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Bess [88]
2 years ago
9

How much energy is required to accelerate a golf ball of mass 0.046 kg initially at rest to a speed of 0.75c?

Physics
1 answer:
Andrew [12]2 years ago
3 0

The required energy to accelerate the golf ball is  0.232875X10^{16} kg-m/sec^2.

<h2>What is Energy?</h2>
  • Energy, which is observable in the execution of labor as well as in the form of light and heat, is the quantitative quality that is transmitted to an object or to a physical phenomenon in physics.
  • Energy is a preserved resource; according to the rule of conservation of energy, energy can only be transformed from one form to another and cannot be created or destroyed.
  • The joule, which is defined as "the energy transmitted to an object by the effort of moving it a length of one metre against a force of one newton," is the standard measure for power in the International System of Units (SI).

Given that,

E=mc^2

=0.046X0.75^2Xc^2

=0.232875X10^{16} kg-m/sec^2 .

Learn more about energy here:

brainly.com/question/13881533

#SPJ4

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Number three. How water evaporates and then forms rain and snow.
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You travel 60 meters to the left in 20 seconds and then you travel 60 meters to the right in 30 seconds; what is your average ve
vovikov84 [41]
Average velocity = displacement / time

In this case, since you move left 60, then right 60, you return to the starting point making the displacement zero.

average velocity = 0 / 60 sec = zero

the average speed is calculated using total distance traveled.

average speed = distance / time

average speed = 120 meters / 60 sec = 2 m/s
4 0
3 years ago
1. A heat engine operates between two reservoirs at T2 = 600 K and T1 = 350 K. It takes in 1.00 x 103 J of energy from the highe
Alex73 [517]

Answer:

\Delta S_u=2.1429\ J.K^{-1}

W_c=416.67\ J

Explanation:

Given:

temperature of source reservoir, T_H=600\ K

temperature of sink reservoir, T_L=350\ K

energy absorbed from the source, Q_{in}=1000\ J

work done, W=250\ J

a.

<u>Now change in entropy of the surrounding:</u>

\Delta S_u=\frac{dQ_L}{T_L}

<em>Since heat engine is a device that absorbs heat from a high temperature reservoir and does some work giving out heat in the universe as the byproduct.</em>

\Delta S_u=\frac{Q_H-W}{T_L}

\Delta S_u=\frac{1000-250}{350}

\Delta S_u=2.1429\ J.K^{-1}

b.

<u>We know Carnot efficiency is given as:</u>

\eta_c=1-\frac{T_L}{T_H}

\eta_c=1-\frac{350}{600}

\eta_c=0.4167

<u>Now the Carnot work done:</u>

W_c=Q_H\times \eta_c

W_c=1000\times 0.4167

W_c=416.67\ J .......................(1)

c.

From eq. (1) we have the Carnot work, so the difference:

\Delta W=W_c-W

\Delta W=416.67-250

\Delta W=166.67\ J

Now, we find:

T_L.\Delta S_u=350\times 2.1429

5 0
3 years ago
If a cup of coffee has temperature 95∘C95∘C in a room where the temperature is 20∘C,20∘C, then, according to Newton's Law of Coo
lina2011 [118]

Answer:

T = 76.39°C

Explanation:

given,

coffee cup temperature = 95°C

Room temperature= 20°C

expression

T( t ) = 20 + 75 e^{\dfrac{-t}{50}}

temperature at t = 0

T( 0 ) = 20 + 75 e^{\dfrac{-0}{50}}

T(0) = 95°C

temperature after half hour of cooling

T( t ) = 20 + 75 e^{\dfrac{-t}{50}}

t = 30 minutes

T( 30 ) = 20 + 75 e^{\dfrac{-30}{50}}

T( 30 ) = 20 + 75 \times 0.5488

T(30) = 61.16° C

average of first half hour will be equal to

T = \dfrac{1}{30-0}\int_0^30(20 + 75 e^{\dfrac{-t}{50}})\ dt

T = \dfrac{1}{30}[(20t - \dfrac{75 e^{\dfrac{-t}{50}}}{\dfrac{1}{50}})]_0^30

T = \dfrac{1}{30}[(20t - 3750e^{\dfrac{-t}{50}}]_0^30

T = \dfrac{1}{30}[(20\times 30 - 3750 e^{\dfrac{-30}{50}} + 3750]

T = \dfrac{1}{30}[600 - 2058.04 + 3750]

T = 76.39°C

4 0
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Help please <br><br> Gravitation
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