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marissa [1.9K]
3 years ago
14

Calculate the work done by an 85.0-kg man who pushes a crate 4.00 m up along a ramp that makes an angle of 20.0° with the horizo

ntal (see below). He exerts a force of 500 N on the crate parallel to the ramp and moves at a constant speed. Be certain to include the work he does on the crate and on his body to get up the ramp.
Physics
1 answer:
Aleks [24]3 years ago
3 0

Answer: Given mass of man = 85kg

Distance crate was pushed = 4m

Force exerted on crate = 500N

Angle inclined = 20°

Let Total force exerted Ft = (mg * sin(20)) + 500

Ft =( 85*9.81* sin20°) + 500

Ft = 758N

Work done Wd = Ft * distance

Wd = 785 * 4 = 3.14 * 10^3J

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Answer:

\mathrm{d.\:Static,\: Sliding,\:Rolling}

Explanation:

Static friction occurs when an object initially starts at rest. When the surfaces of the materials touch, the microscopic unevenness interlock greatest with each other, causing the most friction out of the three.

During sliding friction, an object is already moving or in motion. The microscopic surfaces still interlock, but because the object is in motion, it has a momentum. Therefore, the magnitude of sliding friction is less than that of static friction.

Rolling friction occurs when an object rolls across some surface. Rather than surfaces interlocking, rolling friction is caused by the constant distortion of surfaces. As it rolls, the surfaces of the object are constantly wrapping and changing. This distortion causes the rolling friction. However, it is much less in magnitude when compared to static or sliding friction.

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2 years ago
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3 years ago
Find the magnitude of the free-fall acceleration at the orbit of the Moon (a distance of 60RE from the center of the Earth with
Ede4ka [16]

Answer:

The magnitude of the free-fall acceleration at the orbit of the Moon is 2.728\times 10^{-3}\,\frac{m}{s^{2}} (\frac{2.784}{10000}\cdot g, where g = 9.8\,\frac{m}{s^{2}}).

Explanation:

According to the Newton's Law of Gravitation, free fall acceleration (g), in meters per square second, is directly proportional to the mass of the Earth (M), in kilograms, and inversely proportional to the distance from the center of the Earth (r), in meters:

g = \frac{G\cdot M}{r^{2}} (1)

Where:

G - Gravitational constant, in cubic meters per kilogram-square second.

M - Mass of the Earth, in kilograms.

r - Distance from the center of the Earth, in meters.

If we know that G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}, M = 5.972\times 10^{24}\,kg and r = 382.26\times 10^{6}\,m, then the free-fall acceleration at the orbit of the Moon is:

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6 0
3 years ago
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TiliK225 [7]

Answer:

1-In a uniform electric field, the field lines are straight, parallel, and uniformly spaced this statement is true.

2-Electric field lines near positive point charges radiate outward. this statement is also true.

3-The electric force acting on a point charge is proportional to the magnitude of the point charge. this statement is true as well.

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the electric field created by a point charge is defined by E=KQ/r^2 where k is constant, q is magnitude of charge and r is the distance away from the point charge so the electric filed is distance dependent and can not be constant at all distances.

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Mmm tricky.
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So the opposite of F1 is (-2N)I + (6N)J!

8 0
3 years ago
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