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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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It take my planet amount of time to orbit the sun
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It takes the Earth 365 days to orbit the sun.
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A car came to a stop from a speed of 16m/s in a time of 2.3 seconds. What was the acceleration of the car?
dexar [7]
A=dv/dt. So
a=(0-16)/2.3 = -6.96m/s^2
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A wire 2.80 m in length carries a current of 4.20 A in a region where a uniform magnetic field has a magnitude of 0.260 T. Calcu
klio [65]

Answer:.

F = 3.0576sinθ

For any value of θ < 180

Explanation:

Generally, F = BILsinθ

Where,. F = magnetic force magnitude. B = magnetic Field magnitude.

L = length of wire. I = current

Therefore,

B = 0.260 T, L = 2.80 m

I = 4.20 A

: F = 0.260 × 4.20 × 2.80sinθ

∴ F = 3.0576sinθ

4 0
3 years ago
a 2-kg rifle that is suspended by strings fires a 0.01-kg bullet at 200 m/s. the recoil velocity of the rifle is about
Trava [24]

Explanation:

from Newton's third law of motion we know every action has equal and opposite reaction.

So, the velocity of going out of the rifle = 200 m/s

therefore the recoil is 200 m/s on the backward direction.

3 0
2 years ago
I need help with these questions :<br>(see image )<br>​
lara31 [8.8K]
<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2>_____________________________________</h2><h3>DATA:</h3>

Angle of projection = θ(theta) = 30^0

Initial Velocity = V_0 = 2x10^3

Acceleration due to gravity = g = 9.8 m/s^2

Vertical Velocity = V_Y = ?

Horizontal Velocity = V_X = ?

Range of the Shell = R = ?

Maximum Height = H = ?

<h2>_____________________________________</h2><h3>SOLUTION:</h3>

Vertical Velocity is given by,

                                 V_Y = V_0Cosθ

                                 V_Y = (2x10^3)xCos(30)

                                 V_Y = (2x10)^3x(0.866)\\\\V_Y = 1732.05 \frac{m}{s}

Horizontal Velocity is given by,

                                 V_X = V_0Sinθ

                                 V_X = (2x10^3)xSin(30)\\\\V_X = (2000)x(0.5)\\\\V_X = 1000\frac{m}{s}

Range is given by,

                                R =   \frac{V_0^2}{g}  Sin2θ        

                                R = \frac{(2x10^3)^2}{10} x Sin(60)\\\\R = \frac{4x10^6}{10} x 0.866\\\\R= (4x10^5) x 0.866\\\\R = 346410.16 m                    

Horizontal Velocity is given by,

<h2>                                 H = \frac{V_0^2Sin^2(theta)}{2g}\\\\\\H= \frac{(2x10^3)^2xSin(30)^2}{2x10}\\\\\\H= \frac{(4x10^6)x(0.25)}{20} \\\\H = 50000 m                             _____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2>
7 0
3 years ago
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