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2 years ago

16 En un juego de béisbol, un lanzador logra

Physics

1 answer:

Anuta_ua [19.1K]2 years ago

5 0

Answer:

B transferida.

Explanation:

The kinetic energy of the baseball would be transferred to the bat and to the handler.

The law of conservation of energy states that "energy is neither created nor destroyed but transformed from one form to another".

We know that the baseball posses kinetic energy as it begins to accelerate towards the other player.
Kinetic energy is the energy due to the motion of a body.
As it collides with the bat, it is converted to another form of energy.
The energy form would have been converted to kinetic energy, heat energy and some mechanical energy component.
Therefore, the energy has been transferred from the ball to the bat.

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When vibrational motion in an object increases, which is a true statement?

ss7ja [257]

Hello friend!!

We know that kinetic energy is the energy possessed due to the motion of the object. And we know if the object is in a fast motion then the temperature would be high, whereas if the object is slow in motion then it will have lower temperature. So we know that the kinetic energy is indirectly related to temperature.From our knowledge we can conclude that HIGHER THE TEMPERATURE, HIGHER THE KINETIC ENERGY and LOWER THE TEMPERATURE, LOWER THE KINETIC ENERGY.
Hence, the answer to your question here is,a.kinetic energy, temperature, and thermal energy increase.
Hope it helps!!All the best!!

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3 years ago

How much work is done by 0.070 m3 of gas, when the volume remains constant with pressure of 63 x 105 Pa?

stealth61 [152]

Answer:

W = 0 J

Explanation:

The amount of work done by gas at constant pressure is given by the following formula:

$W = P\Delta V$

where,

W = Work done by the gas

P = Pressure of the gas

ΔV = Change in the volume of the gas

Since the volume of the gas is constant. Therefore, there is no change in the volume of the gas:

$W = P(0\ m^3)\\$

5 0

2 years ago

A 120 kg tackler moving at 3.0 m/s meets head-on (and tackles) a 91 kg halfback moving at 7.5 m/s. What will be their mutual vel

Vesna [10]

Explanation:

It is given that,

Mass of the tackler, m₁ = 120 kg

Velocity of tackler, u₁ = 3 m/s

Mass, m₂ = 91 kg

Velocity, u₂ = -7.5 m/s

We need to find the mutual velocity immediately the collision. It is the case of inelastic collision such that,

$v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}$

$v=\dfrac{120\ kg\times 3\ m/s+91\ kg\times (-7.5\ m/s)}{120\ kg+91\ kg}$

v = -1.5 m/s

Hence, their mutual velocity after the collision is 1.5 m/s and it is moving in the same direction as the halfback was moving initially. Hence, this is the required solution.

3 0

2 years ago

There are competitions in which pilots fly small planes low over the ground and drop weights, trying to hit a target. Part A A p

tamaranim1 [39]

Explanation:

A pilot flying a weight; it takes t = 3.4 s to hit the ground, during which it travels a horizontal distance d = 120 m

h = g/2*t^2

h= 4.903*3.4^2

h=56.701 m

V = d/t

V= 120/3.4

V=35.3 m/sec

Now the pilot does a run at the same height but twice the speed.

How much time (t1) does it take the weight to hit the ground?

t1 = t = 3.4 sec (depending just upon height)

5 0

2 years ago

At a dock, a crane lifts a 2009 kg container 20 m, swings it out over the deck of a freighter, and lowers the container into the

deff fn [24]

Answer:

W = 157.5kJ

Explanation:

Assuming it moves the container at constant speed, the work done by the crane will be equal to the variation of the potential gratitational energy on the container:

$Wc = \Delta E = m*g*(h2 - h1)$ where h2= -8m and h1=0m

Wc = 157.5kJ

6 0

3 years ago