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3 years ago

Altuve hits a ball with a bat. The action force is the impact of the bat pushing against the ball. What is the

Physics

2 answers:

prisoha [69]3 years ago

8 0

Answer:

H. The ball pushing on the bat

Explanation:

The question above is related to "Newton's Third Law of Motion." This law states that if an object exerts a force on another object, the second object will exert the same amount of force on the first object but in an "opposite direction." The force exerted by the first object is known as "action force" while the force exerted by the second object is known as "reaction force."

In the situation above, the ball is exerting an "action force on the bat" by pushing the bat. The bat then reacts to its force by pushing it back with an equal force (reaction force) that is opposing it.

Anuta_ua [19.1K]3 years ago

5 0

Answer: H

Explanation: It’s returning the same force back to the bat

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Which statements accurately describes the waste generated by nuclear reactions ? check all that apply.

Neko [114]

A) its 's hazardous for hundreds to thousand of year B) Some waste can be recycled and used again is the answer

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3 years ago

A 1.0 kg piece of copper with a specific heat of cCu=390J/(kg⋅K) is placed in 1.0 kg of water with a specific heat of cw=4190J/(

Valentin [98]

Answer:

The temperature change of the copper is greater than the temperature change of the water.

Explanation:

deltaQ = mc(deltaT)

Where,

delta T = change in the temperature

m =mass

c = heat capacity

$\frac{(deltaT)_{Cu}}{(deltaT)_{w}}=\frac{4190J/kg.K}{390J/kg.K}\\ \\(deltaT)_{Cu}=10.74(deltaT)_{w}$

The temperature change in the copper is nearly 11 times the temperature change in the water.

So, the correct option is,

The temperature change of the copper is greater than the temperature change of the water.

Hope this helps!

7 0

3 years ago

Discuss two reasons why people find transition between school and university

dangina [55]

Answer:

Is that your answer

6 0

3 years ago

A seesaw has an irregularly distributed mass of 30 kg, a length of 3.0 m, and a fulcrum beneath its midpoint. It is balanced whe

gladu [14]

Answer:

x = 0.9 m

Explanation:

For this exercise we must use the rotational equilibrium relation, we will assume that the counterclockwise rotations are positive

∑ τ = 0

60 1.5 - 78 1.5 + 30 x = 0

where x is measured from the left side of the fulcrum

90 - 117 + 30 x = 0

x = 27/30

x = 0.9 m

In summary the center of mass is on the side of the lightest weight x = 0.9 m

4 0

3 years ago

A kangaroo jumps straight up to a vertical height of 1.66 m. How long was it in the air before returning to Earth? Express your

Nataly [62]

Answer:

The kangaroo was 1.164s in the air before returning to Earth

Explanation:

For this we are going to use the equation of distance for an uniformly accelerated movement, that is:

$x = x_{0} + V_{0}t + \frac{1}{2}at^2$

Where:

x = Final distance

xo = Initial point

Vo = Initial velocity

a = Acceleration

t = time

We have the following values:

x = 1.66m

xo = 0m (the kangaroo starts from the floor)

Vo = 0 m/s (each jump starts from the floor and from a resting position)

a = 9.8 m/s^2 (the acceleration is the one generated by the gravity of earth)

t =This is just the time it takes to the kangaoo reach the 1.66m, we don't know the value.

Now replace the values in the equation

$x = x_{0} + V_{0}t + \frac{1}{2}at^2$

$1.66 = 0 + 0t + \frac{1}{2}9.8t^2$

$1.66 = 4.9t^2$

$\frac{1.66}{4.9} = t^2$

$\sqrt{0.339} = t\\ t = 0.582s$

It takes to the kangaroo 0.582s to go up and the same time to go down then the total time it is in the air before returning to earth is

t = 0.582s + 0.582s

t = 1.164s

The kangaroo was 1.164s in the air before returning to Earth

5 0

4 years ago