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AysviL [449]
3 years ago
7

A box slides down a 31° ramp with an acceleration of 0.99 m/s2. Determine the coefficient of kinetic friction between the box an

d the ramp.μk=______.

Physics
1 answer:
tino4ka555 [31]3 years ago
7 0

Answer:\mu=0.48

Explanation:

Given

inclination \theta =31^{\circ}

Acceleration of object=0.99 m/s^2

Now using FBD

mg\sin \theta -f_r=ma

mg\sin \theta -\mu mg\cos \theta =ma

a=g\sin \theta -\mu g\cos \theta

0.99=5.04-\mu 8.4

\mu 8.4=4.057

\mu =0.48

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How are energy, force, and the motion of objects related?
OLEGan [10]

Answer:

When two objects interact, each one exerts a force on the other that can cause energy to be transferred to or from the object. For example, when energy is transferred to an Earth-object system as an object is raised, the gravitational field energy of the system increases. This energy is released as the object falls; the mechanism of this release is the gravitational force. Likewise, two magnetic and electrically charged objects interacting at a distance exert forces on each other that can transfer energy between the interacting objects.

Explanation:

Even when an object is sitting still, it has energy stored inside that can be turned into kinetic energy (motion). ... A force is a push or pull that causes an object to move, change direction, change speed, or stop. Without a force, an object that is moving will continue to move and an object at rest will remain at rest.

5 0
3 years ago
Read 2 more answers
Consider a space shuttle which has a mass of about 1.0 x 105 kg and circles the Earth at an altitude of about 200.0 km. Calculat
svetlana [45]

Answer:

1.6675×10^-16N

Explanation:

The force of gravity that the space shuttle experiences is expressed as;

g = GM/r²

G is the gravitational constant

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r is the altitude = 200km = 200,000m

Substitute into the formula

g = 6.67×10^-11 × 1.0×10^5/(2×10^5)²

g = 6.67×10^-6/4×10^10

g = 1.6675×10^{-6-10}

g = 1.6675×10^-16N

Hence the force of gravity experienced by the shuttle is 1.6675×10^-16N

6 0
2 years ago
What is the driving force of charge around a circuit
Virty [35]
A force of charge that drive around a circuit is call electeons
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3 years ago
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allochka39001 [22]

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3 0
2 years ago
Given a second class lever with a distance of 5.00 feet from the fulcrum to the effort and a distance of 33.0 inches from the re
leva [86]

Answer:

The correct answer is C. 45.5 lbs.

Explanation:

In a second class lever, the load is located between the point in which the force is exerted and the fulcrum.

The formula for any problem involving a lever is:

F_ed_e=F_ld_l

Where F_e is the effort force, d_e is the total length of the lever, F_l is the load that can be lifted and d_l is the distance between the point of the effort and the fulcrum.

The parameter of the formula that you need is F_l:

F_l=\frac{F_ed_e}{d_l}

The conversion from feet to inches is 1 ft is equal to 12 inches. In this case, 5 ft are equal to 60 inches.

F_l=\frac{25*60}{33}

F_l=45.5 lbs

7 0
3 years ago
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