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3 years ago

1. Rocky lifted a 300 newton rock 2 meters off the ground and carried it 2 -

Physics

1 answer:

sergij07 [2.7K]3 years ago

4 0

Answer: 1200 joules

Explanation:

Work is done when force is applied on an object over a distance. It is measured in joules.

Thus, Work = Force X Distance

Since Force = 300 newton

Total distance = (distance off the ground + distance travelled on the ground)

i.e 2 meters off + 2 more meters = 4metres

Work = Force X Distance

= 300 Newton X 4 metres

= 1200 joules

Thus, Rocky did 1200 joules of work on the

rock

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Force exerted by the bullet = mass * acceleration = 0.013 * 850 = 11.05 Newtons.

the rifle exerts same force in opposite direction so we have

11.05 = 3.5 * a
acceleration = 11.05 / 3.5 = 3.16 m /s^-2

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3 years ago

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أسقط عامل حجرة من سطح بناية سقوطأ حرة. أوجد ما يلي :

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3 years ago

What did scientists predict the big bang should have left ??

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<span>radiation, hydrogen, and helium </span>

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A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provid

Sergeeva-Olga [200]

Answer:

$8.37*10^5$ rpm

Explanation:

Given that rotational kinetic energy = $4.66*10^9J$

Mass of the fly wheel (m) = 19.7 kg

Radius of the fly wheel (r) = 0.351 m

Moment of inertia (I) = $\frac{1}{2} mr ^2$

Rotational K.E is illustrated as $(K.E)_{rt} = \frac{1}{2} I \omega^2$

$\omega = \sqrt{\frac{2(K.E)_{rt}}{I} }$

$\omega = \sqrt{\frac{2(KE)_{rt}}{1/2 mr^2} }$

$\omega = \sqrt{\frac{4(K.E)_{rt}}{mr^2} }$

$\omega = \sqrt{\frac{4*4.66*10^9J}{19.7kg*(0.351)^2} }$

$\omega = 87636.04$

$\omega = 8.76*10^4 rad/s$

Since 1 rpm = $\frac{2 \pi}{60} rad/s$

$\omega = 8.76*10^4(\frac{60}{2 \pi})$

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3 years ago

: The maximum theoretical efficiency of a Carnot engine operating between reservoirs at the steam point and at room temperature

Hunter-Best [27]

Answer:

The value is $\eta = 0.2145$ or 21.45%

Explanation:

From the question we are told that

The first reservoir is at steam point $T_s = 100^o C = 100 + 273 = 373 \ K$

The second reservoir is at room temperature $T_r = 20^o C = 293 \ K$

Generally the maximum theoretical efficiency of a Carnot engine is mathematically evaluated as

$\eta = 1- \frac{T_r}{T_s}$

=> $1 - \frac{ 293}{373}$

=> $\eta = 0.2145$

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3 years ago