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2 years ago

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A woman living in a third-story apartment is moving out. Rather than carrying everything down the stairs, she decides to pack he

r belongings into crates, attach a frictionless pulley to her balcony railing, and lower the crates by rope. How hard must she pull on the horizontal end of the rope to lower a 25 kg crate at steady speed?

1 answer:

KatRina [158]2 years ago

5 0

Answer:

Explanation:

Let the force applied by her be T on the rope . This force will create tension in the rope. It will act in upper direction on the crates . The weight of the crate will act in downward direction .

Net force on the crate = mg - T in downward direction

Applying Newton's law

mg - T = m a

a =0 ( speed is steady )

mg - T = 0

T = mg

= 25 x 9.8

= 245 N

Force to be applied by her

= 245 N

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1) A fan is to accelerate quiescent air to a velocity of 8 m/s at a rate of 9 m3/s. Determine the minimum power that must be sup

azamat

Answer:

$\dot{W} = 339.84 W$

Explanation:

given data:

flow Q = 9 m^{3}/s

velocity = 8 m/s

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2 years ago

A large solar panel on a spacecraft in Earth orbit produces 1.0 kW of power when the panel is turned toward the sun. What power

Mandarinka [93]

Answer:

e*P_s = 11 W

Explanation:

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What power would the solar cell produce if the spacecraft were in orbit around Saturn

Solution:

- We use the relation between the intensity I and distance of light:

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I_s = I_e * ( r_e / r_s ) ^2

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2 years ago

A large crate sits on the floor of a warehouse. Paul and Bob apply constant horizontal forces to the crate. The force applied by

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Answer:

W = -510.98J

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Work done is given as:

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2 years ago

1)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it r

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Answer:

1)

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2)

$y=7.39\ m$

Explanation:

<u>Projectile Motion</u>

When an object is launched near the Earth's surface forming an angle $\theta$ with the horizontal plane, it describes a well-known path called a parabola. The only force acting (neglecting the effects of the wind) is the gravity, which acts on the vertical axis.

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Where $y_o$ is the initial height above the ground level, $v_{oy}$ is the vertical component of the initial velocity and t is the time

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$v_y=v_{oy}-gt$

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$v_{oy}-gt_m=0$

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2 years ago