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4 years ago

1. A mass is on a level plane, it has a weight of 20N. What is the coefficient of kinetic friction if an applied force

Physics

1 answer:

Arturiano [62]4 years ago

4 0

Answer:

0.4

Explanation:

F-Fr=ma where F is applied force, Fr is friction, m is mass and a is acceleration.

Since the mass is moving with a constant velocity, there's no acceleration hence

$F=Fr=\mu N$ where N is the weight of object and \mu is coefficient of kinetic friction.

$F=\mu N and making [tex]\mu$ the subject

$\mu=\frac {F}{N}$

Substituting F for 8 N and N for 20 N

$\mu=\frac {8}{20}=0.4$

Therefore, coefficient of kinetic friction is 0.4

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wo fixed charges, A and B are located at x axis. A is at x = 0 m, B is at x = 4 m. QA = +4.0 μC and QB = -5.0 μC. Calculate the

lys-0071 [83]

Answer:

10250 N/C leftwards

Explanation:

QA = 4 micro Coulomb

QB = - 5 micro Coulomb

AP = 6 m

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A is origin, B is at 4 m and P is at 6 m .

The electric field due to charge QA at P is EA rightwards

$E_{A}=\frac{KQ_{A}}{AP^{2}}=\frac{9\times10^{9}\times4\times10^{-6}}{6^{2}}=1000 N/C (rightwards)$

The electric field due to charge QB at P is EB leftwards

$E_{B}=\frac{KQ_{B}}{BP^{2}}=\frac{9\times10^{9}\times5\times10^{-6}}{2^{2}}=11250 N/C (leftwards)$

The resultant electric field at P due the charges is given by

E = EB - EA

E = 11250 - 1000 = 10250 N/C leftwards

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3 years ago

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3 years ago

A fine-grained sedimentary rock is known as _____.

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3 years ago

Jay pushes the box with a force of 25 Newtons. There are 5 Newtons of friction present. In what direction will the box accelerat

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2 years ago

A ball is thrown upward. The total time the ball remains in the air is 14 s. a) With what speed was the ball thrown upward? b) W

wolverine [178]

Answer:

a) Vi = 137.2 m/s

b) h = 960.4 m

Explanation:

In order to find the initial speed we will use first equation of motion:

Vf = Vi + gt

where,

Vf = Final velocity = 0 m/s (since ball stops at highest point)

Vi = Initial Velocity = ?

g = - 9.8 m/s² (negative sign for upward moyion)

t = time interval = 14 s

Therefore,

0 m/s = Vi + (-9.8 m/s²)(14 s)

Vi = 137.2 m/s

Now, we use second equation of motion to find height (h):

h = Vi t + (1/2)gt²

h = (137.2 m/s)(14 s) + (1/2)(-9.8 m/s²)(14 s)²

h = 1920.8 m - 960.4 m

h = 960.4 m

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3 years ago