Answer:
2.51 m/s
Explanation:
Parameters given:
Angle, A = 33°
Mass, m = 90kg
Inclined distance, D = 2m
Force, F = 600N
Initial speed, u = 2.3m/s
From the relationship between work and kinetic energy, we know that:
Work done = change in kinetic energy
W = 0.5m(v² - u²)
We also know that work done is tẹ product of force and distance, hence, net work done will be the sum of the total work done by the force from the students and gravity.
Hence,
W = F*D*cosA - w*D*sinA
w = m*9.8 = weight
=> W = 600*2*cos33 - 90*9.8*2*sin33
W = 45.7J
=> 45.7 = 0.5*m*(v² - u²)
45.7 = 0.5*90*(v² - 2.3²)
45.7 = 45(v² - 5.29)
=> v² - 5.29 = 1.016
v² = 6.306
v = 2.51 m/s
The final velocity is 2.51 m/s
Answer: v = 3.26 m/s
Explanation:
Force applied (F) = 900 N,
mass of professor and chair (m) = 90kg,
g = acceleration due to gravity = 9.8 m/s²,
Fr = frictional force = 0 ( since from the question, the rollers the chair and professor are moving is frictionless),
θ = angle of inclination = 33°
a = acceleration of object =?
From newton's second law of motion, we have that
F - (mg sinθ +Fr) = ma
Where mg sinθ is the horizontal component of the weight of the mass of professor and chair due to the inclination of the ramp.
But Fr = 0
Hence, we have that
F - mg sinθ = ma
600 - (90×9.8×sin30) = 90 (a)
600 - 480.371= 90a
119.629 = 90a
a = 119.629/ 90
a = 1.33 m/s².
But the body started the morning (at the bottom of the ramp) with a velocity of 2.30 m/s²
Hence u = initial velocity = 2.30 m/s², a = acceleration = 1.33m/s², v = final velocity =?, s = distance covered = 2m
By using equation of motion for a constant acceleration, we have that
v² = u² + 2as
v ² = (2.3)² +2(1.33)×(2)
v² = 5.29 + 5.32
v² = 10.61
v = √10.61
v = 3.26 m/s
A bulb pile is driven to the ground with a 2.5 ton hammer. The drop height is 22 ft and the volume in last batch driven is 4 cu ft. Establish the safe load if the number of blows to drive the last batch is 36, volume of base and plug is 27 cu ft, and the soil K value is 28.
Answer:
159.1 ton
Explanation:
The solution is shown in the attached file
Answer:
The safe load is 159 ton
Explanation:
The safe load is equal to:
[tex]L=\frac{WHBV^{2/3} }{K}[/tex]
Where:
W = weight of hammer = 2.5 ton
H = drop height = 22 ft
B = number of blows used to drive the last batch = 36/4 = 9 ft³
K = dimensionless constant = 28
V = uncompacted volume = 27 ft³
Replacing values:
[tex]L=\frac{2.5*22*9*(27^{2/3}) }{28} =159ton[/tex]
When there is a large difference in two resistor’s sizes, what useful approximations can be used when considering their series and parallel combinations? (This is a handy thing to know in circuit design!)
Answer:
Series circuit approximation = R1 >> R2 then R2 ≈ 0
Parallel circuit approximation = R1 << R2 then R2 ≈ 0
Explanation:
in a series circuit, if there is a large difference between two resistors then we can omit the resistor which has low resistance because the higher value resistor dominates.
R1 >> R2 then R2 ≈ 0
in a parallel circuit, if there is a large difference between two resistors then we can omit the resistor which has high resistance because the lower value resistor dominates.
R1 << R2 then R2 ≈ 0
A circular coil with 169 turns has a radius of 2.6 cm. (a) What current through the coil results in a magnetic dipole moment of 3.4 A·m2? I= A (b) What is the maximum torque that the coil will experience in a uniform field of strength 0.03 T? τmax= Nm (c) If the angle between μ and B is 44.9∘, what is the magnitude of the torque on the coil? τ= Nm (d) What is the magnetic potential energy of coil for this orientation?
Answer with Explanation:
We are given that
Number of turns=n=169
Radius of coil=r=2.6 cm=[tex]2.6\times 10^{-2} m[/tex]
[tex]1 cm=10^{-2}m[/tex]
Area of circular coil=[tex]\pi r^2[/tex]
Where [tex]\pi=3.14[/tex]
Using the formula
Area of circular coil=[tex]3.14\times (2.6\times 10^{-2})^2=21.2\times 10^{-4}m^2[/tex]
a.Magnetic dipole moment=[tex]\mu=3.4 Am^2[/tex]
[tex]I=\frac{\mu}{nA}[/tex]
Using the formula
[tex]I=\frac{3.4}{169\times 21.2\times 10^{-4}}[/tex]
[tex]I=9.5 A[/tex]
b.Magnetic field, B=0.03 T
[tex]\tau_{max}=\mu B[/tex]
Substitute the value
[tex]\tau_{max}=3.4\times \times 0.03=0.102 Nm[/tex]
c.[tex]\theta=44.9^{\circ}[/tex]
[tex]\tau=\tau_{max}sin\theta[/tex]
Substitute the values
[tex]\tau=0.102sin44.9=0.07 Nm[/tex]
Ball B is suspended from a cord of length l attached to cart A, which can roll freely on a frictionless, horizontal track. The ball and the cart have the same mass m. If the cart is given an initial horizontal velocity v0 while the ball is at rest, describe the subsequent motion of the system, specifying the velocities of A and B for the following successive values of the angle θ (assume positive counterclockwise) that the cord will form with the vertical:
(a) θ = θmax
(b) θ = 0
(c) θ = θmin
The motion involves energy conversion between kinetic and potential energy, with ball B reaching maximum potential energy at θmax and θmin, and maximum kinetic energy at θ = 0. Cart A's velocity will be adjusted according to the change in motion of ball B, based on the conservation of momentum.
