio
Answer:
a) Temperatura, b) Temperature, c) Constant , d) None of these , e) Gibbs enthalpy and free energy (G)
Explanation:
a) the expression for ideal gases is PV = nRT
Temperature
b) The internal energy is E = K T
Temperature
c) S = ΔQ/T
In an isolated system ΔQ is zero, entropy is constant
Constant
d) all parameters change when changing status
None of these
e) Gibbs enthalpy and free energy
"With the parents from the previous problem (white WW, black ww) give the phenotype and genotype ratios for this cross."
Explanation:
p1: Ww x ww
gametes: w,W w,w
.......cross......
F1: Ww (2) and ww (2)
(white) (black)
therefore,
genotype ratio - 1: 1
phenotype ratio- 1: 1
The offspring of a cross between a WW (white) and ww (black) parent will all have the genotype Ww and a white phenotype, resulting in a 100% phenotype ratio for white and a 100% genotype ratio for Ww.
When analyzing the cross between a parent with the genotype WW (white) and another with the genotype ww (black), we consider the basic principles of Mendelian genetics. Since white WW is homozygous dominant and black ww is homozygous recessive, all the offspring will have the genotype Ww, making them all white, due to the complete dominance of the white allele.
The phenotype ratio for this cross would be 100% white offspring. The genotype ratio would be 100% heterozygous Ww, because each offspring inherits one W allele from the white parent and one w allele from the black parent.
A pipe is subjected to a tension force of P = 90 kN. The pipe outside diameter is 45 mm, the wall thickness is 5 mm, and the elastic modulus is E = 150 GPa. Determine the normal strain in the pipe. Express the strain in mm/mm.
Final answer:
The normal strain in the pipe subjected to a tension force is calculated using Hooke's Law, which relates stress to strain via the elastic modulus. The cross-sectional area is found by the area difference between outer and inner circles, considering the pipe's diameter and wall thickness. The final strain is the normal stress divided by the elastic modulus.
Explanation:
To determine the normal strain in the pipe due to the tension force, we can use Hooke's Law, which relates stress and strain via the elastic modulus. Normal strain (ε) is calculated by dividing the normal stress (σ) by the elastic modulus (E). The normal stress can be found by dividing the tension force (P) by the cross-sectional area (A) of the pipe.
The cross-sectional area for a pipe is the area of a ring, which can be calculated as the area of the outer circle minus the area of the inner circle. Given the outside diameter (do) is 45 mm and wall thickness (t) is 5 mm, the inside diameter (di) will be do - 2t = 35 mm. So the area is calculated with A = π/4 • (do2 - di2).
After the area is determined, the normal stress is P/A and therefore the strain is ε = σ/E. Keep in mind to convert E to the correct units to match those of stress (N/mm²).
For the provided tension force P = 90 kN and elastic modulus E = 150 GPa, the strain calculation will use these values and the cross-sectional area calculated earlier.
A small glass bead has been charged to + 30.0 nC . A small metal ball bearing 2.60 cm above the bead feels a 1.80×10−2 N downward electric force. Part A What is the charge on the ball bearing? Express your answer with the appropriate units.
Answer:
The charge on the ball bearing 4.507 × 10^-8 C
Explanation:
From Coulomb's law
F = kq1q2/r²
make q2 the subject
q2 = Fr²/kq1
q2 = (1.8×10^-2 × 0.026²) ÷ (9×10^9 × 30×10^-9)
q2 = 4.507 × 10^-8 C
Using Coulomb's Law, the charge on the ball bearing is found to be approximately -1.92 x 10^-8 C. The negative sign indicates that the charges on the bead and the ball bearing are of opposite nature.
Explanation:To determine the charge on the ball bearing, we will use Coulomb's Law, that implies that the product of the two charges divided by their distance from one another is what determines the force between two charges. The formula is: F = k * |q1*q2| / r^2 where F is the electric force, q1 and q2 are the charges, r is the distance between the charges, and k is Coulomb's Constant (8.988x10^9 N * m^2/C^2).
We can rearrange the equation to find the charge on the ball bearing (q2): q2 = F * r^2 / (k * q1)
Substituting the given values: q2 = 1.80x10^-2 N * (2.6x10^-2 m)^2 / (8.988x10^9 N * m^2/C^2 * 3.0x10^-8 C)
With some calculations, we get that the charge on the ball bearing, q2 is approximately -1.92 x 10^-8 C. The negative sign represents the opposite nature of the charges, meaning the bead and ball bear opposite charges.
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In the middle of the night you are standing a horizontal distance of 14.0 m from the high fence that surrounds the estate of your rich uncle. The top of the fence is 5.00 m above the ground. You have taped an important message to a rock that you want to throw over the fence. The ground is level, and the width of the fence is small enough to be ignored. You throw the rock from a height of 1.60 m above the ground and at an angle of 56.0o above the horizontal. (a) What minimum initial speed must the rock have as it leaves your hand to clear the top of the fence? (b) For the initial velocity calculated in part (a), what horizontal distance be-yond the fence will the rock land on the ground?
Answer:
Explanation:
a ) Height to be cleared = 5 - 1.6 = 3.4 m
Horizontal distance to be cleared = 5 m .
angle of throw = 56°
here y = 3.4 , x = 5 , θ = 56
equation of trajectory
y = x tanθ - 1/2 g ( x/ucosθ)²
3.4 = 5 tan56 - 1/2 g ( 5/ucos56)²
3.4 = 7.4 - 122.5 / .3125u²
122.5 / .3125u² = 4
u² = 98
u = 9.9 m /s
Range = u² sin 2 x 56 / g
= 9.9 x 9.9 x .927 / 9.8
= 9.27 m
horizontal distance be-yond the fence will the rock land on the ground
= 9.27 - 5
= 4.27 m
Answer:
a) [tex]u=13.3032\ m.s^{-1}[/tex]
b) [tex]s=3.7597\ m[/tex]
Explanation:
Given:
horizontal distance between the fence and the point of throwing, [tex]r=14\ m[/tex]
height of the fence from the ground, [tex]h_f=5\ m[/tex]
height of projecting the throw above the ground, [tex]h'=1.6\ m[/tex]
angle of projection of throw from the horizontal, [tex]\theta=56^{\circ}[/tex]
Let the minimum initial speed of projection of the throw be u meters per second so that it clears the top of the fence.Now the effective target height, [tex]h=h_f-h'=5-1.6=3.4\ m[/tex]The horizontal component of the velocity that remains constant throughout the motion:
[tex]u_x=u\cos\theta[/tex]
Now the time taken to reach the distance of the fence:
use equation of motion,
[tex]t_f=\frac{r}{u_x}[/tex]
[tex]t_f=\frac{14}{u.\cos56}[/tex] .................................(1)
Now the time taken to reach the fence height (this height must be attained on the event of descending motion of the rock for the velocity to be minimum).
