Answer:
dimensions of the drag coefficient is [tex][M^0 L^0 T^0][/tex]
Drag coefficient is a dimensionless quantity
Explanation:
force is given by[tex]F=\frac{C_{D} \rho V^2 A}{2}[/tex]
we get expression for drag coefficient [tex]C_{D} =\frac{2F}{\rho V^2 A}[/tex]
By substituting the dimensions of the F,V,A and density , we get
[tex]C_{D} =\frac{[F]}{[\rho ][V]^2[A]} \\C_{D} =\frac{[MLT^{-2}]}{[ML^{-3} ][L T^{-1}]^2[L^2]} \\C_{D} =\frac{[MLT^{-2}]}{[ML^{-3} ][L^2 T^{-2}][L^2]} \\C_{D} =\frac{[MLT^{-2}]}{[MLT^{-2}]}\\C_{D}=[M^0 L^0 T^0][/tex]
Drag coefficient is a dimensionless
The dimensions of the drag coefficient, CD, are kg/m.
Explanation:The dimensions of the drag coefficient, CD, can be determined by examining the equation for force, F, of the wind blowing against a building. In this equation, the dimensions for force are mass x acceleration, which are kg x m/s^2. On the other side of the equation, the wind speed, V, has dimensions of m/s, the density, rho, has dimensions of kg/m^3, and the cross-sectional area, A, has dimensions of m^2. Therefore, in order for the equation to be balanced, the dimensions of the drag coefficient must be kg/m.
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In a phasor representation of a transverse wave on a string, what does the length of the phasor represent?
Final answer:
The length of the phasor in a phasor diagram representing a transverse wave on a string indicates the wave's amplitude, which corresponds to the maximum displacement of the medium's particles from their equilibrium position.
Explanation:
In the phasor representation of a transverse wave on a string, the length of the phasor corresponds to the amplitude of the wave. In a phasor diagram, this amplitude represents the maximum displacement of the wave particles from the equilibrium position as the wave propagates through the medium. The phasor's length will rotate in a circular motion at a rate determined by the wave's frequency, and this motion represents the oscillatory nature of the wave at a certain point in space over time. The amplitude is a crucial parameter as it determines the energy carried by the wave, with a larger amplitude indicating a greater energy transfer.
The phasor length is particularly important when analyzing multiple wave forms together, such as voltage and current in electrical circuits, where the ratio of their lengths can denote relative magnitudes, such as resistance in the circuit. In this context, however, we focus on mechanical waves on a string, and the length of the phasor would only represent the wave amplitude, not voltage or current.
A soccer player kicks the ball that travels a distance of 60.0 m on a level field. The ball leaves his foot at an initial speed of (v0) and an angle of 26.0° above the ground. Find the initial speed (v0) of the ball.
Answer:
27.3 m/s
Explanation:
We are given that
Distance travel by ball=x=60 m
[tex]\theta=26^{\circ}[/tex]
We have to find the initial speed([tex]v_0)[/tex] of the ball.
[tex]x=v_0cos\theta t[/tex]
Using the formula
[tex]60=v_0cos 26 t[/tex]
[tex]t=\frac{60}{v_ocos 26}=\frac{60}{v_0\times 0.899}=\frac{66.7}{v_0}[/tex]
The value of y at point of foot of the vertical distance
y=0
[tex]y=v_0sin\theta t-\frac{1}{2}gt^2[/tex]
Using [tex]g=9.8m/s^2[/tex]
Using the formula
[tex]0=v_0sin 26\times \frac{66.7}{v_0}-4.9\times (\frac{66.7}{v_0})^2[/tex]
[tex]4.9\times \frac{(66.7)^2}{v^2_0}=0.44\times 66.7[/tex]
[tex]v^2_0=\frac{4.9\times (66.7)^2}{0.44\times 66.7}[/tex]
[tex]v^2_0=742.8[/tex]
[tex]v_0=\sqrt{742.8}=27.3 m/s[/tex]
Hence, the initial speed of the ball=27.3 m/s
Answer:
27.3 m/s
Explanation:
Horizontal range, R = 60 m
angle of projection, θ = 26°
Let the velocity of projection is vo.
Use the formula of range of the projectile
[tex]R = \frac{u^{2}Sin2\theta} {g}[/tex]
[tex]60 = \frac{v_{0}^{2}Sin52}{9.8}[/tex]
vo = 27.3 m/s
Thus, the velocity of projection is 27.3 m/s.
A 0.73-kg metal sphere oscillates at the end of a vertical spring. As the spring stretches from 0.12 m to 0.23 m (relative to its unstrained length), the speed of the sphere decreases from 7.2 to 4.5 m/s. What is the spring constant of the spring?
The spring constant (k) can be obtained by employing the principle of conservation of energy. Here, the kinetic energy of the metal sphere at the beginning of the motion equals the potential energy at the maximum stretch of the spring. Substituting the given values into the energy equation, solving for 'k' yields the spring constant.
Explanation:The subject of this question lies within the domain of Physics, specifically the domain of mechanics and dynamics dealing with springs and oscillations. The spring constant (k) can be derived from the principle of conservation of energy. Here, we are ignoring friction and air resistance, meaning that the sum of kinetic energy and potential energy remains constant throughout the motion of the metal sphere.
At the beginning, all the energy is kinetic, and at the maximum stretch, all the energy is potential. This can be represented by the equation 0.5*m*v1^2 = 0.5*k*x2^2. By substituting the given values of m (mass = 0.73 kg), v1 (initial velocity = 7.2 m/s), and x2 (maximum displacement = 0.23 m), we can solve for k (spring constant). Here, the calculation would be as follows: k = m*v1^2/x2^2 = (0.73 kg*(7.2 m/s)^2)/(0.23 m)^2. After performing the required calculations, you can obtain the numerical value of the spring constant.
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While David was riding his bike around the circular cul-de-sac by his house, he wondered if the constant circular motion was having any effect on his tires. What would be the best way for David to investigate this?
A.
Measure the circumference of the tire before and after riding.