Explanation:The question relates to the physics concept known as conservation of momentum and energy in the context of pendular motion and collisions within an isolated system. When cart A is given an initial velocity and ball B is at rest, the force that accelerates ball B will be the component of the tension in the string that acts horizontally, as there's no friction resistance on the track. During the motion, the total energy of ball B will be conserved, converting between potential energy when at its highest at θmax and θmin, and kinetic energy when passing through the lowest point at θ = 0. As the system consists of pendular motion, for a given displacement, the velocity of the pendulum and cart at various angles can be determined by using energy conservation.
Given that cart A and ball B have the same mass, the velocities can be tracked by considering the movement as a combination of the cart rolling and the pendulum swinging, respecting the conservation of angular momentum around the axis from which the ball B is suspended.
What is the energy in the spark produced by discharging the second capacitor? 1. The same as the discharge spark of the first capacitor 2. More energetic than the discharge spark of the first capacitor 3. Less energetic than the discharge spark of the first capacitor
Complete Question
The two isolated parallel plate capacitors be- low, one with plate separation d and the other with D > d, have the same plate area A and
are given the same charge Q.
What is the energy in the spark produced by discharging the second capacitor?
1. The same as the discharge spark of the first capacitor
2. More energetic than the discharge spark of the first capacitor
3. Less energetic than the discharge spark of the first capacitor
Answer:
The correct option is 2
Explanation:
The formula for the energy stored in the capacitor is
[tex]U = \frac{Q^2}{2C}[/tex]
And generally the formula for finding the capacitance of a capacitor is
[tex]C = \frac{\epsilon_oA}{d}[/tex]
We can denote the capacitance of the first capacitor as [tex]C_1 = \frac{\epsilon_oA}{d}[/tex]
and denote the capacitance of the second capacitor as [tex]C_2 = \frac{\epsilon_oA}{D}[/tex]
Looking at this formula we can see that C varies inversely with d
as D > d it means that [tex]C_1 > C_2[/tex]
Since the charge is constant
[tex]U\ \alpha\ \frac{1}{C}[/tex] i.e U varies inversely with C
So [tex]C_1 > C_2[/tex] => [tex]U_2 >U_1[/tex]
This means that the energy of spark would be more for capacitor two compared to capacitor one
The correct option is 2
The energy in the spark produced by discharging a capacitor can vary between different capacitors based on their capacitance and potential difference.
When discharging a capacitor, the energy stored in the capacitor is released as a spark, which can be compared between different capacitors.
In this case, if the second capacitor discharges, the spark produced may have varying energy levels compared to the spark produced by the first capacitor, depending on the capacitance and potential difference.
The energy in the spark produced by discharging the second capacitor can be calculated using relevant formulas relating to capacitance, potential difference, and energy stored.
A 55-kg box rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.30, and the coefficient of kinetic friction is 0.20. What horizontal force must be applied to the box to cause it to start sliding along the surface
Answer:
The horizontal force that must be applied to the box to cause it to start sliding along the surface is 162N
Explanation:
To start sliding the box on the surface it must overcome its static frictional force under equilibrium condition
The net force on the box is
F - fs = 0
F = fs = us N = us mg
Force = ( 0.3) x ( 55 kg) x ( 9.8 m/s^2) = 161.7 approximately 162 N
Horizontal force is defined as the forces that equal and opposite in the direction. The horizontal resultant force is always zero.
Given that:
Mass of box = 55 Kg
Coeeficient of Static friction = 0.30
Coefficient of kinetic friction = 0.20
The box if have to slide, then it would have to overcome the static force, such that:
F - Fs = 0
F = Fs
The force will be:
F = [tex]\rm \mu \times mass \times g[/tex]
F = [tex]0.3 \times 55 \times 9.8[/tex]
F = 161.7 Newton
Thus, the force required to slide the box is 161.7 Newton.
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The wheel of a stationary exercise bicycle at your gym makes one rotation in 0.670 s. Consider two points on this wheel: Point P is 10.0 cm from the rotation axis, and point Q is 20.0 cm from the rotation axis. Find the speed of point P on the spinning wheel.
Answer:
0.938 m/s.
Explanation:
Given:
ω = 1 rev in 0.67 s
In rad/s,
1 rev = 2pi rad
ω = 2pi ÷ 0.67
= 9.38 rad/s
Rp = 10 cm
= 0.1 m
V = ω × r
= 9.38 × 0.1
= 0.938 m/s.
A running back with a mass of 86 kg and a velocity of 9 m/s (toward the right) collides with, and is held by, a 129-kg defensive tackle going in the opposite direction (toward the left). What is the velocity of the tackle before the collision for their velocity afterward to be zero
Final answer:
The velocity of the defensive tackle before the collision is -6 m/s (toward the left).
Explanation:
According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. Since the two players stick together after the collision, their velocities will be equal but opposite in direction.
Let's assume that the velocity of the defensive tackle before the collision is v. The total momentum before the collision is given by:
m1 * v1 + m2 * v2 = (m1 + m2) * v
Where:
m1 = mass of the running back = 86 kg
v1 = velocity of the running back = 9 m/s (toward the right)
m2 = mass of the defensive tackle = 129 kg
v2 = velocity of the defensive tackle (to be determined)
Using the above equation, we can solve for v:
(86 kg * 9 m/s) + (129 kg * v2) = (86 kg + 129 kg) * 0
774 kg*m/s + 129 kg * v2 = 0 kg*m/s
v2 = -6 m/s
Therefore, the velocity of the defensive tackle before the collision is -6 m/s (toward the left).