Maximum Height of the projectile:
[tex]v_y^2=u_y^2-2\times g.h_m[/tex]
[tex]h_m=\frac{u_y^2}{19.6}[/tex] ........................(4)
Now the height descended form the maximum height to reach the top of the fence:
[tex]\Delta h=h_m-h'[/tex]
[tex]\Delta h=(\frac{u_y^2}{19.6} -3.4)\ m[/tex]
time taken to descent this height from the top height:
[tex]\Delta h=u_{yt}.t_d+\frac{1}{2} \times g.t_d^2[/tex]
where:
[tex]u_{yt}=[/tex] initial vertical velocity at the top point
[tex]t_d=[/tex] time of descend
[tex](\frac{u_y^2}{19.6} -3.4)=0+0.5\times 9.8\times t_d^2[/tex]
[tex]t_d=\sqrt{(\frac{u_y^2}{96.04} -\frac{3.4}{4.9} )}[/tex]..............................(2)
So we find the time taken by the rock to reach the top of projectile where the vertical velocity is zero:
[tex]v_y=u_y-g.t_t[/tex]
where:
[tex]u_y=[/tex] initial vertical velocity
[tex]v_y=[/tex] final vertical velocity
[tex]t_t=[/tex] time taken to reach the top height of the projectile
[tex]0=u_y-g.t_t[/tex]
[tex]t_t=\frac{u_y}{9.8}\ seconds[/tex] .................................(3)
Now the combined events of vertical and horizontal direction must take at the same time as the projectile is thrown:
So,
[tex]t_f=t_t+t_d[/tex]
[tex]\frac{14}{u.\cos56}=\frac{u_y}{9.8} +\sqrt{(\frac{u_y^2}{96.04} -\frac{3.4}{4.9} )}[/tex]
[tex]\frac{14}{u.\cos56}=\frac{u\sin56}{9.8} +\sqrt{(\frac{(u.\sin56)^2}{96.04} -\frac{3.4}{4.9} )}[/tex]
[tex]\frac{196}{u^2.\cos^2 56} +\frac{u^2\sin^2 56}{96.04} -2.857\times \tan56=\frac{u^2\sin^2 56}{96.04} -0.694[/tex]
[tex]u=13.3032\ m.s^{-1}[/tex]
Max height:
[tex]h_m=\frac{(u.\sin 56)^2}{19.6}[/tex]
[tex]h_m=\frac{(13.3032\times \sin56)^2}{19.6}[/tex]
[tex]h_m=6.2059\ m[/tex]
Now the rock hits down the ground 1.6 meters below the level of throw.
Time taken by the rock to fall the gross height [tex]h_g=h_m+h'[/tex]:
[tex]h_g=u_{yt}.t_g+\frac{1}{2} g.t_g^2[/tex]
[tex]7.8059=0+0.5\times 9.8\times t_g^2[/tex]
[tex]t_g=1.2621\ s[/tex]
Time taken to reach the the top of the fence from the top, using eq. (2):
[tex]t_d=\sqrt{(\frac{u_y^2}{96.04} -\frac{3.4}{4.9} )}[/tex]
[tex]t_d=\sqrt{(\frac{(u.\sin56)^2}{96.04} -\frac{3.4}{4.9} )}[/tex]
[tex]t_d=0.7567\ s[/tex]
Time difference between falling from top height and the time taken to reach the top of fence:
[tex]\Delta t=t_g-t_d[/tex]
[tex]\Delta t=1.2621-0.7567[/tex]
[tex]\Delta t=0.5054\ s[/tex]
b)
Now the horizontal distance covered in this time:
[tex]s=u.\cos56\times\Delta t[/tex]
[tex]s=13.3032\times \cos56\times 0.5054[/tex]
[tex]s=3.7597\ m[/tex] is the horizontal distance covered after crossing the fence.
A supersonic airplane is flying horizontally at a speed of 2570 km/h. What is the centripetal acceleration of the airplane, if it turns from North to East on a circular path with a radius of 80.5 km? Submit Answer Tries 0/12 How much time does the turn take? Submit Answer Tries 0/12 How much distance does the airplane cover during the turn?
The questions deal with centripetal acceleration, speed, and the path of a supersonic airplane making a turn. The key is to use the formulae for centripetal acceleration, circular motion, and circle properties to calculate the desired quantities.
Explanation:The questions pertain to the physics concept of centripetal acceleration. The centripetal acceleration of a body moving in a circular path is given by the equation a = v2/r. In the case of the supersonic airplane, we can plug in the given values of speed (v = 2570 km/h = 713.89 m/s) and radius (r = 80.5 km = 80500 m) into this formula to calculate the centripetal acceleration.
Next, to find the time it would take for the airplane to turn from North to East, we need to understand that the airplane is essentially making a 90-degree turn, or 1/4 of a full circular path. Therefore, the time would be 1/4 of the total time it would take to complete a full circle (T = 2πr/v). The distance covered during the turn would also equivalently be 1/4 of the total circumference of the path, which we can calculate using the formula for the circumference of a circle (C = 2πr).
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An object initially at rest experiences an acceleration of 1.90 m/s² for 6.60 s then travels at that constant velocity for another 8.50 s. What is the magnitude of the object's average velocity over the 15.1 s interval?
Answer:
The magnitude of the object's average velocity is 9.74 m/s (9.80 m/s without any intermediate rounding).
Explanation:
Hi there!
The average velocity is calculated as the displacement of the object divided by the time it takes the object to do that displacement.
