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UNO [17]
3 years ago
5

A hockey pick sliding along a frictional surface strikes a box at rest, after the collision the two objects stick together and m

ove at the same final speed. Which of the following describes the change in momentum and energy of the puck during the collision?
a. puck loses some but not all of its original momentum.
b. one cannot determine
c. puck conserves original momentum, but loses all mechanical energy
d. puck loses some momentum but conserves mechanical energy
e. puck loses conserves all momentum and mechanical energy
f. conserves momentum but loses some mechanical energy
Physics
1 answer:
Hitman42 [59]3 years ago
5 0

Answer:

Explanation:

Option a is correct  

If puck and pick constitute a system then the momentum of the system is conserved but not this may not be valid for the puck .

Option e is correct

If puck and pick is the system then momentum is conserved but because of the presence of friction, mechanical energy is not conserved.

Friction will cause the energy to dissipate in heat.

       

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A sprinter set a high school record in track and field, running 200.0 m in 20.6 s . what is the average speed of the sprinter in
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Answer : The average speed of the sprinter is, 34.95 Km/hr

Solution :

Average velocity : It is defined as the distance traveled by the time taken.

Formula used for average velocity :

v_{av}=\frac{d}{t}

where,

v_{av} = average velocity

d = distance traveled = 200 m

t = time taken = 20.6 s

Now put all the given values in the above formula, we get the average velocity of the sprinter.

v_{av}=\frac{200m}{20.6s}\times \frac{3600}{1000}=34.95Km/hr

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An electron is in motion at 4.0 × 106 m/s horizontally when it enters a region of space between two parallel plates, as shown, s
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Explanation:

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In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

F_e=qE=ma             (1)

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m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

x_{max}=v_o\sqrt{\frac{2d}{a}}             (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

You replace the values of the parameters in the equation (2):

x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm

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