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dexar [7]
3 years ago
6

A 20.0-kg block is initially at rest on a horizontal surface. A horizontal force of 77.0 N is required to set the block in motio

n, after which a horizontal force of 56.0 N is required to keep the block moving with constant speed.Find the coefficients of static and kinetic friction from this information.

Physics
1 answer:
saveliy_v [14]3 years ago
6 0

Answer: The coefficient of static friction is 3.85 and  The coefficient of kinetic friction is 2.8

Explanation:

in the attachment

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If the value of static friction for a couch is 400N, what could be a possible value of its kinetic friction? *
Dafna11 [192]

Answer:

kinetic friction may be greater than 400 N or smaller than 400 N

Explanation:

As we know that maximum value of static friction on the rough surface is known as limiting friction and the formula of this limiting friction is known as

F_s = \mu_s N

now when object is sliding on the rough surface then the friction force on that surface is known as kinetic friction and the formula of kinetic friction is known as

F_k = \mu_k N

now we know that

\mu_k < \mu_s

so here value of limiting static friction force is always more than kinetic friction

also we know that

initially when body is at rest then static friction value will lie from 0 N to maximum limiting friction

and hence kinetic friction may be greater than static friction or if the static friction is maximum limiting friction then kinetic friction is smaller than static friction

so kinetic friction may be greater than 400 N or smaller than 400 N

5 0
3 years ago
A balloon rubbed against denim gains a charge of −7.0 µc. what is the electric force between the balloon and the denim when the
zubka84 [21]
The solution for this problem is computed by through this formula, F = kQq / d²Plugging in the given values above, we can now compute for the answer.
F = 8.98755e9N·m²/C² * -(7e-6C)² / (0.03m)² = -489N, the negative sign denotes attraction.
4 0
3 years ago
If the final speed of an object as is strikes the ground is 77 m/s and it was in the air for 6.5 seconds. What was the initial d
azamat

Answer:

The answer to your question is: 13.2 m/s

Explanation:

final speed (fs) = 77 m/s

t = 6.5 s

gravity (g) = 9.81 m/s2

initial speed (is) = ?

Formula

fs = is + gt     from this equation we clear "is" = fs - gt

Substitution                         is = 77 - (9,81)(6.5)

Process                               is = 77 - 63.8

                                            is = 13.2 m/s

8 0
3 years ago
Bronco the skydiver, whose mass is 100 kg experiences 200 N of air resistance. What is the acceleration of his fall? Show Work
bekas [8.4K]

Explanation:

air resistance (f) =mg

200=100g

g=200/100

g=2m/s^2

5 0
2 years ago
The gravitational force of a star on an orbiting planet 1 is f1. planet 2, which is three times as massive as planet 1 and orbit
Margaret [11]

Let  us consider two bodies having masses m and m' respectively.

Let they are  separated by a distance of r from each other.

As per the Newtons law of gravitation ,the gravitational force between two bodies is given as -  F = G\frac{mm'}{r^{2} }   where G is the gravitational force constant.

From the above we see that F ∝ mm' and F\alpha \frac{1}{r^{2} }

Let the orbital radius of planet  A is r_{1}  = r and mass of planet is m_{1}.

Let the mass of central star is m .

Hence the gravitational force for planet A  is f_{1} =G \frac{m_{1}*m }{r^{2} }

For planet B the orbital radius  r_{2} =2r_{1} and mass m_{2} = 3 m_{1}

Hence the gravitational force f_{2} =G\frac{m m_{2} }{r^{2} }

                                                 f_{2} =G\frac{m*3m_{1} }{[2r_{1}] ^{2} }

                                                 = \frac{3}{4} G\frac{mm_{1} }{r_{1} ^{2} }

Hence the ratio is  \frac{f_{2} }{f_{1} } = \frac{\frac{3}{4}G mm_{1/r_{1} ^2}  }{Gmm_{1}/r_{1} ^2 }

                                      =\frac{3}{4}     [ ans]


                                                 

                           

3 0
3 years ago
Read 2 more answers
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