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Volgvan
3 years ago
15

A car travels along a straight line at a constant speed of 60.0 mi/h for a distance d and then another distance d in the same di

rection at another constant speed. The average velocity for the entire trip is 30.0 mi/h. (a) What is the constant speed with which the car moved during the second distance d
Physics
1 answer:
Makovka662 [10]3 years ago
8 0

Answer:

velocity during second d = 20.0 mi/h

Explanation:

Total distance travelled is 2d, with an average velocity of 30.0 mi/h you can express the time travelled in terms of d:

distance = velocity * time

time = distance / velocity

time = 2d/30.0

The time needed for the first d at 60.0 is:

time = d/60.0

The time in the second d you can get it by substracting both times (total time - time for the first d)

second d time = 2d/30.0 - d/60.0

= 4d/60.0 - d/60.0

= 3d/60.0

and with the time (3d/60.0) and the distance travelled (d) you can get the velocity:

velocity = distance / time

velocity = d / (3d/60.0)

= 60.0/3 = 20.0 mi/h

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Answer:

v₁ = 2.48m/s, v₂ = 0.02m/s

Explanation:

Momentum p must be conserved. p = mv

1) First person throwing the snow ball. The momentum before the throw:

p = (65kg + 0.045kg) * 2.5 m/s

The momentum after the throw:

p = 65kg * v₁ + 0.045kg * 30m/s

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2) Second person catching the ball. The momentum before the catch:

p = 0.045kg * 30m/s + 60kg * 0m/s

The momentum after the catch:

p = (60kg + 0.045kg) * v₂

Solving for velocity v₂ of person 2:

v₂ = 0.045kg * 30m/s / (60kg + 0.045kg) = 0.02 m/s

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3 years ago
A disc of mass m slides with negligible friction along a flat surface with a velocity v. The disc strikes a wall head-on and bou
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Answer:

-v/2

Explanation:

Given that:

  • a disc of mass m
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  • Upon rebounding from the wall its kinetic energy becomes one-fourth of the initial kinetic energy before collision.

<u>We know, kinetic energy is given as:</u>

KE_i=\frac{1}{2}. m.v^2

consider this to be the initial kinetic energy of the body.

<u>Now after collision:</u>

KE_f=\frac{1}{4}\times KE_i

KE_f=\frac{1}{4} \times \frac{1}{2}\times m.v^2

Considering that the mass of the body remains constant before and after collision.

KE_f=\frac{1}{2}\times m.(\frac{v}{2})^2

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Suppose you want to operate an ideal refrigerator with a cold temperature of − 15.5 °C , and you would like it to have a coeffic
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Answer:

15.65 °C

Explanation:

cold temperature (Tc) = -15.5 degree C = 273.15 - 15.5 = 257.65 kelvin

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Th = 288.8 K

Th = 288.8 - 273.15 = 15.65 °C

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