B.
Measure the total distance traveled on his bike and divide this by how long it took him.
C.
Measure the wear on his treads before and after riding a certain number of laps.
D.
Time how long it takes him to ride 5 laps around his cul-de-sac.
Answer:
C.
Measure the wear on his treads before and after riding a certain number of laps.
Answer:
Measure the wear on his treads before and after riding a certain number of laps.
Explanation:
By riding in a circular motion the inside of the tire will be in contact with the road more than the outside of the tire. Thus, to see if the constant circular motion had any effect on his tires David should measure the tread depth on both the inside and the outside of the tires before the experiment and measure the inside and the outside of the tires (at the same location on the tires) after the experiment. Then he can compare the tread loss on the inside of the tire to the tread loss on the outside of the tire.
Given the following frequencies, calculate the corresponding periods. a. 60 Hz b. 8 MHz c. 140 kHz d. 2.4 GHz
The frequency can be defined as the inverse of the period, that is, it can be expressed as
[tex]T = \frac{1}{f}[/tex]
Here,
T = Period
f = Frequency
For each value we only need to replace the value and do the calculation:
PART A)
[tex]T = \frac{1}{f}[/tex]
[tex]T = \frac{1}{60Hz}[/tex]
T = 0.0166s
PART B)
[tex]T = \frac{1}{f}[/tex]
[tex]T = \frac{1}{8*10^6}[/tex]
[tex]T = 1.25*10^{-7} s[/tex]
PART C)
[tex]T = \frac{1}{f}[/tex]
[tex]T = \frac{1}{140*10^{3}}[/tex]
[tex]T = 7.14*10^{-6}s[/tex]
PART D)
[tex]T = \frac{1}{f}[/tex]
[tex]T = \frac{1}{2.4*10^{9}}[/tex]
[tex]T = 4.166*10^{-10}s[/tex]
During a baseball game, a player hits a a ball with a speed of 43m/s at an angle of 25∘ above the horizontal. When the player hit the ball, it was 1m above the ground, and after the hit, the ball flies straight toward the center field fence.
How high above the ground is the ball when it reaches the center field fence, which is a distance of 400ft (122m) away?
Answer:
s_y = 9.82 m
Explanation:
Given:
- Initial velocity v_i = 43 m/s
- Angle with the horizontal Q = 25 degree
- Initial distance s_o = 1 m
- The distance of the center field fence x_f = 122 m
Find:
- How high above the ground is the ball when it reaches the center field fence
Solution:
- The time taken for the ball to reach the fence t_f:
s_x = S(0) + v_x,o*t
122 = 0 + (43*cos(25))*t
t = 122 / (43*cos(25)) = 3.1305 s
- Compute the height of the ball when it reaches the fence:
s_y = S(0) + v_y,o*t + 0.5*g*t^2
s_y = 1 + 43*sin(25)*3.1305 - 0.5*(9.81)*(3.1305)^2
s_y = 9.82 m
A particle moves in a straight line with an initial velocity of 35 m/s and a constant acceleration of 38 m/s2. If at t = 0, x = 0, what is the particle's position (in m) at t = 6 s?
Answer:
d=894 m
Explanation:
Given that
initial velocity ,u= 35 m/s
Acceleration ,a= 38 m/s²
time ,t= 6 s
Given that at t= 0 s ,x= 0 m
We know that
[tex]d=ut+\dfrac{1}{2}at^2 [/tex]
d=Displacement
Now by putting the values
[tex]d=35\times 6+\dfrac{1}{2}\times 38\times 6^2 [/tex]
d=894 m
Therefore the particle position after 6 sec will be 894 m.
Final answer:
The position of the particle at t = 6 seconds, with an initial velocity of 35 m/s and a constant acceleration of 38 m/s², is 894 meters from the start.
Explanation:
The question asks us to calculate the position of a particle moving in a straight line at t = 6 seconds, given an initial velocity of 35 m/s and a constant acceleration of 38 m/s². To find the position, we can use the kinematic equation:
x = v0t + ½at²
where x is the position, v0 is the initial velocity, a is the acceleration, and t is the time. Plugging in our values we get:
x = (35 m/s)(6 s) + ½(38 m/s²)(6 s)²
x = 210 m + ½(38 m/s²)(36 s²)
x = 210 m + 684 m
x = 894 m
Therefore, the position of the particle at t = 6 s is 894 meters from the starting point.
A 1.65 mol sample of an ideal gas for which Cv,m = 3R/2 undergoes the following two-step process:1) from an initial state of the gas described by T = 14.5degrees C and P = 2.00 x 104 Pa, the gas undergoes anisothermal expansion against a constant external pressure of 1.00 x104 Pa until the volume has doubled.2) subsequently the gas is cooled at constant volume. Thetemperature falls to -35.6 degrees C.Calculate q, w, , and for each step and for the overallprocess.
Answer:
W = -1.97KJ, Q = 1.97KJ, Delta U = 0
Delta U = -1.03KJ, Q = -1.03KJ, Delta H = -1.72KJ
Explanation:
The deatiled step by step calculation using the ideal gas equation (Pv =nRT), The first law of thermodynamics ( dQ =dW + dU) as applied is as shown in the attached file.
In the first step, q = -157.29 R mol and w = -2.00 x 10^4 V Pa. In the second step, q = -141.45 R mol and w = 0. The total heat transfer (q_total) is -298.74 R mol and the total work done (w_total) is -2.00 x 10^4 V Pa.
Explanation:The first step in the process is an isothermal expansion. In an isothermal process, the temperature remains constant, which means the change in internal energy (∆U) is zero. Since ∆U = q + w, this means that q = -w. We can calculate q using the equation q = nCv,m∆T, where n is the number of moles, Cv,m is the molar heat capacity at constant volume, and ∆T is the change in temperature. In this case, q = -w = nCv,m∆T = (1.65 mol)(3R/2)(-35.6 + 14.5) = -157.29 R mol.
