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I am Lyosha [343]
3 years ago
7

Two toy cars (m1 = 0.200 kg and m2 = 0.250 kg) are held together rear to rear with a compressed spring between them. When they a

re released, the cars are free to roll away from the ends of the spring.
If you measure the acceleration of the 0.200-kg car to be 2.65 m/s2 to the right, what is the acceleration of the other car?
Physics
1 answer:
timurjin [86]3 years ago
8 0

Answer:

Acceleration of the second particle at that moment is given as

a_2 = 2.12 m/s^2

Explanation:

As we know that both cars are connected by same spring

So on this system of two cars there is no external force

So we will have

F = 0 = m_1a_1 + m_2a_2

now we have

m_1 = 0.200 kg

m_2 = 0.250 kg

a_1 = 2.65 m/s^2

now we have

0.200(2.65) + 0.250a_2 = 0

so we have

a_2 = 2.12 m/s^2

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Natali5045456 [20]

Answer:

A) 37 m

Explanation:

The car is moving of uniformly accelerated motion, so the distance it covers can be calculated by using the following SUVAT equation:

v^2 -u^2 = 2ad (1)

where

v = 0 m/s is the final velocity of the car

u = 24 m/s is the initial velocity

a is the acceleration

d is the length of the skid

We need to find the acceleration first. We know that the force responsible for the (de)celeration is the force of friction, so:

F=ma=-\mu mg

where

m = 1000 kg is the mass of the car

\mu = 0.80 is the coefficient of friction

a is the deceleration of the car

g = 9.8 m/s^2 is the acceleration due to gravity

The negative sign is due to the fact that the force of friction is against the motion of the car, so the sign of the acceleration will be negative because the car is slowing down. From this equation, we find:

a=-\mu g

And we can substitute it into eq.(1) to find d:

v^2 -u^2 = 2(\-mu g) d\\d= \frac{v^2-u^2}{-2 \mu g}=\frac{0-(24 m/s)^2}{-2(0.80)(9.8 m/s^2)}=36.7 m \sim 37 m

7 0
3 years ago
What does it mean if a paint sample "matches" a known sample from a vehicle? Does this indicate the same source? Why or why not?
Mandarinka [93]
It wouldn't indicate the same source if the paint was from a different place. If the paint was the same brand and color it would be but if not then no
4 0
3 years ago
A stone is dropped from a tower 100 meters above the ground. The stone falls past ground level and into a well. It hits the wate
lilavasa [31]

Take the stone's position at ground level to be the origin, and the downward direction to be negative. Then its position in the air y at time t is given by

y=100\,\mathrm m-\dfrac g2t^2

Let d be the depth of the well. The stone hits the bottom of the well after 5.00 s, so that

-d=100\,\mathrm m-\dfrac g2(5.00\,\mathrm s)^2\implies d=\boxed{22.6\,\mathrm m}

7 0
3 years ago
skateboarder, starting from rest, rolls down a 13.5 m ramp. When she arrives at the bottom of the ramp her speed is 7.37 m/s. If
scZoUnD [109]

Answer:

1.7 m/s²

Explanation:

d = length of the ramp = 13.5 m

v₀ = initial speed of the skateboarder = 0 m/s

v = final speed of the skateboarder = 7.37 m/s

a = acceleration

Using the equation

v² = v₀² + 2 a d

7.37² = 0² + 2 a (13.5)

a = 2.01 m/s²

θ = angle of the incline relative to ground = 29.9

a' = Component of acceleration parallel to the ground

Component of acceleration parallel to the ground is given as

a' = a Cosθ

a' = 2.01 Cos29.9

a' = 1.7 m/s²

7 0
3 years ago
Rock composed of many thin layers. This image appears in an Earth science magazine with this caption: "A non-foliated rock found
blondinia [14]

Answer:

its B

Explanation:

4 0
3 years ago
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