A lead block drops its temperature by 5.90 degrees celsius when 427 J of heat are removed from it. what is the mass of the block?(unit=kg)
Mass of the block is 0.557 kg
Solution:
The quantity of heat (q) liberated during any process is equal to the product of the mass of the block (m), specific heat (C) and the change in temperature (ΔT). Specific heat of lead is 130 J/kg °C.
We have to rearrange the equation to find the mass of the block of lead as,
[tex]\begin{array}{l}\boldsymbol{q}=\boldsymbol{m} \times \boldsymbol{C} \times \boldsymbol{\Delta} \mathbf{T}\\ \\\boldsymbol{m}=\frac{\mathbf{q}}{\boldsymbol{c} \times \mathbf{\Delta} \mathbf{T}}\\ \\\mathbf{m}=\frac{427 \mathbf{J}}{13 \mathbf{0}_{\mathbf{4} *}^{\prime} \mathbf{r} \times \mathbf{s} \mathbf{9} \mathbf{0}^{*} \mathbf{c}}=\mathbf{0} . \mathbf{5} \mathbf{5} \mathbf{7} \mathbf{k} \mathbf{g}\end{array}[/tex]
Thus the mass (m) = 0.577 kg
Answer:
the correct answer is 0.57
Explanation:
A transverse sine wave with an amplitude of 2.50 mm and a wavelength of 1.80 m travels from left to right along a long, horizontal stretched string with a speed of 36.0 m/s. Take the origin at the left end of the undisturbed string. At time t = 0, the left end of the string has its maximum upward displacement.
(a) What are the frequency and angular frequency of the wave?
(b) What is the wave number of the wave?
Answer:
(a)
[tex]f=20Hz\\And \\w=125.7s^{-1}[/tex]
(b)
[tex]K=3.490m^{-1}[/tex]
Explanation:
We know that the speed of any periodic wave is given by:
v=f×λ
The wave number k is given by:
K=2π/λ
Given data
Amplitude A=2.50mm
Wavelength λ=1.80m
Speed v=36 m/s
For Part (a)
For the wave frequency we plug our values for v and λ.So we get:
v=f×λ
[tex]36.0m/s=(1.80m)f\\f=\frac{36.0m/s}{1.80m}\\ f=20Hz\\[/tex]
And the angular speed we plug our value for f so we get:
[tex]w=2\pi f\\w=2\pi (20)\\w=125.7s^{-1}[/tex]
For Part (b)
For wave number we plug the value for λ.So we get
K=2π/λ
[tex]K=\frac{2\pi }{1.80m}\\ K=3.490m^{-1}[/tex]
The answers for this question are
a.) frequency of the wave is 20 Hz.
b.) angular frequency of the wave is [tex]125.71428\ s^{-1}[/tex]
c.) wave number of the wave is [tex]k=3.4920\ m^{-1}[/tex].
Given to us:
Amplitude A= 2.50 mm,
Wavelength λ= 1.80 m,
Velocity V= 36.0 m/sec,
a.) To find out frequency and angular frequency of the wave,we know
[tex]frequency=\dfrac{Velocity}{Wavelength}[/tex]
[tex]f=\dfrac{V}{\lambda}[/tex]
Putting the numerical value,
[tex]f=\dfrac{36}{1.80}\\\\f= 20\ Hz[/tex]
Therefore, the frequency of this wave is 20 Hz.
Also, for angular frequency
[tex]f=2\pi \omega[/tex]
Putting the numerical value,
[tex]\omega=2\pi f \\\\\omega=2\times\pi \times 20\\\\\omega= 125.71428\ s^{-1}[/tex]
Therefore, the angular frequency of this wave is [tex]125.71428\ s^{-1}[/tex].
b.) To find out the wave number of the wave (k),
The wave number for an EM field is equal to 2π divided by the wavelength(λ) in meters.
[tex]k=\dfrac{2\pi}{\lambda}[/tex]
Putting the numerical value,
[tex]k=\dfrac{2\pi}{\lambda}\\\\k=\dfrac{2\pi}{1.80}\\\\k=3.4920\ m^{-1}[/tex]
Therefore, the wave number of the wave is [tex]k=3.4920\ m^{-1}[/tex].
Hence, for this wave
a.) frequency of the wave is 20 Hz.
b.) angular frequency of the wave is [tex]125.71428\ s^{-1}[/tex]
c.) wave number of the wave is [tex]k=3.4920\ m^{-1}[/tex].
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At a carnival, you can try to ring a bell by striking a target with a 7.76-kg hammer. In response, a 0.372-kg metal piece is sent upward toward the bell, which is 4.87 m above. Suppose that 29.7 percent of the hammer's kinetic energy is used to do the work of sending the metal piece upward. How fast must the hammer be moving when it strikes the target so that the bell just barely rings?
Answer:
3.93 m/s
Explanation:
Let the kinetic energy of hammer be 'K' and speed of hitting the target be 'v'.
Given:
Mass of the hammer (M) = 7.76 kg
Mass of the metal piece (m) = 0.372 kg
Kinetic energy of the hammer used by the metal piece = 29.7% of 'K' = 0.297K
Vertical height traveled by the metal piece (h) = 4.87 m
From conservation of energy, the kinetic energy used by the metal piece is transformed to the gravitational potential energy when it reaches the height of the bell.
Gravitational potential energy of the piece is given as:
[tex]U=mgh\\\\U=0.372\times 9.8\times 4.87=17.754\ J[/tex]
Now, as per question:
[tex]0.297K=17.754\ J\\\\K=\frac{17.754}{0.297}\\\\K=59.78\ J[/tex]
Therefore, the kinetic energy of the hammer is 59.78 J.