The displacement is calculated as the distance between the final position of the object and the initial position. In this problem, the displacement is equal to the traveled distance because the object travels only in one direction:
a.v = Δx/t
Where:
a.v = average velocity.
Δx = displacement = final position - initial position
t = time
So, let's find the distance traveled while the object was accelerating. For that, we will use this equation:
x = x0 + v0 · t + 1/2 · a · t²
Where:
x = position of the object at time t.
x0 = initial position.
v0 = initial velocity.
t = time.
a = acceleration.
In this case, since the object is initially at rest, v0 = 0. If we place the origin of the frame of reference at the point where the object starts moving, then x0 = 0. So, the equation of the position of the object after a time t will be:
x = 1/2 · a · t²
x = 1/2 · 1.90 m/s² · (6.60 s)²
x = 41.4 m
The object traveled 41.4 m during the first 6.60 s.
Now, let's find the rest of the traveled distance.
When the velocity is constant, a = 0. Then, the equation of position will be:
x = x0 + v · t
Let's place now the origin of the frame of reference at the point where the object starts traveling at constant velocity so that x0 = 0:
x = v · t
The velocity reached by the object during the acceleration phase is calculated as follows:
v = v0 + a · t (v0 = 0 because the object started from rest)
v = 1.90 m/s² · 6.60 s
v = 12.5 m/s
Then, the distance traveled by the object at a constant velocity will be:
x = 12.5 m/s · 8.50 s
x = 106 m
The total traveled distance in 15.1 s is (106 m + 41.4 m) 147 m.
Then the displacement will be:
Δx = final position - initial position
Δx = 147 m - 0 = 147 m
and the average velocity will be:
a.v = Δx/t
a.v = 147 m / 15.1 s
a.v = 9.74 m/s
The magnitude of the object's average velocity is 9.74 m/s (9.80 m/s without any intermediate rounding).
At the proportional limit, a 2 in. gage length of a 0.500 in. diameter alloy rod has elongated 0.0035 in. and the diameter has been reduced by 0.0003 in. The total tension force on the rod was 5.45 kips. Determine the following properties of the material: (a) the proportional limit. (b) the modulus of elasticity. (c) Poisson’s ratio
Answer:
a)Proportional limit = 27.756ksi
b) The modulus of elasticity = 15860ksi
c)Poisson ratio = 0.343
Explanation:
The detailed steps and calculation is as shown in the attached file.
A wave has the mathematical form . What is the displacement of a particle at the origin after a time ?
The given question is incomplete. The complete question is as follows.
A wave has the mathematical form y = [tex](0.2 m) sin(2\pi ft - \frac{2 \pi x}{\lambda}[/tex]. What is the displacement of a particle at the origin after a time [tex]t = \frac{1}{8T}[/tex]?
Explanation:
Let us assume that at origin, x = 0 and the value of t is given as [tex]\frac{1}{8T}[/tex].
Therefore, we will calculate the value of displacement as follows.
y = [tex]0.2 \times sin[\frac{2 \pi}{T} \times \frac{1}{8}T - \frac{2 \pi}{\lambda (0)}][/tex]
= [tex]0.2 sin [\frac{\pi}{4} - 0][/tex]
= 0.1414 m
= 0.14 (approx)
Thus, we can conclude that displacement of a particle at the origin after a time [tex]t = \frac{1}{8T}[/tex] is 0.14.
A major leaguer hits a baseball so that it leaves the bat at a speed of 30.0 m/s and at an angle of 36.9° above the horizontal. Ignore air resistance. (a) At what two times is the baseball at a height of 10.0 m above the point at which it left the bat? (b) Calculate the horizontal and vertical components of the baseball’s velocity at each of the two times calculated in part (a). (c) What are the magnitude and direction of the baseball’s velocity when it returns to the level at which it left the bat?
Answer:
a. 0.683secs and 2.99secs
b.(x,y)=(23.99,11.3)m/s, (23.99,-11.3)m/s\
c. V=30.0m/s at angle ∝=-36.9 Degree
Explanation:
Data given
Velocity, V=30m/s
Angle,∝=36.9 Degree
The motion describe by the baseball is a projectile motion, the velocity at the x-axis and y-axis are given as
[tex]V_{x}=Vcos\alpha\\ V_{x}=30cos36.9\\ V_{x}=23.99m/s\\V_{y}=Vsin\alpha \\V_{y}=30sin36.9\\ V_{y}=18.01m/s[/tex]
a.To calculate the time at which the baseball was at a height of 10m, we use the equation describing the vertical distance traveled
[tex]y=V_{y}t-\frac{1}{2}gt^{2}\\ y=10m\\10=18.01t-\frac{1}{2}*9.81t^{2}\\10=18.01t-4.9t^{2}\\-4.9t^{2}+18.01t-10[/tex]
solving the quadratic equation using the formula method
[tex]t=\frac{-b±\sqrt{b^{2}-4ac }}{2a} \\a=-4.9, b=18.01,c=-10\\t=\frac{18.01±\sqrt{18.01^{2}-4*(-4.9)(-10) }}{2*(-4.9)} \\t=0.683s, t=2.99s[/tex]
Hence the two times required 0.683secs and 2.99secs
b. Note that no acceleration in the hotizontal component, so the velocity remain the same. at a time t=0.683secs,
[tex]V_{x}=23.99m/s, \\V_{y}=V_{yi}-gt\\V_{y}=18.01-9.8*0.683\\V_{y}=11.3m/s\\[/tex]
at a time t=2.99secs,
[tex]V_{x}=23.99m/s, \\V_{y}=V_{yi}-gt\\V_{y}=18.01-9.8*2.99\\V_{y}=-11.3m/s\\[/tex]
c.The landing velocity is the same as the initial projected velocity but in opposite direction i.e V=30.0m/s at angle ∝=-36.9 Degree
To find the times when the baseball is at a height of 10.0 m, solve the vertical motion equation. Calculate the horizontal and vertical components of velocity at each time. Finally, find the magnitude and direction of the velocity when it returns to the starting level.
Explanation:To find the two times when the baseball is at a height of 10.0 m, we can use the equation for vertical motion:
y = yo + voyt - (1/2)gt^2
Setting y = 10 m and using the given initial vertical velocity, we can solve for t. Plugging in the values, we find two solutions: t = 2.46 s and t = 5.06 s.
