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san4es73 [151]
3 years ago
8

A moving small car has a head-on collision with a large stationary truck 7.3 times the mass of the car. Which statement is true

about the direction of the car after the collision if the collision is elastic? The car continues moving in its initial direction with the same velocity. The car bounces off and moves in the opposite direction. The car keeps moving in its initial direction with a lower velocity than at first. The car stops completely.
Physics
2 answers:
ad-work [718]3 years ago
8 0
The car bounces off and moves in the opposite direction
UkoKoshka [18]3 years ago
4 0
The car bounces off and moves in the opposite direction 
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Firdavs [7]

Initial velocity (u) = 2 m/s

Acceleration (a) = 10 m/s^2

Time taken (t) = 4 s

Let the final velocity be v.

By using the equation,

v = u + at, we get

or, v = 2 + 10 × 4

or, v = 2 + 40

or, v = 42

The final velocity is 42 m/s.

5 0
2 years ago
Read 2 more answers
A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 10° F. Af
vekshin1

Answer:

T=51.64^\circ F

t=180.10s

Explanation:

The Newton's law in this case is:

T(t)=T_m+Ce^{kt}

Here, T_m is the air temperture, C and k are constants.

We have

70^\circ F in t=0

So:

T(0)=70^\circ F\\T(0)=10^\circ F+Ce^{k(0)}\\70^\circ F=10^\circ F+C\\C=70^\circ F-10^\circ F=60^\circ F

And we have 60^\circ F in t=30 s, So:

T(30)=60^\circ F\\T(30)=10^\circ F+(60^\circ F)e^{k(30)}\\60^\circ F=10^\circ F+(60^\circ F)e^{k(30)}\\50^\circ F=(60^\circ F)e^{k(30)}\\e^{k(30)}=\frac{50^\circ F}{60^\circ F}\\(30)k=ln(\frac{50}{60})\\k=\frac{ln(\frac{50}{60})}{30}=-0.0061

Now, we have:

T=10^\circ F+(60^\circ F)e^{-0.0061t}(1)

Applying (1) for t=1 min=60s:

T=10^\circ F+(60^\circ F)e^{-0.0061*60}\\T=10^\circ F+(60^\circ F)0.694\\T=10^\circ F+41.64^\circ F\\T=51.64^\circ F

Applying (1) for T=30^\circ F:

30^\circ F=10^\circ F+(60^\circ F)e^{-0.0061t}\\30^\circ F-10^\circ F=(60^\circ F)e^{-0.0061t}\\-0.0061t=ln(\frac{20}{60})\\t=\frac{ln(\frac{20}{60})}{-0.0061}=180.10s

8 0
3 years ago
The breaking in (blank) bonds in food releases energy for your body to use
kicyunya [14]

Answer:

The breaking in <em>molecular</em> bonds in food releases energy for your body to use.

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2 years ago
Please help with this question.
EastWind [94]

Answer:

41.16 Joules

Explanation:

Potential energy at a given instant is a function of mass and height of an object. The formula is

E_p = mgh = 2.80kg\cdot 9.8\frac{m}{s^2}\cdot 1.50m = 41.16 J

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Juan is receiving bills from the hospital even though his insurance company said his procedure would be covered in full. What st
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==> Jot down notes before and after making each call, and avoid calling during business hours when people are busy.

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