We know that,
Kinetic energy = [tex]\frac{1}{2}mv^2[/tex]
So, [tex]K=\frac{1}{2}mv^2[/tex]
Expressing in terms of 'v', we get:
[tex]mv^2=2K\\\\v^2=\frac{2K}{m}\\\\v=\sqrt{\frac{2K}{m}}[/tex]
Plug in the given values and solve for 'v'. This gives,
[tex]v=\sqrt{\frac{2\times 59.78}{7.76}}\\\\v=3.93\ m/s[/tex]
Therefore, the hammer must move with a speed of 3.93 m/s when it strikes the target so that the bell just barely rings.
The particle starts from rest at t=0. What is the magnitude p of the momentum of the particle at time t? Assume that t>0. Express your answer in terms of any or all of m, F, and t.
Answer:
Ft
Explanation:
We are given that
Initial velocity=u=0
We have to find the magnitude of p of the momentum of the particle at time t.
Let mass of particle=m
Applied force=F
Acceleration, [tex]a=\frac{F}{m}[/tex]
Final velocity , [tex]v=a+ut[/tex]
Substitute the values
[tex]v=0+\frac{F}{m}t=\frac{F}{m}t[/tex]
We know that
Momentum, p=mv
Using the formula
[tex]p=m\times \frac{F}{m}t=Ft[/tex]
The magnitude of the momentum of the particle after time, t is Ft = mv.
The given parameters:
initial velocity of the particle, u = 0The magnitude of the momentum of the particle after time, t is calculated as follows;
[tex]P =mv[/tex]
The force applied to the particle is calculated as;
[tex]F = ma = \frac{mv}{t} \\\\Ft = mv[/tex]
Thus, the magnitude of the momentum of the particle after time, t is Ft = mv.
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An object with total mass mtotal = 15.8 kg is sitting at rest when it explodes into three pieces. One piece with mass m1 = 4.5 kg moves up and to the left at an angle of θ1 = 18° above the –x axis with a speed of v1 = 27.5 m/s. A second piece with mass m2 = 5.4 kg moves down and to the right an angle of θ2 = 23° to the right of the -y axis at a speed of v2 = 21.4 m/s. 1)What is the magnitude of the final momentum of the system (all three pieces)?
Answer:
Explanation:
total mass, M = 15.8 kg
initial velocity, u = 0 m/s
m1 = 4.5 kg
m2 = 5.4 kg
v1 = 27.5 m/s at an angle 18°
v2 = 21.4 m/s at an angle 23°
As there is no external force acts on the system, so the moemntum of the system is conserved.
Momentum before collision = momentum after collision
as the momentum before collision is zero, so the momentum after collision is also zero.
A cube has one corner at the origin and the opposite cornerat the point (L,L,L). The sides of the cube are parallel to thecoordinate planes. The electric field in and around the cube isgiven by E=(a+bx)x+cy.Part A Find the total electric flux phi E through the surface of the cube. Express your answer in terms of a,b,c and L. Phi E=Part B This part will be visible after youcomplete previous item(s). Part C What is the net charge q inside the cube? Express your answer in terms of a,b,c ,L,and epsilon. q=
Answer:
Explanation:
in this case, flux and area vectors are parallel . therefore, flux is just the dot product of electric field and area vector.
(flux=EAcos0 =EA)
flux through the face x face
Physics homework question answer, step 1, image 1
the flux through -x face
Physics homework question answer, step 1, image1
Step 2
the flux through y face,
Physics homework question answer, step 2, image 2
the flux through -y face,
Physics homework question answer, step 2, image 2
the flux through the z faces are zero,
therefore, the net flux is,
Physics homework question answer, step 2, image 3
...
Final answer:
The total electric flux IS ΦE = aL2 + (a + bL)L2 + cL3 net charge is q = ε0 ( aL2 + (a + bL)L2 + cL3 )
Explanation:
The question asks to find the total electric flux ΦE through the surface of a cube with an electric field given by E = (a + bx)x + cy, and to determine the net charge q inside the cube.
Part A: Total Electric Flux through the Cube's Surface
To determine the electric flux through the cube, we need to use Gauss's Law, which relates the electric flux through a closed surface to the charge enclosed by the surface.
The electric flux ΦE through a surface is defined by the integral of the electric field E dotted with the differential area dA over the entire surface. Since the sides of the cube are parallel to the coordinate planes, the electric field components perpendicular to the x, y, and z faces will contribute to the flux.
In this case, only the x-component of the electric field (a + bx)x contributes to the flux through the surfaces perpendicular to the x-axis, which are at x = 0 and x = L.
The y-component cy of the electric field contributes to the flux through the surfaces perpendicular to the y-axis, which are at y = 0 and y = L. There is no z-component to the electric field, so the surfaces perpendicular to the z-axis will have no flux.
The total electric flux through the cube is thus:
ΦE = ∫( (a + bx) x + cy ) · dA
For the two faces perpendicular to the x-axis:
ΦE, x=0 = ∫( a x ) · dA = aL2 (at x = 0)
ΦE, x=L = ∫( (a + bL)x ) · dA = (a + bL)L2 (at x = L)
For the two faces perpendicular to the y-axis:
ΦE, y=0 = 0 (since cy is 0 at y = 0)
ΦE, y=L = ∫( cy ) · dA = cL3 (at y = L)
Summing these up:
ΦE = aL2 + (a + bL)L2 + cL3
Part C: Net Charge Inside the Cube
The net charge q inside the cube can be found using Gauss's Law, which states "the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity of free space (epsilon_0)."
q = ΦE ε0
Substituting the expression for ΦE we have:
q = ε0 ( aL2 + (a + bL)L2 + cL3 )
A bullet is fired with a horizontal velocity of 1500 ft/s through a 6-lb block A and becomes embedded in a 4.95-lb block B. Knowing that blocks A and B start moving with velocities of 5 ft/s and 9 ft/s, respectively, determine (a) the weight of the bullet, (b) its velocity as it travels from block A to block B.