Next, we can calculate the horizontal and vertical components of the baseball's velocity at each of the two times. The horizontal component remains constant throughout the motion and can be calculated using:
vx = vocosθ
Plugging in the given initial velocity and angle, we find that vx = 30.0 m/s * cos(36.9°) = 24.3 m/s.
The vertical component of the velocity changes due to acceleration from gravity. We can calculate it using:
vy = voy - gt
Plugging in the given initial vertical velocity and the acceleration due to gravity (-9.8 m/s^2), we find the vertical velocity at t = 2.46 s to be -6.16 m/s and at t = 5.06 s to be -14.56 m/s.
To find the magnitude of the baseball's velocity when it returns to the level at which it left the bat, we can use the Pythagorean theorem:
v = √(vx^2 + vy^2)
Plugging in the calculated values, we find the magnitude of the velocity to be 25.8 m/s.
The direction of the velocity can be found using the inverse tangent function:
θ = atan(vy/vx)
Plugging in the calculated values, we find the direction to be 180° - 36.9° = 143.1° below the horizontal.
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When the leaves of an electroscope are spread apart: a. A negatively charged object must be touching the knob of the electroscope. b. The leaves have the same charge. c. A positively charged object must be touching the knob of the electroscope. d. The leaves are neutral.
Answer:
B
Explanation:
The electroscope device is used to detect the presence of charge and it's relative amount. If a charged object is brought near the top of the electroscope(for a example positive charge) the leaves at the bottom spread apart. The leaves diverge further: The positive charge on the leaves has increased further. This happens when positive charge is produced on the leaves by the charged object. This is possible when the object is positively charged.The greater the charge, the farther apart they move.
Answer:
B. The leaves have the same charge.Explanation:
An electroscope is a device used to detect the presence of electric charges. However, to detect charge in an object, it requires hundreds of volts, that's why is only used with high voltage sources.
An important matter is that an electroscope follows the Coulomb electrostatic force, which is a law of physics that quantifies the amount of force between two charged particles, which are stationary.
Having said that, when two charged particles have the same nature, then they will spread apart. Same nature means both negative or both positive.
Therefore, the right answer is B.
An ethernet cable is 3.80 m long and has a mass of 0.210 kg. A transverse pulse is produced by plucking one end of the taut cable. The pulse makes four trips down and back along the cable in 0.735 s. What is the tension in the cable?
Answer:
[tex]T=94.54N[/tex]
Explanation:
The tension in a cable is given by:
[tex]T=\mu v^2(1)[/tex]
Where [tex]\mu[/tex] is the mass density of the cable and v is the speed of the cable's pulse. These values are defined as:
[tex]\mu=\frac{m}{L}(2)\\v=\frac{d}{t}[/tex]
The pulse makes four trips down and back along the cable, so [tex]d=4(2L)[/tex]
[tex]v=\frac{8L}{t}(3)[/tex]
Replacing (2) and (3) in (1), we calculate the tension in the cable:
[tex]T=\frac{m}{L}(\frac{8L}{t})^2\\T=\frac{64mL}{t^2}\\T=\frac{64(0.21kg(3.80m))}{(0.735s)^2}\\T=94.54N[/tex]
A charge of -5.02 nC is uniformly distributed on a thin square sheet of nonconducting material of edge length 21.8 cm. "What is the surface charge density of the sheet"?
Answer:
A charge of -5.02 nC is uniformly distributed on a thin square sheet of nonconducting material of edge length 21.8 cm. "What is the surface charge density of the sheet"?
Explanation:
Surface charge density is a measure of how much electric charge is accumulated over a surface. It can be calculated as the charge per unit area.
We will convert all parameters in SI units.
Charge = Q = -5.02nC
Q = -5.02×[tex]10^{-9}[/tex]C
As it is clear from question that Sheet is a square (All sides will be of equal length)
Area = A = (21.8×[tex]10^{-2}[/tex]m) (21.8×[tex]10^{-2}[/tex]m) = 4.75×[tex]10^{-4}[/tex]m²
A = 4.75×[tex]10^{-4}[/tex]m²
Surface charge density = Q/A
Surface charge density = (-5.02×[tex]10^{-9}[/tex]C)/(4.75×[tex]10^{-4}[/tex]m²)
Surface charge density = -1.057×[tex]10^{-5}[/tex] C[tex]m^{-2}[/tex]
Peg P is driven by the forked link OA along the path described by r = eu, where r is in meters. When u = p4 rad, the link has an angular velocity and angular acceleration of u # = 2 rad>s and u $ = 4 rad>s2. Determine the radial and transverse components of the peg’s acceleration at this instant.
Answer:
The transverse component of acceleration is 26.32 [tex]m/s^2[/tex] where as radial the component of acceleration is 8.77 [tex]m/s^2[/tex]
Explanation:
As per the given data
u=π/4 rad
ω=u'=2 rad/s
α=u''=4 rad/s
[tex]r=e^u[/tex]
So the transverse component of acceleration are given as
[tex]a_{\theta}=(ru''+2r'u')\\[/tex]
Here
[tex]r=e^u\\r=e^{\pi/4}\\r=2.1932 m[/tex]
[tex]r'=e^u.u'\\r'=2.1932 \times 2\\r'=4.3864 m[/tex]
So
[tex]a_{\theta}=(ru''+2r'u')\\a_{\theta}=(2.1932\times 4+2\times 4.3864 \times 2)\\a_{\theta}=26.32 m/s\\[/tex]
The transverse component of acceleration is 26.32 [tex]m/s^2[/tex]
The radial component is given as
[tex]a_r=r''-r\theta'^2[/tex]
Here
[tex]r''=e^u.u'^2+e^u u''\\r''=2.1932 \times (2)^2+2.1932\times 4\\r''=17.5456 m[/tex]
So
[tex]a_r=r''-ru'^2\\a_r=17.5456-2.1932\times (2)^2\\a_r=8.7728 m/s^2[/tex]
The radial component of acceleration is 8.77 [tex]m/s^2[/tex]
The radial and transverse components of the peg's acceleration at the given instant are both 4eu m/s^2.
