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galina1969 [7]
2 years ago
6

If mass increases ,what must happen to the force in order to achieve the same change in motion?

Physics
1 answer:
igor_vitrenko [27]2 years ago
8 0

Answer:

increase

Explanation:

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The law of conservation of energy and describe the energy transformation that occur as you coast down a long hill on a bicycle a
Phantasy [73]

As you coast down a long hill on your bicycle, potential energy from your height is converted to kinetic energy as you and your bike are pulled downward by gravity along the slope of the hill. While there is air resistance and friction slowing you down by a little bit, your speed increases gradually until you apply the brakes, causing enough friction to slow yourself and the bike to a stop at the bottom

8 0
3 years ago
Standing and holding a barbell that has a mass of 100kg at a height of 2m involves being done on the barbell to maintain
Basile [38]

Answer:

1960Joules

Explanation:

Since we are not told what too find, we can as well find the Gravitational Potential Energy.

GPE = mass * acceleration due to gravity * height

GPE = 100*9.8 * 2

GPE = 980*2

GPE = 1960Joules

Hence the gravitational potential Energy is 1960Joules

6 0
3 years ago
In an elastic collision, a 400-kg bumper car collides directly from behind with a second, identical bumper car that is traveling
kirza4 [7]

Answer:

v₁ = 3.5 m/s

v₂ = 6.4 m/s

Explanation:

We have the following data:

m₁ = mass of trailing car = 400 kg

m₂ = mass of leading car = 400 kg

u₁ = initial speed of trailing car = 6.4 m/s

u₂ = initial speed of leading car = 3.5 m/s

v₁ = final speed of trailing car = ?

v₂ = final speed of leading car = ?

The final speed of the leading car is given by the following formula:

v_2=\frac{2m_1}{m_1+m_2}u_1-\frac{m_1-m_2}{m_1+m_2}u_2\\\\v_2=\frac{(2)(400\ kg)}{400\ kg+400\ kg}(6.4\ m/s)-\frac{400\ kg-400\ kg}{400\ kg + 400\ kg}(3.5\ m/s)

<u>v₂ = 6.4 m/s</u>

The final speed of the leading car is given by the following formula:

v_1=\frac{m_1-m_2}{m_1+m_2}u_1+\frac{2m_2}{m_1+m_2}u_2\\\\v_1=\frac{400\ kg-400\ kg}{400\ kg + 400\ kg}(6.4\ m/s)+\frac{(2)(400\ kg)}{400\ kg+400\ kg}(3.5\ m/s)

<u>v₁ = 3.5 m/s</u>

4 0
3 years ago
Which of the following best describes a noncontact force
Sonja [21]

Answer:

a force that is able to act at a distance

Explanation:

:)

7 0
3 years ago
A stunt driver drives a car horizontally off the edge of a cliff at 3.8m/s and reaches the water below 2.5s later.
andreyandreev [35.5K]
A. The cliff was 30.7 m high
B. I also got 9.5 as the horizontal distance

Here is my work, I find making charts like this one to find knowns and unknowns can be helpful

4 0
3 years ago
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