Answer:
weight of the bullet is 0.0500 lb
velocity as it travels from block A to block B is 900 ft/s
Explanation:
given data
horizontal velocity = 1500 ft/s
mass block A = 6-lb
mass block B = 4.95 lb
blocks A velocity = 5 ft/s
blocks B velocity = 9 ft/s
solution
we apply here law of conservation of momentum that is
m × v(o) + m1 × (0) + m2 × (0) = m × v(2) + m1 × v1 + m2 × v2 ................1
put here value and we get m
m = [tex]\frac{m1 \times v1 +m2 \times v2}{v(o) - v2}[/tex] ..............2
m = [tex]\frac{6 \times 5 + 4.95 \times 9}{1500 - 9}[/tex]
m = 0.0500 lb
and
when here bullet is pass through the A block then moment is conserve that is
m × v(o) + m × v1 + m1 × v1 ............3
v1 = [tex]\frac{m\times v(o)- m1\times v1}{m}[/tex]
v1 = [tex]\frac{0.0500\times 1500 - 61\times 5}{0.05}[/tex]
v1 = 900 ft/s
A 10 kg turkey, He kicks the 0.5 kg ball with a force of 50N for 0.2 seconds and the ball flies straight away horizontally from turkey.?
What is the velocity of the ball after the kick?
What force did the turkey feel?
Why doesn’t the turkey go flying backward since the ball flies forward? Is momentum conserved?
Answer:
a. 20m/s
b.50N
c. Turkey has a larger mass than the ball. Neglible final acceleration and therefore remains stationery.
Explanation:
a. Given the force as 50N, times as 0.2seconds and the weight of the ball as 0.5 kg, it's final velocity can be calculated as:
[tex]F\bigtriangleup t=m\bigtriangleup v\\\\50N\times 0.2s=0.5kg\times \bigtriangleup v\\\\\bigtriangleup v=2(50N\times0.2)\\\\=20m/s[/tex]
Hence, the velocity of the ball after the kick is 20m/s
b.The force felt by the turkey:
#Applying Newton's 3rd Law of motion, opposite and equal reaction:
-The turkey felt a force of 50N but in the opposite direction to the same force felt by the ball.
c. Using the law of momentum conservation:
-Due to ther external forces exerted on the turkey, it remains stationery.
-The turkey has a larger mass than the ball. It will therefore have a negligible acceleration if any and thus remains stationery.
-Momentum is not conserved due to these external forces.
The velocity of the ball after the kick is 200 m/s. The turkey feels a force of 50N. The turkey doesn't go flying backward because of its greater mass compared to the ball.
Explanation:The velocity of the ball after the kick can be calculated using the equation:
velocity = force / mass * time
Substituting the values, we get:
velocity = 50N / 0.5kg * 0.2s = 200 m/s
The force that the turkey feels can be determined using Newton's third law, which states that for every action, there is an equal and opposite reaction. So, the force the turkey feels will be the same as the force of the kick, which is 50N.
The turkey doesn't go flying backward because the turkey has a greater mass than the ball. According to Newton's second law, the acceleration of an object is inversely proportional to its mass. Since the turkey has a greater mass, it experiences less acceleration compared to the ball. Conservation of momentum is upheld in this situation, as the momentum of the turkey and ball together remains the same before and after the kick.
A comet orbits a star in a strongly elliptical orbit. The comet and star are far from other massive objects. As the comet travels away from the star, how does the kinetic energy and potential energy of the system change?
Answer:
the kinetic energy decreases and the potential energy Increases.
Explanation:
as the comet travels away from the star it gains an energy which it posses because of its position or state.once the comet moves away form the star the kinetic decreases until its lost all together to where the potential energy starts increasing.
Answer:
The potential energy increase and the kinetic energy decrease.
The Acorn Insurance Company charges $3.50 for each unit of coverage under a block of 1-year term life insurance policies. The annual premium amount for a $50,000 policy in this block would be equal to ____.
Answer:
$175
Explanation:
Insurance premium is expressed as a rate $1000
($3.50 per $1000)
Therefore;
Annual premium= $50000x$3.50/$1000
= $175
Final answer:
The annual premium for a $50,000 policy at the rate of $3.50 per unit of coverage from the Acorn Insurance Company would be $175.00.
Explanation:
The Acorn Insurance Company charges $3.50 for each unit of coverage under a 1-year term life insurance policy. To calculate the annual premium for a $50,000 policy, we'll need to divide the total coverage amount by the unit price and then multiply by the cost per unit. Here's the math:
Total Coverage Needed: $50,000
Cost Per Unit of Coverage: $3.50
Number of Units: $50,000 / $1,000 = 50
Annual Premium: 50 units * $3.50 = $175.00
So, the annual premium amount for a $50,000 policy in this block would be $175.00.
A chamber fitted with a piston can be controlled to keep the pressure in the chamber constant as the piston moves up and down to increase or decrease the chamber volume. The chamber contains an ideal gas at 296 K and 1.00 atm.What is the work done on the gas as the piston compresses it from 1.00 L to 0.633 L ?
Express your answer with the appropriate units.
W =
J
Answer:
W = 37.2J
Explanation:
See attachment below.