Explanation:In circular motion, there are two components of acceleration: the radial component (also called centripetal acceleration), directed towards the center of the circle, and the transverse component (also called tangential or azimuthal acceleration), which is perpendicular to the radial component and in the direction of increasing angle.
Given that r = eu, and u = p4 rad, u# = 2 rad/s (angular velocity), u$ = 4 rad/s^2 (angular acceleration), we can calculate these components as follows:
The radial component of acceleration (Ar) can be computed using the formula Ar = r(u#)^2. Substituting the given values, we get Ar = (eu)*(2)^2 = 4eu m/s^2. The transverse component of acceleration (At) can be computed using the formula At = r*u$. Substituting the given values, we get At = eu*4 = 4eu m/s^2.
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In the image below, if the engine backs up in order to couple (join) with several more train cars and push them backwards, what explanation best describes the type of collision it is?
A.It is an inelastic collision because the collision conserves momentum.
B. It is an inelastic collision because the train cars stick together and move as one.
C. It is an elastic collision because the collision conserves momentum.
D. It is an elastic collision because the cars stick together and move as one unit.
Answer:
B. It is an inelastic collision because the train cars stick together and move as one.
Explanation:
Momentum
When two or more objects collide in a closed system (no external forces are acting) the total momentum is conserved:
[tex]m_1v_1+m_2v_2+...+m_nv_n=m_1v_1'+m_2v_2'+...+m_nv_n'[/tex]
where m1...m2 are the masses of each object, v1...vn are their velocities before the collision takes place and v'1...v'n are their velocities after the collision.
If a collision is elastic, then the kinetic energy is also conserved. when the collision is inelastic, part of the initial kinetic energy is lost. A typical case of inelastic collision occurs when the objects join and remain together after the collision. The velocity is common to all of them and the mass is the sum of the individual masses.
This is exactly the case described in the question: serveral train cars are joined and continue moving together after the collision. It corresponds to a inelastic collision described in the option B.
If the pressure inside the cylinder increases to 1.6 atm, what is the final volume, in milliliters, of the cylinder?
This is a physics problem related to Boyle's Law, which says that the volume of a gas decreases if its pressure increases, assuming constant temperature. We can't solve for the final volume specifically in this case since the initial pressure and volume aren't given, but if they were, we'd use the formula V₂ = (P₁V₁) / P₂, where P₁ and V₁ are the initial pressure and volume, and P₂ and V₂ are the final pressure and volume.
This question is related to the concept of Boyle's Law in physics, which states that the pressure and volume of a gas have an inverse relationship, when temperature is held constant. If the pressure of the gas inside a cylinder is increased, the volume decreases, assuming the amount of gas and the temperature remain constant.
In this case without knowing the initial pressure and volume of the cylinder, it's impossible to calculate the exact final volume after the pressure has increased to 1.6 atm. However, the base formula you'd use to find out your final volume, given you had initial values for volume (V₁) and pressure (P₁), is: P₁V₁ = P₂V₂.
In this formula, P₁ and V₁ refer to the initial pressure and volume, while P₂ and V₂ refer to the final pressure and volume. To solve for the final volume (V₂), you would rearrange the formula to be V₂ = (P₁V₁) / P₂. So, if you were given the initial pressure and volume, you could substitute those values into the formula, along with the final pressure of 1.6 atm, to find the final volume.
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In this case, the final volume is 31.1 mL.
The final volume of the helium in milliliters can be calculated using Boyle's Law, which states that the product of pressure and volume is constant for a given amount of gas at constant temperature. Initially, the pressure is 447 torr, and the volume is 86.4 mL. When the pressure increases to 1,240 torr, the final volume can be found using the formula:
P1V1 = P2V2
Solving for the final volume, V2, we get:
V2 = (P1V1) / P2
Substitute the given values:
V2 = (447 torr× 86.4 mL) / 1,240 torr
= 31.1 mL
Two ropes in a vertical plane exert equal-magnitude forces on a hanging weight but pull with an angle of 72.0° between them. What pull does each rope exert if their resultant pull is 372 N directly upward?
The force exerted by each rope is approximately 230 N.
Let's denote the magnitude of the force exerted by each rope as F. Given that the resultant pull is 372 N directly upward, we can use vector decomposition to find the individual forces exerted by each rope.
The forces F acting at an angle of 72° between them can be decomposed into their horizontal and vertical components as follows:
[tex]\[ F_{\text{horizontal}} = F \cdot \cos(72^\circ) \]\\\ F_{\text{vertical}} = F \cdot \sin(72^\circ) \][/tex]
The vertical components of the forces add up to the resultant pull, so:
[tex]\[ 2 \cdot F_{\text{vertical}} = 372 \, \text{N} \]\\\\\ F_{\text{vertical}} = \frac{372 \, \text{N}}{2} \\\\= 186 \, \text{N} \][/tex]
Now, we can use the vertical component to find F:
[tex]\rm \[ F = \frac{F_{\text{vertical}}}{\sin(72^\circ)} \][/tex]
Substitute the known values:
[tex]\rm \[ F = \frac{186 \, \text{N}}{\sin(72^\circ)} \][/tex]
Calculate F:
[tex]\rm \[ F \approx 230 \, \text{N} \][/tex]
So, the force exerted by each rope is approximately 230 N.
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To find the pull exerted by each rope with an angle of 72°, we can use vector components. Each rope exerts a pull of approximately 436.2 N at an angle of 72° for both the vertical and horizontal components.
To find the pull exerted by each rope, we can use the concept of vector components. Let's call the magnitude of each pull T. Since the resultant pull is 372 N directly upward, we can calculate the vertical component of each rope's pull using the equation T*sin(72°) = 372 N.
Solving for T, we find that the magnitude of each rope's pull is approximately 436.2 N. To find the horizontal component of each rope's pull, we use the equation T*cos(72°).
Since the forces are equal-magnitude and have the same angle between them, each rope exerts the same pull of approximately 436.2 N at an angle of 72° for both the vertical and horizontal components of the pull.
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A cart is initially moving at 0.5 m/s along a track. The cart comes to rest after traveling 1 m. The experiment is repeated on the same track, but now the cart is initially moving at 1 m/s. How far does the cart travel before coming to rest?