The work done during an expansion or contraction process can be calculated using the equation w = -P∆V, where P is the external pressure and ∆V is the change in volume. In this case, the volume doubles, so ∆V = 2V, and the pressure is constant at 1.00 x 10^4 Pa. Therefore, w = -P∆V = -(1.00 x 10^4 Pa)(2V) = -2.00 x 10^4 V Pa.
In the second step, the gas is cooled at constant volume, so no work is done (w = 0). The heat transfer (q) can be calculated using the same equation as before, q = nCv,m∆T. In this case, q = (1.65 mol)(3R/2)(-35.6 - 14.5) = -141.45 R mol.
Putting it all together, for the first step, q = -w = -157.29 R mol and for the second step, q = -141.45 R mol. The total heat transfer for the overall process is the sum of the heat transfers for each step, so q_total = q1 + q2 = (-157.29 R mol) + (-141.45 R mol) = -298.74 R mol. As for the total work done (w_total), it is the sum of the work done in the first step and the work done in the second step, so w_total = w1 + w2 = (-2.00 x 10^4 V Pa) + 0 = -2.00 x 10^4 V Pa.
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Three parachutists have the following masses: A: 50 kg, B: 40 kg, C: 75 kg Which one has the greatest terminal velocity?
Answer:
A: 50 kg
Explanation:
"Stop to Think 16.1" on page 423 of your textbook. Also, for situation (a), descibe what happens to the speed of the wave, the frequency, and the wavelength when you start moving your hand up and down at a faster rate.
Answer:
wave speed= constant
frequency = increase
wavelength = decrease
Explanation:
Solution:
- The three basic parameters of a wave are speed, frequency and wavelength. These three parameters are related to each other by an expression:
v = f * λ
Where,
- v is the speed of the wave in m/s.
- f frequency of the wave in Hz.
- λ wavelength of the wave in m
- We are asked how would each of these parameter change if we move the hand up and down faster. The hand moves from a crest to trough faster than before and back again. We can see that the time between a cycle has decreased; hence, frequency f increases. Consequently, we can see that wave speed v remains constant - the medium of transfer of wave energy - remains same. Then from our relation above if we hold speed constant and increase f then the wavelength λ would have to decrease.
When looking at the top of a building 450 m away, the angle between the top of the building and your eye level is 30°. If your eyes are 1.5 m above the ground, how tall is the building? ANSWER IN 3 DECIMALS (###.###) You might need to use your calculator's sin,cos or tan
Answer:
261.307 m
Explanation:
b = Base of triangle = 450 m
p = Perpendicular of the triangle
[tex]\theta[/tex] = Angle of the triangle = [tex]30^{\circ}[/tex]
From trigonometry
[tex]tan\theta=\dfrac{p}{b}[/tex]
[tex]\Rightarrow p=btan\theta[/tex]
[tex]\Rightarrow p=450\times tan30[/tex]
[tex]\Rightarrow p=259.807\ m[/tex]
Height of the building = 1.5+259.807 = 261.307 m
Two equally charged tiny spheres of mass 1.0 g are placed 2.0 cm apart. When released, they begin to accelerate away from each other at What is the magnitude of the charge on each sphere, assuming only that the electric force is present? (k = 1/4πε0 = 9.0 × 109 N ∙ m2/C2)
Answer:
[tex]1.36\times 10^{-7} C[/tex]
Explanation:
We are given that
Mass of charged tine spheres=m=1 g=[tex]\frac{1}{1000}=0.001 kg[/tex]
1 kg=1000g
The distance between charged tine spheres=r=2 cm=[tex]\frac{2}{100}=0.02 m[/tex]
1 m=100 cm
Acceleration =[tex]a =414 m/s^2[/tex]
Let q be the charge on each sphere.
[tex]k=9\times 10^9Nm^2/C^2[/tex]
The electric force between two charged particle
[tex]F=\frac{kq_1q_2}{r^2}[/tex]
Using the formula
The force between two charged tiny spheres=[tex]F_e=\frac{kq^2}{(0.02)^2}[/tex]
According to Newton's second law , the net force
[tex]F=ma[/tex]
[tex]F=F_e[/tex]
[tex]0.001\times 414=\frac{9\times 10^9\times q^2}{(0.02)^2}[/tex]
[tex]q^2=\frac{0.001\times 414\times (0.02)^2}{9\times 10^9}[/tex]
[tex]q=\sqrt{\frac{0.001\times 414\times (0.02)^2}{9\times 10^9}}[/tex]
[tex]q=1.36\times 10^{-7} C[/tex]
Hence, the magnitude of charge on each tiny sphere=[tex]1.36\times 10^{-7} C[/tex]
What would the force be if the separation between the two charges in the top window was adjusted to 8.19 ✕10-11 m? (The animation will not adjust that far--you will have to calculate the answer).
q1 = q2 = 1.00 ✕ e
The electrostatic force between the two charges is [tex]3.4\cdot 10^{-8}N[/tex]
Explanation:
The electrostatic force between two charges is given by Coulomb's law:
[tex]F=k\frac{q_1 q_2}{r^2}[/tex]
where:
[tex]k=8.99\cdot 10^9 Nm^{-2}C^{-2}[/tex] is the Coulomb's constant
[tex]q_1, q_2[/tex] are the two charges
r is the separation between the two charges
In this problem, we have the following data:
[tex]q_1 = q_2 = 1.00e[/tex] is the magnitude of the two charges, where
[tex]e=1.6\cdot 10^{-19}C[/tex] is the fundamental charge
[tex]r=8.19\cdot 10^{-11}m[/tex] is the separation between the two charges
Substutiting into the equation, we find the force:
[tex]F=(8.99\cdot 10^9)\frac{(1.00\cdot 1.6\cdot 10^{-19})^2}{(8.19\cdot 10^{-11})^2}=3.4\cdot 10^{-8}N[/tex]
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The surface tension of a liquid is to be measured using a liquid film suspended on a U-shaped wire frame with an 12-cm-long movable side. If the force needed to move the wire is 0.096 N, determine the surface tension of this liquid in air.
