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Flura [38]
3 years ago
14

How does kinetic energy affect the stopping distance of a small vehicle compared to a large vehicle?

Physics
1 answer:
zhannawk [14.2K]3 years ago
6 0
The equation for kinetic energy is KE=1/2mv^2
So if the small vehicle AND the large vehicle both had the same speed, the large vehicle would have more kinetic energy since it has greater mass, resulting in a much more larger stopping distance than the smaller car.
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A penny rolls along the table for a distance of 1.3 meters. Jackie pushes it 40 centimeters further in the same direction.
brilliants [131]
40 meters times 1 meter over 100 centimeters equals 0.4 meters. 1.3 meters + 40 centimeters =. 1.3 m + 0.4 m = 1.7 m. The answer is 1.7 meters
7 0
3 years ago
What is the net charge of a copper atom if it gains 2 electrons?
Alex_Xolod [135]
If an atom gains electrons, it develops a negative charge equal to the number of electrons gained.
So the net charge on the copper atom which gained 2 electrons will be -2.
7 0
3 years ago
Consider the plot below describing motion along a straight line with an initial position of 10 m. −4 −3 −2 −1 0 1 2 3 4 5 6 1 2
Ugo [173]

Answer:

+1 m/s^2

Explanation:

Acceleration is given by

a=\frac{\Delta v}{\Delta t}

where

\Delta v is the change in velocity

\Delta t is the time interval in which the change in velocity occurs

To find the acceleration at 1 second, we can take the data at t = 1 s and t = 2. We find:

\Delta t = -3 -(-4) = 1 s

\Delta v = 2 - (1) = +1 m/s

So, the acceleration is

a=\frac{+1}{1}=+1 m/s^2

4 0
3 years ago
You're driving a vehicle of mass 1350 kg and you need to make a turn on a flat road. The radius of curvature of the turn is 71 m
sergey [27]

Answer:

v=12.65\ m.s^{-1}

Explanation:

Given:

  • mass of vehicle, m=1350\ kg
  • radius of curvature, r=71\ m
  • coefficient of friction, \mu=0.23

<u>During the turn to prevent the skidding of the vehicle its centripetal force must be equal to the opposite balancing frictional force:</u>

m.\frac{v^2}{r} =\mu.N

where:

\mu= coefficient of friction

N= normal reaction force due to weight of the car

v= velocity of the car

1350\times \frac{v^2}{71} =0.23\times (1350\times 9.8)

v=12.65\ m.s^{-1} is the maximum velocity at which the vehicle can turn without skidding.

5 0
3 years ago
A school bus has a mass of 18,200 kg. The bus moves at 13.5 m/s. How fast must a 0.142-kg baseball move in order to have the sam
nikitadnepr [17]

Answer:

bus momentum

p_bus= m_bus x v_bus

=18,200 x 16.5

basball momentum

pball=mball x vball

=0.142 x v

p_bus = pball

18200 x 16.5 = 0.142 x v

v=(18200 x 16.5)/0.142

v is the answer for baseball

Explanation:

⚠️not my answer tryna be honest here⚠️

3 0
2 years ago
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