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Degger [83]
3 years ago
8

A go-kart and rider have a mass of 14 kg. If the cart accelerates at 6 m/s² during a 40 m sprint in 100 seconds, how much power

did the cart do?
Physics
1 answer:
Nutka1998 [239]3 years ago
5 0

Answer:

<em>P=33.6 \ W</em>

Explanation:

<u>Mechanical Work and Power</u>

Work is the amount of energy transferred by a force. It's a scalar quantity, with SI units of joules.

Being \vec F the force vector and \vec s the displacement vector, the work is calculated as:

W=\vec F\cdot \vec s

If both the force and displacement are parallel, then we can use the equivalent scalar formula:

W=F.s

Mechanical Power is the amount of energy transferred or converted per unit of time. The SI unit of power is the watt, equal to one joule per second.

The power can be calculated as:

\displaystyle P=\frac {W}{t}

Where W is the work and t is the time.

If an object of mass m has an acceleration a, the net force is:

F=m.a

The go-kart and rider have a mass of m=14 kg and accelerate at 6 m/s^2, thus the net force applied is:

F=14\cdot 6 = 84\ N

The work done by the cart when traveling d= 40 m is:

W=84\cdot 40

W=3,360\ J

Finally, the power for t= 100 seconds is:

\displaystyle P=\frac {3,360}{100}

P=33.6 \ W

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Given p=mv, with m being the electron mass and v its velocity, we can find the electron's velocity:
v= \frac{p}{m}= \frac{1.94 \cdot 10^{-26} kgm/s}{9.1 \cdot 10^{-31} kg}=  2.13 \cdot 10^4 m/s

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3 years ago
A spacecraft is moving past the earth at a constant speed of 0.60 times the speed of light. The astronaut measures the time inte
Afina-wow [57]

Answer:

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Explanation:

Given the data in the question;

speed of the spacecraft as it moves past the is 0.6 times the speed of light

we know that speed of light c = 3 × 10⁸ m/s

so speed of spacecraft v = 0.6 × c = 0.6c

time interval between ticks of the spacecraft clock Δt₀ = 3.2 seconds

Now, from time dilation;

t = Δt₀ / √( 1 - ( v² / c² ) )

t = Δt₀ / √( 1 - ( v/c )² )

we substitute

t = 3.2 / √( 1 - ( 0.6c / c )² )

t = 3.2 / √( 1 - ( 0.6 )² )

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t = 4 seconds

Therefore, the time interval that an earth observer measures is 4 seconds

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Answer:

Explanation:

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