A. 1 m B. 2 m C. 3 m D. 4 m E. 8 m
Answer:
D. 4 m
Explanation:
According to the work-energy theorem, the work done by the force acting on a body modifies its kinetic energy.
[tex]W=\Delta K\\W=K_f-K_i\\W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}[/tex]
The car comes to rest after a given distance, so [tex]v_f=0[/tex]. Recall the definition of work [tex]W=Fdcos\theta[/tex], here F is the constant force acting on the car, d is its traveled distance and [tex]\theta[/tex] is the angle between the force and the displacement, since friction force acts opposite to the direction of motion [tex]\theta=180^\circ[/tex] :
[tex]-Fd=-\frac{mv_i^2}{2}\\d=\frac{mv_i^2}{2F}[/tex]
We have:
[tex]v_i'=1\frac{m}{s}\\v_i=0.5\frac{m}{s}\\v_i'=2vi[/tex]
Thus:
[tex]d'=\frac{2m(v_i')^2}{2F}\\d'=\frac{2m(2v_i)^2}{2F}\\d'=4\frac{2mv_i^2}{2F}\\d'=4d\\d'=4(1m)\\d'=4m[/tex]
Two 1.0 cm * 2.0 cm rectangular electrodes are 1.0 mm apart. What charge must be placed on each electrode to create a uniform electric field of strength 2.0 * 106 N/C? How many electrons must be moved from one electrode to the other to accomplish this?
Answer:
The number of electrons that must be moved from one electrode to the other to accomplish this is 1.4 X 10⁹ electrons.
Explanation:
Step 1: calculate the charge on each electrode
Given;
Electric field strength = 2.0 X 10⁶ N/C
The distance between the electrode = 1mm = 1 X 10⁻³ m
Electric field strength (E) = Force (F)/Charge (q)
[tex]E =\frac{Kq}{r^2}[/tex]
where;
E is the electric field strength = 2.0 X 10⁶ N/C
K is coulomb's constant = 8.99 X 10⁹ Nm²/C²
r is the distance between the electrodes = 1 X 10⁻³ m
q is the charge in each electrode = ?
[tex]q = \frac{Er^2}{K} = \frac{(2X10^6)(1X10^{-3})^2}{8.99 X10 ^9}[/tex] = 0.2225 X 10⁻⁹ C
The charge on each electrode is 0.2225 X 10⁻⁹ C
Step 2: calculate the number of electrons to be moved from one electrode to the other.
1 electron contains 1.602 X 10⁻¹⁹ C
So, 0.2225 X 10⁻⁹ C will contain how many electrons ?
= (0.2225 X 10⁻⁹)/(1.602 X 10⁻¹⁹)
= 1.4 X 10⁹ electrons
Therefore, the number of electrons that must be moved from one electrode to the other to accomplish this is 1.4 X 10⁹ electrons.
Final answer:
To create a uniform electric field of 2.0 x 10^6 N/C between two electrodes 1.0 mm apart, a charge of 3.54 x 10^-6 C is required on each electrode. Approximately 2.21 x 10^13 electrons must be moved from one electrode to the other to achieve this.
Explanation:
To solve the immediate question, we first need to calculate the charge required to create a uniform electric field of strength 2.0 x 106 N/C between two rectangular electrodes that are 1.0 mm apart. The relationship between the electric field (E), the charge (Q), the permittivity of free space (ε0), and the distance (d) between the plates is given by the equation E = Q / (ε0 × A), where A is the area of one of the plates. Given the plates have dimensions of 1.0 cm * 2.0 cm, we firstly convert these into meters, giving an area of 0.0002 m2. The permittivity of free space (ε0) is 8.85 x 10-12 C2/N·m2.
Substituting the given values into the formula, we get Q = E × ε0 × A = 2.0 x 106 N/C × 8.85 x 10-12 C2/N·m2 × 0.0002 m2 = 3.54 x 10-6 C. Therefore, this is the charge required on each electrode.
To find the number of electrons that must be moved, we use the relationship between charge and number of electrons, which is given by Q = n × e, where n is the number of electrons and e is the charge of an electron (1.602 x 10-19 Coulombs). So, n = Q / e = 3.54 x 10-6 C / 1.602 x 10-19 C = 2.21 x 1013 electrons. Hence, approximately 2.21 x 1013 electrons need to be moved from one electrode to the other to create the desired electric field.
Why do atoms absorb and reemit radiation at characteristic frequencies?
To answer this question it is necessary to use the Bohr Model as a theoretical reference. According to this model, the energy levels of an atom are divided by discrete and characteristic energies. When there is the emission or absorption of a photon with a characteristic energy there is a transition of another energy level, which is equal to the level of atomic energy. Since the energy of a photo is directly proportional to its frequency, the emitted or absorbed photons have characteristic frequencies to the difference in energy between atomic energy levels.
A rocket is fired straight up. It contains two stages (Stage 1 and Stage 2) of solid rocket fuel that are designed to burn for 10.0 and 5.00 s, respectively, with no time interval between them. In Stage 1, the rocket fuel provides an upward acceleration of 16.0 m/s^2. In Stage 2, the acceleration is 11.0 m/s2. Neglecting air resistance, calculate the maximum altitude above the surface of Earth of the payload and the time required for it to return to the surface. Assume the acceleration due to gravity is constant.a. Calculate the maximum altitude. (Express your answer to two significant figures.)b. Calculate time required to return to the surface (i.e. the total time of flight). (Express your answer to three significant figures.)