The surface tension of the liquid in air is 0.8 N/m.
Explanation:To determine the surface tension of the liquid, we need to use the formula F = yL, where F is the force needed to move the wire, y is the surface tension, and L is the length of the wire. In this case, F = 0.096 N and L = 12 cm. We can rearrange the formula to solve for y: y = F / L. Plugging in the values, we get y = 0.096 N / 0.12 m = 0.8 N/m. So, the surface tension of the liquid in air is 0.8 N/m.
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A man pushes his lawnmower with a velocity of +0.75 m/s relative to the ground. A girl rides by on her bike with a velocity of +6.5 m/s relative to the ground. What is the velocity of the girl relative to the lawnmower? A. 0 m/s B. +5.75 m/s C. +6.5 m/s D. +7.25 m/s
Answer:
B. +5.75 m/s
Explanation:
When there are two bodies, a and b, whose velocities measured by a third observer (in this case, the ground) are [tex]V_a[/tex] and [tex]V_b[/tex] respectively, the relative velocity of B with respect to A is given by:
[tex]V_{ba}=V_b-V_a[/tex]
Thus, the velocity of the girl relative to the lawnmower is:
[tex]V_{ba}=6.5\frac{m}{s}-0.75\frac{m}{s}\\V_{ba}=5.75\frac{m}{s}[/tex]
I took the test and got B) +5.75 m/s correct
A has the magnitude 14.4 m and is angled 51.6° counterclockwise from the positive direction of the x axis of an xy coordinate system. Also, B = ( 14.3 m )i + (8.52 m )j on that same coordinate system. We now rotate the system counterclockwise about the origin by 20.0° to form an x'y' system. On this new system, what are (a)Ã and (b) B, both in unit-vector notation? (a) Number i 4.545346 It i 13.66381 Î Units m (b) Number i 10.52359 î+ i 12.89707 Units its
Final answer:
To find the transformed vector representations in a rotated coordinate system, the angle of vector A is adjusted by the rotation angle, and the components are calculated using trigonometric functions. Vector B's components in the rotated system are found using a rotation matrix.
Explanation:
The provided question pertains to transforming the representation of vectors in a rotated coordinate system in the subject of physics. The coordinate system is rotated counterclockwise, and the goal is to find the new representations of vectors A and B in unit-vector notation on the x'y' system. Given the initial magnitude and direction angle of vector A and the Cartesian components of vector B on the xy coordinate system, we can calculate their components on the rotated x'y' coordinate system.
The original vector A has a magnitude of 14.4 m and an angle of 51.6° from the positive x-axis. After rotation by 20°, the new angle becomes 51.6° - 20.0° = 31.6° from the new x'-axis. Using the formulas Ax' = A cos θ' and Ay' = A sin θ', where θ' is the new angle, we can find the rotated components of A.
The vector B is already given in Cartesian coordinates as ( 14.3 m )i + (8.52 m )j. To find the components of B in the rotated system, we use a rotation matrix, giving us new components Bx' and By'.
In conclusion, to find the transformed vectors in the rotated system, we apply the rotation to both the magnitude and angle of A, and use a rotation matrix for the components of B.
Explain how astronomers might use spectroscopy to determine the composition and temperature of a star.
Final answer:
Astronomers utilize spectroscopy to analyze the spectrum of a star, identifying unique absorption lines corresponding to different elements, which reveals the star's composition. Spectral lines' broadening indicates temperature and pressure, and shifts in these lines help measure a star's motion, including radial and rotational velocities.
Explanation:
Understanding Stellar Spectroscopy
Astronomers use spectroscopy as a powerful tool to determine various characteristics of stars, including their composition and temperature. When light from a star passes through a prism or diffraction grating, it spreads out into a spectrum of colors. This spectrum contains dark lines known as absorption lines, which are unique to the elements present in the star's atmosphere, as different chemical elements absorb light at specific wavelengths. Therefore, by analyzing these lines, astronomers can identify the elements that make up a star.
Analyzing the broadening of spectral lines can inform us about a star's temperature and pressure. Warmer temperatures and higher pressures in a star's atmosphere tend to broaden the spectral lines. Additionally, the pressure can give clues about the star's size, as stars with lower atmospheric pressure tend to be larger, or giant stars.
Motions of the Stars are also revealed through spectroscopy. The Doppler effect causes spectral lines to shift towards the red end of the spectrum if the star is moving away from us (redshift) or towards the blue end if it is approaching (blueshift). This allows astronomers to measure the star's radial velocity. Spectral line broadening can also indicate the star's rotational velocity, while proper motion is deduced from the movement of the lines over time across the spectrum.
Huck Finn walks at a speed of 0.70 m/sm/s across his raft (that is, he walks perpendicular to the raft's motion relative to the shore). The raft is traveling down the Mississippi River at a speed of 1.60 m/sm/s relative to the river bank. What is Huck's velocity (speed and direction) relative to the river bank?
Answer:
Explanation:
Given
Velocity of Huck w.r.t to raft [tex]v_{H,raft}=0.7\ m/s[/tex]
Perpendicular to the motion of raft
Velocity of Raft in the river [tex]v_{raft,river}=1.6\ m/s[/tex]
As Huck is traveling Perpendicular to the raft so he possess two velocities i.e. vertical velocity and horizontal velocity of River when observed from bank
[tex]v_{Huck,river\ bank}=0.7\hat{j}+1.6\hat{i}[/tex]
So magnitude of velocity is given by
[tex]|v|=\sqrt{0.7^2+1.6^2}[/tex]
[tex]|v|=\sqrt{0.49+2.56}[/tex]
[tex]|v|=\sqrt{3.05}[/tex]
[tex]|v|=1.74\ m/s[/tex]
For direction [tex]\tan =\frac{0.7}{1.6}=0.4375[/tex]
[tex]\theta =23.63^{\circ}[/tex] w.r.t river bank
Daring Darless wishes to cross the Grand Canyon of the Snake River by being shot from a cannon. She wishes to be launched at 56° relative to the horizontal so she can spend more time in the air waving to the crowd. With what minimum speed must she be launched to cross the 520-m gap?