A 0.9 µF capacitor is charged to a potential difference of 10.0 V. The wires connecting the capacitor to the battery are then disconnected from the battery and connected across a second, initially uncharged, capacitor. The potential difference across the 0.9 µF capacitor then drops to 2 V. What is the capacitance of the second capacitor?
Answer:
3.6μF
Explanation:
The charge on the capacitor is defined by the formula
q = CV
because the charge will be conserved
q₁ = C₁V₂
q₂ = C₂V₂ where C₂ V₂ represent the charge on the newly connected capacitor and the voltage drop across the two capacitor will be the same
q = q₁ + q₂ = C₁V₂ + C₂V₂
CV = CV₂ + C₂V₂
CV - CV₂ = C₂V₂
C ( V - V₂) = C₂V₂
C ( V/ V₂ - V₂ /V₂) = C₂
C₂ = 0.9 ( 10 /2) - 1) = 0.9( 5 - 1) = 3.6μF
A white billiard ball with mass mw = 1.43 kg is moving directly to the right with a speed of v = 3.39 m/s and collides elastically with a black billiard ball with the same mass mb = 1.43 kg that is initially at rest. The two collide elastically and the white ball ends up moving at an angle above the horizontal of θw = 38° and the black ball ends up moving at an angle below the horizontal of θb = 52°. 1)What is the final speed of the white ball? m/s 2)What is the final speed of the black ball? m/s 3)What is the magnitude of the final total momentum of the system? kg-m/s 4)What is the final total energy of the system?
Answer: a) VW = 1.28m/s
b) Vb = 3.86m/s
c) p = 5.82kgm/s
d) E = 11.84J
Explanation: To solve this question, we make use of explosion formula in linear momentum concept.
Please find the attached file for the solution
A 6.0-cm-diameter, 11-cm-long cylinder contains 100 mg of oxygen (O2) at a pressure less than 1 atm. The cap on one end of the cylinder is held in place only by the pressure of the air. One day when the atmospheric pressure is 100 kPa, it takes a 173 N force to pull the cap off.
Explanation:
The given data is as follows.
Mass of oxygen present = 100 mg = [tex]100 \times 10^{-3}[/tex] g
So, moles of oxygen present are calculated as follows.
n = [tex]\frac{100 \times 10^{-3}}{32}[/tex]
= [tex]3.125 \times 10^{-3}[/tex] moles
Diameter of cylinder = 6 cm = [tex]6 \times 10^{-2}[/tex] m
= 0.06 m
Now, we will calculate the cross sectional area (A) as follows.
A = [tex]\pi \times \frac{(0.06)^{2}}{4}[/tex]
= [tex]2.82 \times 10^{-3} m^{2}[/tex]
Length of tube = 11 cm = 0.11 m
Hence, volume (V) = [tex]2.82 \times 10^{-3} \times 0.11[/tex]
= [tex]3.11 \times 10^{-4} m^{3}[/tex]
Now, we assume that the inside pressure is P .
And, [tex]P_{atm}[/tex] = 100 kPa = 100000 Pa,
Pressure difference = 100000 - P
Hence, force required to open is as follows.
Force = Pressure difference × A
= [tex](100000 - P) \times 2.82 \times 10^{-3}[/tex]
We are given that force is 173 N.
Thus,
[tex](100000 - P) \times 2.82 \times 10^{-3}[/tex] = 173
Solving we get,
P = [tex]3.8650 \times 10^{4} Pa[/tex]
= 38.65 kPa
According to the ideal gas equation, PV = nRT
So, we will put the values into the above formula as follows.
PV = nRT
[tex]38.65 \times 3.11 \times 10^{-4} = 3.125 \times 10^{-3} \times 8.314 \times T[/tex]
T = 462.66 K
Thus, we can conclude that temperature of the gas is 462.66 K.
A thin aluminum rod lies along the x-axis and has current of I = 16.0 A running through it in the +x-direction. The rod is in the presence of a uniform magnetic field, perpendicular to the current. There is a magnetic force per unit length on the rod of 0.113 N/m in the −y-direction.
(a) What is the magnitude of the magnetic field (in mT) in the region through which the current passes?
(b) What is the direction of the magnetic field in the region through which the current passes?
Answer:
a) The magnitude of the magnetic field = 7.1 mT
b) The direction of the magnetic field is the +z direction.
Explanation:
The force, F on a current carrying wire of current I, and length, L, that passes through a magnetic field B at an angle θ to the flow of current is given by
F = (B)(I)(L) sin θ
F/L = (B)(I) sin θ
For this question,
(F/L) = 0.113 N/m
B = ?
I = 16.0 A
θ = 90°
0.113 = B × 16 × sin 90°
B = 0.113/16 = 0.0071 T = 7.1 mT
b) The direction of the magnetic field will be found using the right hand rule.
The right hand rule uses the first three fingers on the right hand (the thumb, the pointing finger and the middle finger) and it predicts correctly that for current carrying wires, the thumb is in the direction the wire is pushed (direction of the force; -y direction), the pointing finger is in the direction the current is flowing (+x direction), and the middle finger is in the direction of the magnetic field (hence, +z direction).
The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 volt potential difference is suddenly applied to the initially uncharged plates through a 1025 ohm resistor in series with the deflection plates. How long does it take for the potential difference between the deflection plates to reach 55 volts?