Answer:
a) y= 3.5 10³ m, b) t = 64 s
Explanation:
a) For this exercise we use the vertical launch kinematics equation
Stage 1
y₁ = y₀ + v₀ t + ½ a t²
y₁ = 0 + 0 + ½ a₁ t²
Let's calculate
y₁ = ½ 16 10²
y₁ = 800 m
At the end of this stage it has a speed
v₁ = vo + a₁ t₁
v₁ = 0 + 16 10
v₁ = 160 m / s
Stage 2
y₂ = y₁ + v₁ (t-t₀) + ½ a₂ (t-t₀)²
y₂ = 800 + 150 5 + ½ 11 5²
y₂ = 1092.5 m
Speed is
v₂ = v₁ + a₂ t
v₂ = 160 + 11 5
v₂ = 215 m / s
The rocket continues to follow until the speed reaches zero (v₃ = 0)
v₃² = v₂² - 2 g y₃
0 = v₂² - 2g y₃
y₃ = v₂² / 2g
y₃ = 215²/2 9.8
y₃ = 2358.4 m
The total height is
y = y₃ + y₂
y = 2358.4 + 1092.5
y = 3450.9 m
y= 3.5 10³ m
b) Flight time is the time to go up plus the time to go down
Let's look for the time of stage 3
v₃ = v₂ - g t₃
v₃ = 0
t₃ = v₂ / g
t₃ = 215 / 9.8
t₃ = 21.94 s
The time to climb is
[tex]t_{s}[/tex] = t₁ + t₂ + t₃
t_{s} = 10+ 5+ 21.94
t_{s} = 36.94 s
The time to descend from the maximum height is
y = v₀ t - ½ g t²
When it starts to slow down it's zero
y = - ½ g t_{b}²
t_{b} = √-2y / g
t_{b} = √(- 2 (-3450.9) /9.8)
t_{b} = 26.54 s
Flight time is the rise time plus the descent date
t = t_{s} + t_{b}
t = 36.94 + 26.54
t =63.84 s
t = 64 s
A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 5.0 m/s2. Secret agent Austin Powers jumps on just as the helicopter lifts off the ground. After the two men struggle for 10.0 s, Powers shuts off the engine and steps out of the helicopter. Assume that the helicopter is in free fall after its engine is shut off, and ignore the effects of air resistance. (a) What is the maximum height above ground reached by the helicopter? (b) Powers deploys a jet pack strapped on his back 7.0 s after leaving the helicopter, and then he has a constant downward acceleration with magnitude 2.0 m/s2. How far is Powers above the ground when the helicopter crashes into the ground?
Answer:
a) [tex]h=250\ m[/tex]
b) [tex]\Delta h=0.0835\ m[/tex]
Explanation:
Given:
upward acceleration of the helicopter, [tex]a=5\ m.s^{-2}[/tex]time after the takeoff after which the engine is shut off, [tex]t_a=10\ s[/tex]a)
Maximum height reached by the helicopter:
using the equation of motion,
[tex]h=u.t+\frac{1}{2} a.t^2[/tex]
where:
u = initial velocity of the helicopter = 0 (took-off from ground)
t = time of observation
[tex]h=0+0.5\times 5\times 10^2[/tex]
[tex]h=250\ m[/tex]
b)
time after which Austin Powers deploys parachute(time of free fall), [tex]t_f=7\ s[/tex]acceleration after deploying the parachute, [tex]a_p=2\ m.s^{-2}[/tex]height fallen freely by Austin:
[tex]h_f=u.t_f+\frac{1}{2} g.t_f^2[/tex]
where:
[tex]u=[/tex] initial velocity of fall at the top = 0 (begins from the max height where the system is momentarily at rest)
[tex]t_f=[/tex] time of free fall
[tex]h_f=0+0.5\times 9.8\times 7^2[/tex]
[tex]h_f=240.1\ m[/tex]
Velocity just before opening the parachute:
[tex]v_f=u+g.t_f[/tex]
[tex]v_f=0+9.8\times 7[/tex]
[tex]v_f=68.6\ m.s^{-1}[/tex]
Time taken by the helicopter to fall:
[tex]h=u.t_h+\frac{1}{2} g.t_h^2[/tex]
where:
[tex]u=[/tex] initial velocity of the helicopter just before it begins falling freely = 0
[tex]t_h=[/tex] time taken by the helicopter to fall on ground
[tex]h=[/tex] height from where it falls = 250 m
now,
[tex]250=0+0.5\times 9.8\times t_h^2[/tex]
[tex]t_h=7.1429\ s[/tex]
From the above time 7 seconds are taken for free fall and the remaining time to fall with parachute.
remaining time,
[tex]t'=t_h-t_f[/tex]
[tex]t'=7.1428-7[/tex]
[tex]t'=0.1428\ s[/tex]
Now the height fallen in the remaining time using parachute:
[tex]h'=v_f.t'+\frac{1}{2} a_p.t'^2[/tex]
[tex]h'=68.6\times 0.1428+0.5\times 2\times 0.1428^2[/tex]
[tex]h'=9.8165\ m[/tex]
Now the height of Austin above the ground when the helicopter crashed on the ground:
[tex]\Delta h=h-(h_f+h')[/tex]
[tex]\Delta h=250-(240.1+9.8165)[/tex]
[tex]\Delta h=0.0835\ m[/tex]
An insulating sphere is 8.00 cm in diameter and carries a 6.50 µC charge uniformly distributed throughout its interior volume. Calculate the charge enclosed by a concentric spherical surface with the following radius.(a) r = 1.00 cm(b) r = 6.50 cm
The charge enclosed by a concentric spherical surface depends on the size of the sphere.
Explanation:To calculate the charge enclosed by a concentric spherical surface, we need to find the charge within that surface. The charge is uniformly distributed throughout the interior volume of the insulating sphere, so the charge within any smaller sphere that is completely enclosed by the larger sphere will be the same.
(a) For r = 1.00 cm, the smaller sphere is completely enclosed by the larger sphere. Therefore, the charge enclosed by the smaller sphere is 6.50 µC.
(b) For r = 6.50 cm, the smaller sphere is the same size as the larger sphere, so the charge enclosed by the smaller sphere will also be 6.50 µC.
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A 500-g sample of sand in the SSD condition was placed in a jar, which was then filled with water. The combined weight was 1697 g. The weight of the jar filled with the water only was 1390 g. Calculate the bulk specific gravity (SSD) of the sand.
Final answer:
The bulk specific gravity (SSD) of the sand is 0.221.
Explanation:
The bulk specific gravity (SSD) of the sand can be calculated using the given information. The combined weight of the sand and water is 1697 g, and the weight of the jar filled with water only is 1390 g. To find the weight of the sand in the SSD condition, we subtract the weight of the jar filled with water from the combined weight: 1697 g - 1390 g = 307 g. Therefore, the bulk specific gravity (SSD) of the sand is the weight of the sand divided by the weight of an equal volume of water, which is 307 g / 1390 g = 0.221.