Answer:
She must be launched with a speed of 74.2 m/s.
Explanation:
Hi there!
The equations of the horizontal component of the position vector and the vertical component of the velocity vector are the following:
x = v0 · t · cos θ
vy = v0 · sin θ + g · t
x = horizontal distance traveled at time t.
v0 = initial velocity.
t = time.
θ = launching angle.
vy = vertical component of the velocity vector at time t.
g = acceleration due to gravity (-9.8 m/s²).
To just cross the 520-m gap, the maximum height of the flight must be reached halfway of the gap at 260 m horizontally (see attached figure).
When she is at the maximum height, her vertical velocity is zero. So, when x = 260 m, vy = 0. Using both equations we can solve the system for v0:
x = v0 · t · cos θ
Solving for v0:
v0 = x/ (t · cos θ)
Replacing v0 in the second equation:
vy = v0 · sin θ + g · t
0 = x/(t·cos(56°)) · sin(56°) + g · t
0 = 260 m · tan (56°) / t - 9.8 m/s² · t
9.8 m/s² · t = 260 m · tan (56°) / t
t² = 260 m · tan (56°) / 9.8 m/s²
t = 6.27 s
Now, let's calculate v0:
v0 = x/ (t · cos θ)
v0 = 260 m / (6.27 s · cos(56°))
v0 = 74.2 m/s
She must be launched with a speed of 74.2 m/s.
Answer:
it must be launched at a speed of 74.2 m/s
Explanation:
I really hope this helps
A small object of mass 3.82 g and charge -16.5 µC is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground. What are the magnitude and direction of the electric field?
The question deals with the calculation of the magnitude and direction of an electric field necessary to keep a charged object motionless. The two forces acting on the object, namely the gravitational force and the electric force, cancel out making it motionless. The electric field direction is upward as it must counteract the gravitational pull.
Explanation:In this question, we're examining an object that stays motionless in a uniform electric field. This can be resolved using the equilibrium of forces acting on the object. Given that the object stays motionless, the gravitational force and the electric force on the object should balance each other.
The gravitational force (Fg) experienced by the object is the object mass (m) times the acceleration due to gravity (g), which equals 3.82g * 9.81 m/s². The electric force (Fe) is equal to the charge (q) times the electric field (E), which equals -16.5µC * E.
To find the electric field E, we equate these forces - this gives us
E = Fg / |q|,
where |q| means the absolute value of the charge. The direction of the electric field is taken as the direction of the force that a positive test charge would experience.
Thus, the electric field direction is upwards since the force needed to balance gravity must act against it.
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A 2.0 m × 4.0 m flat carpet acquires a uniformly distributed charge of −10 μC after you and your friends walk across it several times. A 6.0 μg dust particle is suspended in midair just above the center of the carpet.
What is the charge on the dust particle?
The charge on the dust particle is [tex]-2.07 x 10^{-14} C.[/tex]
The dust particle will obtain a charge due to the electric field produced by the charged carpet. Be that as it may, calculating its correct charge requires a few presumptions and steps:
1. Charge density:
To begin with, we ought to calculate the charge density [tex]\sigma[/tex] of the carpet:
[tex]\sigma[/tex] = total Charge / Range = -10 μC / (2.0 m x 4.0 m) = -2.5 μC/m²
2. Electric Field:
The charge thickness creates an electric field (E) over the carpet. Ready to utilize the equation:
E = [tex]\sigma[/tex] / ϵ0
here, ϵ0 is the permittivity of free space which is equal to [tex]8.85 x 10^{-12[/tex]F/m
Electric field, E = (-2.5 μC/m² / [tex]8.85 x 10^{-12} F/m[/tex]) = ([tex]-2.83 x 10^{5} N/C[/tex])
3. dust particle Charge:
The dust particle will involve an electrostatic force due to the electric field. Since the molecule is suspended, the net force on it must be zero. This implies the electrostatic force must balance the gravitational force acting on the molecule.
Suspicions:
The dust particle could be a circle with uniform charge dissemination.
Discussing resistance is unimportant.
Calculations:
Tidy molecule mass (m): 6.0 μg = [tex]6.0 x 10^{-9} kg[/tex]
Gravitational force (Fg): Fg = m * g (where g is increasing speed due to gravity,= 9.81 m/s²)
Electrostatic force (Fe): Fe = q * E (where q is the charge of the dust particle)
Likening the powers:
Fg = Fe
m * g = q * E
Tackling for q:
q = Fg / E = (m * g) / E = [tex](6.0 x 10^{-9} kg[/tex] * 9.81 m/s²) / ([tex]-2.83 x 10^{5} N/C[/tex]) = [tex]-2.07 x 10^{-14}[/tex] C
Subsequently, the charge on the dust particle is[tex]-2.07 x 10^{-14}[/tex] C.
The ultimate normal stress in members AB and BC is 350 MPa. Find the maximum load P if the factor of safety is 4.5. AB has an outside diameter of 250mm and BC has an outside diameter of 150mm. Both pipes have a wall thickness of 8mm
Answer:
P_max = 278 KN
Explanation:
Given:
- The ultimate normal stress S = 350 MPa
- Thickness of both pipes t = 8 mm
- Pipe AB: D_o = 250 mm
- Pipe BC: D_o = 150 mm
- Factor of safety FS = 4.5
Find:
Find the maximum load P_max
Solution:
- Compute cross sectional areas A_ab and A_bc:
A_ab = pi*(D_o^2 - (D_o - 2t)^2) / 4
A_ab = pi*(0.25^2 - 0.234^2) / 4
A_ab = 6.08212337 * 10^-3 m^2
A_bc = pi*(D_o^2 - (D_o - 2t)^2) / 4
A_bc = pi*(0.15^2 - 0.134^2) / 4
A_bc = 3.568212337 * 10^-3 m^2
- Compute the Allowable Stress for each pipe:
sigma_all = S / FS
sigma_all = 350 / 4.5
sigma_all = 77.77778 MPa
- Compute the net for each member P_net,ab and P_net,bc:
P_net,ab = sigma_all * A_ab
P_net,ab = 77.77778 MPa*6.08212337 * 10^-3
P_net,ab = 473054.0399 N
P_net,bc = sigma_all * A_bc
P_net,bc = 77.77778 MPa*3.568212337 * 10^-3
P_net,bc = 277577.1721 N
- Compute the force P for each case:
P_net,ab = P + 50,000
P = 473054.0399 - 50,000
P = 423 KN
P_net,bc = P = 278 KN
- P_max allowed is the minimum of the two load P:
P_max = min (423, 278) = 278 KN
A plane flies 125 km/hr at 25 degrees north of east with a wind speed of 36 km/hr at 6 degrees south of east. What is the resulting velocity of the plane (in km/hr)?
