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3 years ago

A man wishes to pull a crate 15m across a rough floor by exerting a force of 100 N. The

Physics

1 answer:

Darina [25.2K]3 years ago

3 0

Answer:

option (E) is correct.

Explanation:

Work done is defined as the product of force and the distance in the direction of force.

force, f = 100 N

Coefficient of friction, = 0.25

distance = 15 m

So, net force F = f - friction force

F = 100 - 0.25 x m g

Work = (100 - 0.25 mg) x d cosθ

For minimum work, the angle should be maximum.

So, the value of θ is 76°.

thus, option (E) is correct.

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2 years ago

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Sketch a position-time graph for a bear starting

Dmitrij [34]

Explanation:

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3 years ago

A satellite orbits the earth a distance of 1.50 × 107 m above the planet's surface and takes 8.65 hours for each revolution abou

kupik [55]

Answer:

The acceleration of the satellite is $0.87 m/s^{2}$

Explanation:

The acceleration in a circular motion is defined as:

$a = \frac{v^{2}}{r}$ (1)

Where a is the centripetal acceleration, v the velocity and r is the radius.

The equation of the orbital velocity is defined as

$v = \frac{2 \pi r}{T}$ (2)

Where r is the radius and T is the period

For this particular case, the radius will be the sum of the high of the satellite ( $1.50x10^{7} m$ ) and the Earth radius ( $6.38x10^{6} m$ ) :

$r = 1.50x10^{7} m+6.38x10^{6}m$

$r = 21.38x10^{6}m$

Then, equation 2 can be used:

$T = 8.65 hrs \cdot \frac{3600 s}{1hrs}$ ⇒ $31140 s$

$v = \frac{2 \pi (21.38x10^{6}m)}{31140s}$

$v = 4313 m/s$

Finally equation 1 can be used:

$a = \frac{(4313m/s)^{2}}{21.38x10^{6}m}$

$a = 0.87 m/s^{2}$

Hence, the acceleration of the satellite is $0.87 m/s^{2}$

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3 years ago

5p - 14 = 8 p + 4 find p

Digiron [165]

Answer:

Explanation:

5p - 14 = 8p + 4

5p = 8p + 18 <-- Moving constants to one side; add the same number of +14 to both sides.

-3p = 18. <-- The same thing with the variable itself.

p = -6 <-- Divide both sides by negative 3.

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3 years ago

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Answer:

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2 years ago