Explanation:
Given data:
Area A = 10 cm×2 cm = 20×10⁻⁴ m²
Distance d between the plates = 1 mm = 1×10⁻³m
Voltage of the battery is emf = 100 V
Resistance = 1025 ohm
Solution:
In RC circuit, the voltage between the plates is related to time t. Initially the voltage is equal to that of battery V₀ = emf = 100V. But After time t the resistance and capacitor changes it and the final voltage is V that is given by
[tex]V = V_{0}(1-e^{\frac{-t}{RC} } )\\\frac{V}{V_{0} } = 1-e(^{\frac{-t}{RC} }) \\e^{\frac{-t}{RC} } = 1- \frac{V}{V_{0} }[/tex]
Taking natural log on both sides,
[tex]e^{\frac{-t}{RC} } = 1- \frac{V}{V_{0} } \\\frac{-t}{RC} = ln(1-\frac{V}{V_{0} } )\\t = -RCln(1 - \frac{V}{V_{0} })[/tex]
[tex]t = -RC ln (1-\frac{V}{V_{0} })[/tex] (1)
Now we can calculate the capacitance by using the area of the plates.
C = ε₀A/d
= [tex]\frac{(8.85*10^{-12))} (20*10^{-4}) }{1*10^{-3} }[/tex]
= 18×10⁻¹²F
Now we can get the time when the voltage drop from 100 to 55 V by putting the values of C, V₀, V and R in the equation (1)
[tex]t = -RC ln (1-\frac{V}{V_{0} })[/tex]
= -(1025Ω)(18×10⁻¹² F) ln( 1 - 55/100)
= 15×10⁻⁹s
= 15 ns
An electric motor is plugged into a standard wall socket and is running at normal speed. Suddenly, some dirt prevents the shaft of the motor from turning quite so rapidly. What happens to the back emf of the motor, and what happens to the current that the motor draws from the wall socket?
a.) The back emf decreases, and the current drawn from the socket decreases.
b.) The back emf increases, and the current drawn from the socket decreases.
c.) The back emf decreases, and the current drawn from the socket increases.
d.) The back emf increases, and the current drawn from the socket increases.
Final answer:
The back emf of the motor decreases due to reduced speed caused by dirt in the shaft, which, in turn, causes the current drawn from the wall socket to increase.
Explanation:
When dirt prevents the shaft of an electric motor from turning rapidly, the back emf produced by the motor decreases. Since the back emf is proportional to the motor's angular velocity, any decrease in speed results in a decrease in back emf. Because the back emf opposes the applied voltage from the wall socket, a decrease in back emf implies that the voltage across the motor's coil will increase, causing the motor to draw a larger current from the socket. Therefore, the current that the motor draws from the wall socket will increase. The correct answer to the question is (c) The back emf decreases, and the current drawn from the socket increases.
A charge Q is transfered from an initially uncharged plastic ball to an identical ball 12 cm away. The force of attraction is then 17 mN. How many electrons were transfered from one ball to the other
Answer:
Explanation:
The ball from which electrons are transferred will acquire Q charge and the ball receiving electrons will acquire - Q charge . force of attraction between them
= k Q² / d²
9 x 10⁹ x Q² / .12² = 17 X 10⁻³
Q² = .0272 X 10⁻¹²
Q = .1650 X 10⁻⁶ C
No of electrons = .1650 x 10⁻⁶ / 1.6 x 10⁻¹⁹
= .103125 x 10¹³
1.03125 x 10¹²
The dimension of a room is 10 feet high, 40 feet long and 30 feet wide. If ¼ point of pyridine is allowed to volatilize in this room at NTP with nor ventilation, what is the concentration in mg/m3 and ppm for pyridine in this room? Pyridine s.g. = 0.98
Answer:
Concentration in mg/m^3 = 3845.39mg/m^3
in PPM = 3.84539 ppm
Explanation:
From the dimensions, we compute the volume (V)
V = 10 x 40 x 30 = 12000ft^3
Volatilize volume = 1/4 *12000 = 3000ft^3 at NTP
Volume of pyridine in the room = ( 12000 - 3000) = 9000ft^3
converting to meters = 9000 * (0.3048)^3m^3
= 254. 85m^3
Computing concentration(C) = mass/s.g/volume
S.g converted to mg per meter cube = 0.98x1000= 0.98 * 10^3g/m^3 = 0.98 * 10^6mg/m^3
Hence concentration = 0.98 * 10^6/254.85 = 3845.39 mg/m^3
Parts per million denotes low concentration of a solution
1ppm = 1mg per liter , and 1 liter = 0.001m^3
Hence, this = 3845.39/1000 = 3.84539 PPM
Two identical wires A and B are subject to tension. The tension in wire A is 3 times larger than that in wire B. Find the ratio of the frequencies of the first harmonic in these two wires, fA1 / fB1.
Answer:
1.732
Explanation:
Let
Tension in wire B=T
Tension in wire A=3 T
We have to find the ratio of the frequencies of the first harmonic in these two wires.
When two wires are identical then the length of both wires are same.
Suppose, the length of each wire=l
Frequency=[tex]\frac{1}{2l}\sqrt{\frac{T}{\mu}}[/tex]
Where [tex]\mu=[/tex]Mass per unit length
Mass per unit length of both wires are same because the two wires are identical.
[tex]\mu_A=\mu_B[/tex]
[tex]\frac{f_A}{f_B}=\frac{\frac{1}{2l}\sqrt{\frac{T_A}{\mu_A}}}{\frac{1}{2l}\sqrt{\frac{T_B}{\mu_B}}}[/tex]
[tex]\frac{f_A}{f_B}=\frac{\frac{1}{2l}\sqrt{\frac{T_A}{\mu_A}}}{\frac{1}{2l}\sqrt{\frac{T_B}{\mu_A}}}=\sqrt{\frac{T_A}{T_B}}=\sqrt{\frac{3T}{T}}[/tex]
[tex]\frac{f_A}{f_B}=\sqrt 3=1.732[/tex]
The ratio of the frequencies of the first harmonic in these two wires is [tex]\sqrt{3}[/tex].
