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Fynjy0 [20]
3 years ago
14

A hockey puck of mass m traveling along the x axis at 4.5 m/s hits another identical hockey puck at rest. If after the collision

the second puck travels at a speed of 3.5 m/s at an angle of 30° above the x axis, is this an elastic collision?
A) Yes, since momentum is conserved.
B) No, since momentum is not conserved.
C) Yes, since kinetic energy is conserved.
D) No, since kinetic energy is not conserved.
E) Not enough information.
Physics
1 answer:
TEA [102]3 years ago
5 0

Answer:

D) No, since kinetic energy is not conserved.

Explanation:

Since momentum is always conserved in all collision

so in Y direction we can say

0 = m(3.5 sin30) - mv_y

v_y = 1.75 m/s

Now similarly in X direction we will have

m(4.5) = m(3.5 cos30 ) + mv_x

v_x = 1.47 m/s

now final kinetic energy of both puck after collision is given as

KE_f = \frac{1}{2}m(3.5^2) + \frac{1}{2}m(1.75^2 + 1.47^2)

KE_f = 8.73 m

initial kinetic energy of both pucks is given as

KE_i = \frac{1}{2}m(4.5^2) + 0

KE_i = 10.125 m

since KE is decreased here so it must be inelastic collision

D) No, since kinetic energy is not conserved.

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kozerog [31]

Answer:

t should be 3.57 second

Explanation:

Formula used is v = u+at

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Substitute each of the info given into the formula and calculate.

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Answer:

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i = o * f / (o - f)

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Image is real and inverted and 16.6 / 54.2 * 6 = 1.94 cm tall

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3 years ago
A cylinder with a moving piston expands from an initial volume of 0.250 L against an external pressure of 2.00 atm. The expansio
Step2247 [10]

The final volume of the gas is 144.25 L

Explanation:

For an ideal gas kept at constant pressure, the work done by the gas on the surroundings is given by

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V_i is the initial volume

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For the gas in the cylinder in this problem,

p = 2.00 atm

V_i = 0.250 L

And we also know the work done,

W = 288 J

So we can solve the equation for V_f, the final volume:

V_f = V_i + \frac{W}{p}=0.250 + \frac{288}{2.00}=144.25 L

Learn more about ideal gases:

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f_{beat} =| f_2\pm f_1 |

The reference frequency in our case would be 392Hz, and since there is the possibility of the upper and lower range for the amount of beats per second that the two possible frequencies are heard would be

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Therefore the two possible frequencies the piano wire is vibrating at, would be 396Hz and 388Hz

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