Answer:
V = 156.85 Km/h
Explanation:
Speed of plane = 125 Km/h
angle of plane= 25° N of E
Speed of wind = 36 Km/h
angle of plane = 6° S of W
Horizontal component of the velocity
V_x = 125 cos 25° + 36 cos 6°
V_x = 149 Km/h
Vertical component of the velocity
V_y = 125 sin 25° - 36 sin 6°
V_y = 49 Km/h
Resultant of Velocity
[tex]V = \sqrt{V_x^2 + V_y^2}[/tex]
[tex]V = \sqrt{149^2 + 49^2}[/tex]
V = 156.85 Km/h
the resulting velocity of the plane is equal to V = 156.85 Km/h
How many times does a typical person blink her eyes in a lifetime?
Answer:
415,224,000
Explanation:
a person blinks 10 times per minute ,60 minutes in a hour so 600 per hour,24 hours per day so 14,400 blinks per day and there are 365 days in a year so 5,256,000 blinks per year and an average person lives to 79 years so 415224000 in an average lifetime
The information on a can of soda indicates that the can contains 355 mL. The mass of a full can of soda is 0.369 kg, while an empty can weighs 0.153 N. Determine the specific weight, density, and specific gravity of the soda and compare your results with the corresponding values for water at 20 oC. Express your results in SI units.
Answer:
[tex]\rho=995.50\ kg.m^{-3}[/tex]
[tex]\bar w=9765.887\ N.m^{-3}[/tex]
[tex]s=0.9955[/tex]
Explanation:
Given:
volume of liquid content in the can, [tex]v_l=0.355\ L=3.55\times 10^{-4}\ L[/tex]mass of filled can, [tex]m_f=0.369\ kg[/tex]weight of empty can, [tex]w_c=0.153\ N[/tex]So, mass of the empty can:
[tex]m_c=\frac{w_c}{g}[/tex]
[tex]m_c=\frac{0.153}{9.81}[/tex]
[tex]m_c=0.015596\ kg[/tex]
Hence the mass of liquid(soda):
[tex]m_l=m_f-m_c[/tex]
[tex]m_l=0.369-0.015596[/tex]
[tex]m_l=0.3534\ kg[/tex]
Therefore the density of liquid soda:
[tex]\rho=\frac{m_l}{v_l}[/tex] (as density is given as mass per unit volume of the substance)
[tex]\rho=\frac{0.3534}{3.55\times 10^{-4}}[/tex]
[tex]\rho=995.50\ kg.m^{-3}[/tex]
Specific weight of the liquid soda:
[tex]\bar w=\frac{m_l.g}{v_l}=\rho.g[/tex]
[tex]\bar w=995.5\times 9.81[/tex]
[tex]\bar w=9765.887\ N.m^{-3}[/tex]
Specific gravity is the density of the substance to the density of water:
[tex]s=\frac{\rho}{\rho_w}[/tex]
where:
[tex]\rho_w=[/tex] density of water
[tex]s=\frac{995.5}{1000}[/tex]
[tex]s=0.9955[/tex]
Explanation:
The given data is as follows.
Volume of pop in can, V = [tex]355 \times 10^{-6} m^{3}[/tex]
Mass of a full can of pop is as follows.W = mg
= [tex]0.369 \times 9.81[/tex]
= 3.6198 N
Weight of empty can, [tex]w_{1}[/tex] = 0.153 N
Now, weight of pop in the can is calculated as follows.[tex]w_{2} = W - w_{1}[/tex]
= 3.6198 - 0.153
= 3.467 N
Calculate the specific weight of the liquid as follows.[tex]\gamma = \frac{\text{weight of liquid}}{\text{volume of liquid}}[/tex]
= [tex]\frac{3.467}{355 \times 10^{-6}}[/tex]
= 9766.197 [tex]N/m^{3}[/tex]
Density of the fluid is calculated as follows.[tex]\rho = \frac{\gamma}{g}[/tex]
= [tex]\frac{9766.197}{9.81}[/tex]
= 995.535 [tex]kg/m^{3}[/tex]
Now, specific gravity of the fluid is calculated as follows.S.G = [tex]\frac{\text{density of liquid}}{\text{density of water}}[/tex]
= [tex]\frac{\rho}{\rho_{w}}[/tex]
= [tex]\frac{995.535}{1000}[/tex]
= 0.995
In a mixture of the gases oxygen and helium, which statement is valid: (a) the helium molecules will be moving faster than the oxygen molecules, on average; (b) both kinds of molecules will be moving at the same speed; (c) the oxygen molecules will, on average, be moving more rapidly than the helium molecules; (d) the kinetic energy of the helium will exceed that of the oxygen; (e) none of the above.
Answer:
a
Explanation:
Given:
In a mixture of the gases oxygen and helium, which statement is valid:
(a) the helium molecules will be moving faster than the oxygen molecules, on average
(b) both kinds of molecules will be moving at the same speed
(c) the oxygen molecules will, on average, be moving more rapidly than the helium molecules
(d) the kinetic energy of the helium will exceed that of the oxygen
(e) none of the above
Solution:
- We will use Boltzmann distribution to answer this question. The root mean square speed of molecules of a gas gives the average speed as follows:
V_rms = sqrt ( 3 k T / m )
- Where, k is the Boltzmann constant, T is the temperature and m is the mass of a single molecule of a gas.