The given parameters;
tension in wire A = Ttension in wire B = 3TThe frequency of first harmonic of each wire is calculated as follows;
[tex]F_ A = \frac{1}{2l} \sqrt{\frac{3T}{\mu} } \\\\F_B = \frac{1}{2l} \sqrt{\frac{T}{\mu} }[/tex]
where;
l is the length of the wiresT is the tension on the wireμ is the mass per unit lengthThe ratio of the two frequencies is calculated as follows;
[tex]\frac{F_A}{F_B} = \frac{\frac{1}{2l} \sqrt{\frac{3T}{\mu} } }{\frac{1}{2l} \sqrt{\frac{T}{\mu} } } \\\\\frac{F_A}{F_B} = \sqrt{\frac{3T}{T} } \\\\\frac{F_A}{F_B} = \sqrt{3}[/tex]
Thus, the ratio of the frequencies of the first harmonic in these two wires is [tex]\sqrt{3}[/tex].
Learn more here:https://brainly.com/question/14149129
The potential energy for a certain mass moving in one dimension is given by U(x)=(2.0J/m3)x3−(15J/m2)x2+(36J/m)x−23JU(x)=(2.0J/m3)x3−(15J/m2)x2+(36J/m)x−23JU(x) = (2.0 {\rm J/m}^{3})x^{3}- (15 {\rm J/m}^{2})x^{2}+ (36 {\rm J/m})x - 23 {\rm J}. Find the location(s) where the force on the mass is zero.
Answer:x=2 and x=3
Explanation:
Given
Potential Energy for a certain mass is
[tex]U(x)=2x^3-15x^2+36x-23[/tex]
and we know force is given by
[tex]F=-\frac{\mathrm{d} U}{\mathrm{d} x}[/tex]
[tex]F=-(2\times 3x^2-15\times 2x+36)[/tex]
For Force to be zero F=0
[tex]\Rightarrow 6x^2-30x+36=0[/tex]
[tex]\Rightarrow x^2-5x+6=0[/tex]
[tex]\Rightarrow x^2-2x-3x+6=0[/tex]
[tex]\Rightarrow (x-2)(x-3)=0[/tex]
Therefore at x=2 and x=3 Force on particle is zero.
The block has a weight of 75 lblb and rests on the floor for which μkμk = 0.4. The motor draws in the cable at a constant rate of 6 ft/sft/s. Neglect the mass of the cable and pulleys.
The given question is incomplete. The complete question is as follows.
The block has a weight of 75 lb and rests on the floor for which [tex]\mu k[/tex] = 0.4. The motor draws in the cable at a constant rate of 6 ft/sft/s. Neglect the mass of the cable and pulleys.
Determine the output of the motor at the instant [tex]\theta = 30^{o}[/tex].
Explanation:
We will consider that equilibrium condition in vertical direction is as follows.
[tex]\sum F_{y} = 0[/tex]
N - W = 0
N = W
or, N = 75 lb
Again, equilibrium condition in the vertical direction is as follows.
[tex]\sum F_{x} = 0[/tex]
[tex]T_{2} - F_{k}[/tex] = 0
[tex]T_{2} = \mu_{k} N[/tex]
= [tex]0.4 \times 75 lb[/tex]
= 30 lb
Now, the equilibrium equation in the horizontal direction is as follows.
[tex]\sum F_{x} = 0[/tex]
[tex]T Cos (30^{o}) + T Cos (30^{o}) = T_{2}[/tex]
[tex]2T Cos (30^{o}) = T_{2}[/tex]
or, T = [tex]\frac{T_{2}}{2 Cos (30^{o})}[/tex]
= [tex]\frac{30}{2 Cos (30^{o})}[/tex]
= [tex]\frac{30}{1.732}[/tex]
= 17.32 lb
Now, we will calculate the output power of the motor as follows.
P = Tv
= [tex]17.32 lb \times 6[/tex]
= [tex]103.92 \times \frac{1}{550} \times \frac{hp}{ft/s}[/tex]
= 0.189 hp
or, = 0.2 hp
Thus, we can conclude that output of the given motor is 0.2 hp.
Answer:
The out put power is 0.188 hp.
Explanation:
Given that,
Weight = 75 lb
Coefficient of friction = 0.4
Rate = 6 ft/s
Suppose, Determine the output of the motor at the instant θ = 30°.
For block,
We need to calculate the force in vertical direction
Using balance equilibrium equation in vertical
[tex]\sum{F_{y}}=0[/tex]
[tex]N-W=0[/tex]
[tex]N=W[/tex]
Put the value into the formula
[tex]N=75\ lb[/tex]
Using balance equilibrium equation in horizontal
[tex]\sum{F_{x}}=0[/tex]
[tex]T_{2}-f_{k}=0[/tex]
[tex]T_{2}=\mu_{k}N[/tex]
Put the value into the formula
[tex]T_{2}=0.4\times75[/tex]
[tex]T_{2}=30\ lb[/tex]
For pulley,
We need to calculate the force
Using balance equilibrium equation in horizontal
[tex]\sum{F_{x}}=0[/tex]
[tex]T\cos\theta+T\cos\theta=T_{2}[/tex]
[tex]2T\cos30=T_{2}[/tex]
[tex]T=\dfrac{T_{2}}{2\cos30}[/tex]
Put the value into the formula
[tex]T=\dfrac{30}{2\times\cos30}[/tex]
[tex]T=17.32\ lb[/tex]
We need to calculate the out put power
Using formula of power
[tex]P=Tv[/tex]
Put the value into the formula
[tex]P=17.32\times6[/tex]
[tex]P=103.92\ lb.ft/s[/tex]
[tex]P=0.188\ hp[/tex]
Hence, The out put power is 0.188 hp.