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frutty [35]
3 years ago
10

A heavy object and a light object are dropped from the same height. If we neglect air resistance, which will hit the ground firs

t?
Physics
1 answer:
Maksim231197 [3]3 years ago
6 0

Answer:

None, both objects will hit ground at the same time.

Explanation:

  • Assuming no air resistance present, and that both objects start from rest, we can apply the following kinematic equation for the vertical displacement:

        \Delta h = \frac{1}{2}*g*t^{2}  (1)

  • As the left side in (1) is the same for both objects, the right side will be the same also.
  • Since g is constant close to the surface of the Earth, it's also the same for both objects.
  • So, the time t must be the same for both objects also.
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Varg sees a spring that has a spring constant of 4 N/m that is stretched 5 m. He stretches the spring an additional 5 m. Conside
AysviL [449]

Answer:

Elastic potential energy, E = 200 J

Explanation:

It is given that,

Spring constant, K = 4 N/m

initial stretching in the spring, x = 5 m

Finally, it is stretched an additional 5 m i.e. x' = 5 m        

Let E is the elastic energy in the spring after Varg stretches the spring. it is given by :

E=\dfrac{1}{2}k(x+x')^2

E=\dfrac{1}{2}\times 4\times (10)^2

E = 200 J

So, the elastic energy in the spring after Varg stretches the spring is 200 J. hence, this is the required solution.

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A baseball catcher puts on an exhibition by catching a 0.15-kg ball dropped from a helicopter at a height of 101 m. What is the
yaroslaw [1]

Answer:

The speed of the ball 1.0 m above the ground is 44 m/s (Answer A).

Explanation:

Hi there!

To solve this problem, let´s use the law of conservation of energy. Since there is no air resistance, the only energies that we should consider is the gravitational potential energy and the kinetic energy. Because of the conservation of energy, the loss of potential energy of the ball must be compensated by a gain in kinetic energy.

In this case, the potential energy is being converted into kinetic energy as the ball falls (this is only true when there are no dissipative forces, like air resistance, acting on the ball). Then, the loss of potential energy (PE) is equal to the increase in kinetic energy (KE):

We can express this mathematically as follows:

-ΔPE = ΔKE

-(final PE - initial PE) = final KE - initial KE

The equation of potential energy is the following:

PE = m · g · h

Where:

PE = potential energy.

m = mass of the ball.

g = acceleration due to gravity.

h = height.

The equation of kinetic energy is the following:

KE = 1/2 · m · v²

Where:

KE = kinetic energy.

m = mass of the ball.

v = velocity.

Then:

-(final PE - initial PE) = final KE - initial KE          

-(m · g · hf - m · g · hi) = 1/2 · m · v² - 0     (initial KE = 0 because the ball starts from rest)  (hf = final height, hi = initial height)

- m · g (hf - hi) = 1/2 · m · v²

2g (hi - hf) = v²

√(2g (hi - hf)) = v

Replacing with the given data:

√(2 · 9.8 m/s²(101 m - 1.0 m)) = v

v = 44 m/s

The speed of the ball 1.0 m above the ground is 44 m/s.

3 0
3 years ago
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