- In general, a mixture has a constant equilibrium temperature T_eq.
- So the v_rms is governed by the mass of a single molecule.
- We know that mass of single molecule of Oxygen is higher than that of Helium molecule. Hence, the relation of mass is inversely proportional to square of root mean speed. So the helium molecules will be moving faster than the oxygen molecules.
- Note: The kinetic energy of the mixture remains constant because it is due to the interaction of the molecules within i.e oxygen and helium. Which makes the kinetic energy independent of mass.
E_k = 0.5*m*v_rms^2
E_k = 0.5*m*(3*k*T/ m )
E_k = 0.5*3*k*T
Hence, E_k is only the function of Temperature which we already established to remain constant at equilibrium.
OPTION A.
In a mixture of oxygen and helium, the helium molecules move faster on average due to their lighter weight, though the average kinetic energy of both gases remains the same at a given temperature.
Explanation:In a mixture of gases, the speeds of the molecules of different gases are primarily dependent on their masses. For gases at a given temperature, all have the same average kinetic energy (KEavg) for their molecules. However, gases made up of lighter molecules, such as helium, have more high-speed particles and an average speed (Urms) that is higher than gases composed of heavier molecules, like oxygen.
Therefore, in a mixture of oxygen and helium, statement (a) the helium molecules will be moving faster than the oxygen molecules, on average, is valid. This is due to helium molecules being lighter than oxygen molecules. Moreover, the average kinetic energy of both gases, helium and oxygen, would be the same at a given temperature, meaning statement (d) the kinetic energy of the helium will exceed that of the oxygen, would not be valid.
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Compute the ratio of the rate of heat loss through a single-pane window with area 0.15 m2 to that for a double-pane window with the same area. The glass of a single pane is 4.5 mm thick, and the air space between the two panes of the double-pane window is 6.60 mm thick. The glass has thermal conductivity 0.80 W/m⋅K. The air films on the room and outdoor surfaces of either window have a combined thermal resistance of 0.15 m2⋅K/W. Express your answer using two significant figures.
Answer:
2.80321285141
Explanation:
[tex]L_g[/tex] = Thickness of glass = 4.5 mm
[tex]k_g[/tex] = Thermal conductivity of glass = 0.8 W/mK
[tex]R_0[/tex] = Combined thermal resistance = [tex]0.15\times m^2K/W[/tex]
[tex]L_a[/tex] = Thickness of air = 6.6 mm
[tex]k_a[/tex] = Thermal conductivity of air = 0.024 W/mK
The required ratio is the inverse of total thermal resistance
[tex]\dfrac{2(L_g/k_g)+R_0+(L_a/k_a)}{(L_g/k_g)+R_0}\\ =\dfrac{2(4.5\times 10^{-3}/0.8)+0.15+(6.6\times 10^{-3}/0.024)}{(4.5\times 10^{-3}/0.8)+0.15}\\ =2.80321285141[/tex]
The ratio is 2.80321285141
Answer:
[tex]\frac{\dot Q}{\dot Q'} =2.6668[/tex]
Explanation:
Given:
area of the each window panes, [tex]A=0.15\ m^2[/tex]thickness of each pane, [tex]t_g=4.5\times 10^{-3}\ m[/tex]air gap between the two pane of a double pane window, [tex]t_a=6.6\times 10^{-3}\ m[/tex]thermal conductivity of glass, [tex]k_g=0.8\ W.m^{-1}.K^{-1}[/tex]thermal resistance of the air on the either sides of double pane window, [tex]R_{th}=0.15\ m^2.K.W^{-1}[/tex]Heat loss through single pane window:
Using Fourier's law of conduction,
[tex]\dot Q=A.dT\div (R_{th}+\frac{t_g}{k} )[/tex]
[tex]\dot Q=0.15\times dT\div (0.15+\frac{4.5\times 10^{-3}}{0.8})[/tex]
[tex]\dot Q=0.9638\ dT\ [W][/tex]
Heat loss through double pane window:
[tex]\dot Q'=dT\times A\div(R_{th}+2\times \frac{t_g}{k}+\frac{t_a}{k_a} )[/tex]
where:
[tex]dT=[/tex] change in temperature
[tex]k_a=[/tex] coefficient of thermal conductivity of air [tex]= 0.026\ W.m^{-1}.K^{-1}[/tex]
[tex]\dot Q'=dT\times 0.15\div (0.15+2\times \frac{4.5\times 10^{-3}}{0.8}+\frac{6.6\times 10^{-3}}{0.026})[/tex]
[tex]\dot Q'=0.3614\ dT\ [W][/tex]
Now the ratio:
[tex]\frac{\dot Q}{\dot Q'} =\frac{0.9638(dT)}{0.3614(dT)}[/tex]
[tex]\frac{\dot Q}{\dot Q'} =2.6668[/tex]
If our eyes could see a slightly wider region of the electromagnetic spectrum, we would see a fifth line in the Balmer series emission spectrum. Calculate the wavelength λλlambda associated with the fifth line.
Answer:
λ = 397 nm
Explanation:
given,
Rydberg wavelength equation for Balmer series
[tex]\dfrac{1}{\lambda}=R(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2})[/tex]
R is the Rydberg constant, R = 1.097 x 10⁷ m⁻¹
n_i = initial energy level
n_f = final energy level
where as for Balmer series n_f = 2
n_i = 7
[tex]\dfrac{1}{\lambda}=(1.097\times 10^7)(\dfrac{1}{2^2}-\dfrac{1}{7^2})[/tex]
[tex]\dfrac{1}{\lambda}=(1.097\times 10^7)(\dfrac{1}{2^2}-\dfrac{1}{7^2})[/tex]
[tex]\dfrac{1}{\lambda}=2.5186\times 10^6[/tex]
[tex]\lambda = 3.97\times 10^{-7}[/tex]
Hence, the wavelength is equal to λ = 397 nm
The froghopper, a tiny insect, is a remarkable jumper. Suppose a colony of the little critters is raised on the Moon, where the acceleration due to gravity is only 1.62 m/s 2 , whereas gravity on Earth is g = 9.81 m/s 2 . If on Earth a froghopper's maximum jump height is h and its maximum horizontal jump range is R , what would its maximum jump height and range be on the Moon in terms of h and R ? Assume the froghopper's takeoff velocity is the same on the Moon and Earth.
