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White raven [17]
3 years ago
15

If an object is not accelerating what can you determine about the sum of all the forces on the object

Physics
1 answer:
Nataliya [291]3 years ago
6 0
The forces are equal to each other, if there is no net force then there is no acceleration taking place. This object could be moving at constant velocity or just be still in place
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1. A 12 kilogram block is sitting on a platform 24 m high. How much Potential energy does
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The amount of potential energy the block contains is 2,822.4 Joules

<u>Given the following data:</u>

  • Mass of block = 12 kg
  • Height of platform = 24 meters.

We know that the acceleration due to gravity (g) of an object on planet Earth is equal to 9.8 m/s^2.

To determine the amount of potential energy the block contains:

Mathematically, potential energy (P.E) is given by the formula;

P.E = mgh

Where:

  • m is the mass of object.
  • g is the acceleration due to gravity.
  • h is the height of an object.

Substituting the parameters into the formula, we have;

P.E = 12 \times 9.8 \times 24

Potential energy (P.E) = 2,822.4 Joules

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Why are fossil fuels like coal, oil, and natural gas<br> called nonrenewable resources?
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4 years ago
a 282 kg bumper car moving right at 3.50 m/s collides with a 155 kg bumper car moving 1.88 m/s left. afterwards, the 282 kg car
Vaselesa [24]

The momentum of the 155 kg car afterwards is 469.7 kg m/s to the right

Explanation:

We can solve the problem by using the law of conservation of momentum: the total momentum of the system is conserved before and after the collision, so we can write:

p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

where:

m_1 = 282 kg is the mass of the bumper car

u_1 = 3.50 m/s is the initial velocity of the bumper car (we take the right as positive direction)

v_1 = 0.800 m/s is the final velocity of the bumper car

m_2 = 155 kg is the mass of the second bumper car

u_2 = -1.88 m/s is the initial velocity of the second car (moving to the left)

v_2 is the final velocity of the second car

Solving for v_2,

v_2 = \frac{m_1 u_1+m_2 u_2 - m_1 v_1}{m_2}=\frac{(282)(3.50)+(155)(-1.88)-(282)(0.800)}{155}=3.03 m/s

where the positive sign means the direction is to the right.

And now we can find the momentum of the 155 kg afterwards, which is

p_2 = m_2 v_2 = (155)(3.03)=469.7 kg m/s (to the right)

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3 years ago
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