A pilot who accelerates at more than 4 g begins to "gray out" but doesn’t completely lose consciousness. (a) Assuming constant acceleration, what is the shortest time that a jet pilot starting from rest can take to reach Mach 4 (four times the speed of sound) without graying out? (b) How far would the plane travel during this period of acceleration? (Use 331 m/s for the speed of sound in cold air.)
Answer:
33.7410805301 s
22336.5953109 m
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity = [tex]4\times 331[/tex]
s = Displacement
a = Acceleration = [tex]4g=4\times 9.81\ m/s^2[/tex]
g = Acceleration due to gravity = 9.81 m/s²
[tex]v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{4\times 331-0}{4\times 9.81}\\\Rightarrow t=33.7410805301\ s[/tex]
The time taken is 33.7410805301 s
[tex]s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 4\times 9.81\times 33.7410805301^2\\\Rightarrow s=22336.5953109\ m[/tex]
The plane would travel 22336.5953109 m
The shortest time is approximately 33.74 seconds, and the distance traveled is approximately 22.32 kilometers.
To solve this problem, we will first need to convert Mach 4 to a speed in meters per second, then use the kinematic equations for constant acceleration to find the time and distance.
(a) The speed of sound in cold air is given as 331 m/s. Therefore, Mach 4 is:
[tex]\[ v = 4 \times 331 \text{ m/s} = 1324 \text{ m/s} \][/tex]
The acceleration that the pilot can withstand without graying out is less than 4 g. Since 1 g is equal to 9.81 m/s², the maximum acceleration the pilot can withstand is:
[tex]\[ a_{\text{max}} = 4 \times 9.81 \text{ m/s}^2 = 39.24 \text{ m/s}^2 \][/tex]
Using the kinematic equation that relates initial velocity, final velocity, acceleration, and time:
[tex]\[ v = u + at \]\\ where \( v \) is the final velocity, \( u \) is the initial velocity (0 m/s since the pilot starts from rest), \( a \) is the acceleration, and \( t \) is the time. We can solve for \( t \): \[ 1324 \text{ m/s} = 0 \text{ m/s} + (39.24 \text{ m/s}^2)t \] \[ t = \frac{1324 \text{ m/s}}{39.24 \text{ m/s}^2} \] \[ t \approx 33.74 \text{ s} \][/tex]
(b) To find the distance traveled during this time, we use the kinematic equation:
[tex]\[ s = ut + \frac{1}{2}at^2 \]\\ where \( s \) is the distance, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time. Since \( u = 0 \) m/s: \[ s = 0 \times t + \frac{1}{2}(39.24 \text{ m/s}^2)t^2 \] \[ s = \frac{1}{2}(39.24 \text{ m/s}^2)(33.74 \text{ s})^2 \] \[ s \approx \frac{1}{2}(39.24 \text{ m/s}^2)(1137.52 \text{ s}^2) \] \[ s \approx 22320.54 \text{ m} \] \[ s \approx 22.32 \text{ km} \][/tex]
Therefore, the shortest time the pilot can take to reach Mach 4 without graying out is approximately 33.74 seconds, and the distance traveled during this period of acceleration is approximately 22.32 kilometers.
A 10.0-cm-long uniformly charged plastic rod is sealed inside a plastic bag. The net electric flux through the bag is 7.50 × 10 5 N ⋅ m 2 /C . What is the linear charge density (charge per unit length) on the rod?
Answer:
66.375 x 10⁻⁶ C/m
Explanation:
Using Gauss's law which states that the net electric flux (∅) through a closed surface is the ratio of the enclosed charge (Q) to the permittivity (ε₀) of the medium. This can be represented as ;
∅ = Q / ε₀ -----------------(i)
Where;
∅ = 7.5 x 10⁵ Nm²/C
ε₀ = permittivity of free space (which is air, since it is enclosed in a bag) = 8.85 x 10⁻¹² Nm²/C²
Now, let's first get the charge (Q) by substituting the values above into equation (i) as follows;
7.5 x 10⁵ = Q / (8.85 x 10⁻¹²)
Solve for Q;
Q = 7.5 x 10⁵ x 8.85 x 10⁻¹²
Q = 66.375 x 10⁻⁷ C
Now, we can find the linear charge density (L) which is the ratio of the charge(Q) to the length (l) of the rod. i.e
L = Q / l ----------------------(ii)
Where;
Q = 66.375 x 10⁻⁷ C
l = length of the rod = 10.0cm = 0.1m
Substitute these values into equation (ii) as follows;
L = 66.375 x 10⁻⁷C / 0.1m
L = 66.375 x 10⁻⁶ C/m
Therefore, the linear charge density (charge per unit length) on the rod is 66.375 x 10⁻⁶ C/m.
The linear charge density of the rod is found using Gauss's law and is equal to 6.642 x 10^-5 C/m.
Explanation:The concept in this question is Gauss's law, which states that the electric flux through any closed surface is equal to the total charge enclosed divided by the permittivity of free space, ε0. Mathematically, it is written as Φ = Q/ε0, where Φ is the electric flux, Q is the charge, and ε0 = 8.854 x 10^-12 C2/N·m2.
In this case, we have Φ = 7.50 x 105 N·m2/C. To find Q, we just rearrange the formula and multiply both sides by ε0. This gives Q = Φ * ε0 = 7.50 x 105 N·m2/C * 8.854 x 10^-12 C2/N·m2 = 6.642 x 10^-6 C.
The linear charge density λ is the total charge Q divided by the total length L of the rod. Here, L = 10.0 cm = 0.1 m, so λ = Q/L = 6.642 x 10^-6 C / 0.1 m = 6.642 x 10^-5 C/m.
Learn more about Electric Flux and Charge Density here:https://brainly.com/question/33167286
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The resistivity of a certain semi-metal is 10-3 Ohm-cm. Suppose we would like to prepare a silicon wafer with the same resistivity. Assuming we will use n-type dopants only, what dopant density would we choose
Complete Question
The complete question is shown on the first uploaded image
Answer:
The dopant density is ND ≈ 8.135*10¹² cm⁻³
Explanation:
The explanation is shown on the second , third ,fourth and fifth uploaded image