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2 years ago

How would decreasing the time it takes you to run a certain distance affect your speed? Explain your reasoning.

Physics

1 answer:

Tom [10]2 years ago

5 0

Explanation:

The speed is determined by the distance that is traveled by an object during a given amount of time. If the time is decreased, the speed would increase. The reason for this is that time is inversely proportional to the speed of an object.

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If the weightlifter's force is considered the action, what is the magnitude of the

Nookie1986 [14]

Answer:

Explanation:

I hope that helps I try

5 0

3 years ago

You can analogize the photoelectric energy to

Anon25 [30]

Answer:

a useful sporting analogy ... Teachers of science routinely use analogies to help ... balls over the bar and you will probably find that.

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4 years ago

Consider a 2-kg bowling ball sits on top of a building that is 40 meters tall. It falls to the ground. Think about the amounts o

likoan [24]

Answer:

1) At the highest point of the building.

2) The same amount of energy.

3) The kinetic energy is the greatest.

4) Potential energy = 784.8[J]

5) True

Explanation:

Question 1

The moment when it has more potential energy is when the ball is at the highest point in the building, that is when the ball is at a height of 40 meters from the ground. It is taken as a point of reference of potential energy, the level of the soil, at this point of reference the potential energy is zero.

$E_{p} = m*g*h\\E_{p} = 2*9.81*40\\E_{p} = 784.8[J]$

Question 2)

The potential energy as the ball falls becomes kinetic energy, in order to be able to check this question we can calculate both energies with the input data.

$E_{p}=m*g*h\\ E_{p} = 2*9.81*20\\ E_{p} = 392.4[J]\\$

And the kinetic energy will be:

$E_{k}=0.5*m*v^{2}\\ where:\\v = velocity = 19.8[m/s]\\E_{k}=0.5*2*(19.8)^{2}\\ E_{k}=392.04[J]$

Therefore it is the ball has the same potential energy and kinetic energy as it is half way through its fall.

Question 3)

As the ball drops all potential energy is transformed into kinetic energy, therefore being close to the ground, the ball will have its maximum kinetic energy.

$E_{k}=E_{p}=m*g*h = 2*9.81*40\\ E_{k} = 784.8[J]\\ E_{k} = 0.5*2*(28)^{2}\\ E_{k} = 784 [J]$

Question 4)

It can be easily calculated using the following equation

$E_{p} =m*g*h\\E_{p}=2*9.81*40\\E_{p} =784.8[J]$

Question 5)

True

The potential energy at 20[m] is:

$E_{p}=2*9.81*20\\ E_{p}= 392.4[J]\\The kinetic energy is:\\E_{k}=0.5*2*(19.8)^{2} \\E_{k}=392[J]$

3 0

3 years ago

The chart show the masses and velocities of two colliding objects that stick together after a collision.

finlep [7]

Answer:

1,500 kg.m/s

Explanation:

First, arrange the information in a table:

Object Mass (kg) Velocity (m/s)

A 200 15

B 150 - 10

After the collision, the two objects are stick together, thus you talk aobut one object and one momentum.

According to the law of convervation of momentum, the momentum after the collision is equal to the momentum before the collision.

Momentum before the collision, P₁:

$P_1=m_{A,1}\times v_{A,1}+m_{B,1}\times v_{B,1}$

$P_1=200kg\times 15m/s+150kg\times (-10m/s)\\\\ P_1=3000kg.m/s-1500kg.m/s=1500kg.m/s$

Momentum after the collision:

As stated, it es equal to the momentum before the collision: 1,500 kg . m/s

6 0

4 years ago

Which is the term used for a change in the magnitude of the velocity

Alborosie

Rate of change of velocity is called acceleration. When this rate of change is constant, it is called uniform acceleration. Like the change of velocity of an object falling freely under gravity. Its velocity increases by 9.8 m every second. This acceleration is denoted as 9.8m/s^2

8 0

3 years ago