Answer:
hₘₒₒₙ = 6.05 h
Rₘₒₒₙ = 6.05 R
Explanation:
Let θ be the angle of jump.
Let h and R be maximum height and horizontal range attained on earth respectively.
Let hₘₒₒₙ and Rₘₒₒₙ be the maximum height and horizontal range on the moon respectively
The range for a projectile is given as
R = v₀(x)T = v₀ cos(θ) T
T = (2v₀ sinθ)/g
Range, R = (v₀ cos θ)(2v₀ sinθ)/g = v₀²(2sinθcosθ)/g = v₀² (sin2θ)/g
The maximum range occurs at θ = 45°
Maximum range R = v₀²/g = v₀²/9.8 = 0.102v₀²
On the moon, g = 1.62 m/s²
Maximum range, Rₘₒₒₙ = v₀²/gₘₒₒₙ = v₀²/1.62 = 0.617v₀²
Rₘₒₒₙ = 6.05 R
Maximum Height of a projectile is given as = (v₀² Sin²θ)/2g
θ = 45°; sin 45° = (√2)/2; sin²45° = 2/4 = 1/2
h = v₀²(1/2)/2g = v₀²/4g
On earth, g = 9.8 m/s²
h = v₀²/(4×9.8) = v₀²/39.2 = 0.0255v₀²
On the moon, gₘₒₒₙ = 1.62 m/s²
hₘₒₒₙ = v₀²/(4×1.62) = v₀²/6.48 = 0.154v₀²
hₘₒₒₙ = 6.05 h
An alpha particle (atomic mass 4.0 units) experiences an elastic head-on collision with a gold nucleus (atomic mass 197 units) that is originally at rest. What is the fractional loss of kinetic energy for the alpha particle
Answer:
0.08
Explanation:
The alpha particle suffers a head-on collision with the gold nucleus, so it retraces it path after the collision.
Let us take the masses of the particles in atomic mass units.
The initial momentum and kinetic energy of the gold nucleus is 0(since it is stationary). So, applying conservation of momentum and energy, we get the following two equations:
[tex]m_{1}u_{1}=m_{1}v_{1}+m_{2}v_{2}[/tex] ..........(1)
[tex]\frac{1}{2}m_{1}u_{2}^{2}=\frac{1}{2} m_{1} v_{1}^{2} +\frac{1}{2} m_{2} v_{2}^{2}[/tex] ..........(2)
where,
[tex]m_{1}[/tex] = mass of the alpha particle = 4 units
[tex]m_{2}[/tex] = mass of the gold nucleus = 197 units
[tex]u_{1}[/tex] = initial velocity of the alpha particle
[tex]v_{1}[/tex] = final velocity of the alpha particle
[tex]v_{2}[/tex] = final velocity of the gold nucleus
Now, we shall substitute the value of [tex]v_{2}[/tex] from equation (1) in equation (2). After some simplifications, we get,
[tex]u_{1}^{2}=v_{1}^{2}+\frac{m_{1}}{m_{2}} (u_{1}^{2}+v_{1}^{2}-2u_{1}v_{1})[/tex]
Dividing both sides by [tex]u_1^2[/tex] and substituting [tex]x=\frac{v_1}{u_1}[/tex] and [tex]k=\frac{m_1}{m_2}[/tex] , we get,
[tex]1=x^2+k(1+x^2-2x)\\[/tex]
or, [tex]x^2(k+1)-2kx+(k-1)=0[/tex]
Here, [tex]k=\frac{m_1}{m_2}=\frac{4}{197}=0.02[/tex]
Therefore, [tex]x=\frac{2(0.02)\pm\sqrt{(2\times0.02)^2-(4\times1.02\times-0.98)} }{2\times1.02}[/tex]
or, [tex]x = 1, -0.96[/tex]
Our required solution is -0.96 because the final velocity([tex]v_1[/tex]) of the alpha particle will be a little less the initial velocity([tex]u_1[/tex]). The negative sign comes as the alpha particle reverses it's direction after colliding with the gold nucleus.
Fractional change in kinetic energy is given by,
[tex]\delta E=\frac{\frac{1}{2} m_1u_1^2-\frac{1}{2}m_1v_1^2 }{\frac{1}{2}m_1u_1^2 }=1-x^2=0.078\approx0.08[/tex]
The alpha particle can lose a significant amount of its kinetic energy in a head-on elastic collision with a gold nucleus due to the gold nucleus's much larger mass. The original kinetic energy of the alpha particle is converted to potential energy before being transferred mostly to the gold nucleus. Specific loss would depend upon the alpha particle's original kinetic energy.
Explanation:The question pertains to the concept of elastic collisions, specifically between an alpha particle and a gold nucleus. In an elastic collision, both momentum and kinetic energy are conserved. However, while total energy is conserved, individual kinetic energies of colliding particles may change. Since the gold nucleus, which was initially at rest, is significantly more massive (197 units) than the alpha particle (4.0 units), the alpha particle can lose a significant amount of its kinetic energy in a head-on collision.
To calculate the fractional loss of kinetic energy for the alpha particle in this instance, we would use the principle of conservation of kinetic energy and momentum. The kinetic energy of an alpha particle before the collision is transformed into both kinetic and potential energy during the collision as it approaches the gold nucleus until its original energy is converted to potential energy.
Upon collision, a good proportion of this energy is transferred to the gold atom, given its much larger mass. However, necessary calculations would require specific knowledge of the kinetic energy of the alpha particle before the collision, which may vary depending upon the specific nuclear decay process involved.
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