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3 years ago

8

the police department determined that the force required to drag a 130 N(29 lb) car tire across the pavement at a constant veloc

ity is called 100N (23 lb ) specifications from the trucks manufacturer claim that the effecrive coefficient of kinetic friction between the tires and road for both new car and truck??

1 answer:

ki77a [65]3 years ago

4 0

The force of friction is given by:
f = μR, where μ is the friction coefficient and R is the reaction force, which will be equal to the weight.
100 = μ x 130
μ = 0.77

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A 125 W motor accelerates a block along a level, frictionless surface at an average speed of 5.0

Lelu [443]

Answer:

25N

Explanation:

Given parameters:

Power of the motor = 125W

Average speed = 5m/s

Unknown:

Force supplied to the motor = ?

Solution:

Work done is the force applied to move a body through a particular distance;

Work done = Force x distance

Also,

Work done = Power x time

So;

Force x distance = Power x time

since force is the unknown;

Force = $\frac{Power x time}{distance}$

Speed = $\frac{distance }{time}$

Force = $\frac{Power}{speed}$

Now solve;

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Four charges of magnitude +q are placed at the corners of a square whose sides have a length d. What is the magnitude of the tot

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Answer:

$F = \frac{4kqQb}{(b^2 + \frac{d^2}{2})^{1.5}}$

Explanation:

Since all the four charges are equidistant from the position of Q

so here we can assume this charge distribution to be uniform same as that of a ring

so here electric field due to ring on its axis is given as

$E = \frac{k(4q)x}{(x^2 + R^2)^{1.5}}$

here we have

x = b

and the radius of equivalent ring is given as the distance of each corner to the center of square

$R = \frac{d}{\sqrt2}$

now we have

$E = \frac{4kq b}{(b^2 + \frac{d^2}{2})^{1.5}}$

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$F = QE$

$F = \frac{4kqQb}{(b^2 + \frac{d^2}{2})^{1.5}}$

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3 years ago

With that push, they have accelerated their masses, which means they have _____ of their bodies.

nexus9112 [7]

As per Definition of acceleration we know that rate of change in velocity of object is known as acceleration

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a.) changed the velocity

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Bingel [31]

Answer:

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Explanation:

Given that,

There are three lines, so you can calculate three different values of the function at one time.

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Given table is,

Time t = 0, 1, 1,02

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Using equation of motion

$h=h_{0}+ut-\dfrac{1}{2}gt^2$

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g = acceleration due to gravity

At t = 0,

Put the value into the formula

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Put the value in the equation

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So, velocity at 1.02 sec is given

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$h=ut-\dfrac{1}{2}gt^2$

On differentiating w.r.to t

$h'(t)=u-\dfrac{1}{2}g(2t)$

Put the value into the formula

$h'(t)=22-\times32\times(1.02)$

$h'(t)=-10.64\ feet/sec$

Hence, The rate of change of height with respect to time is -10.64 feet/sec.

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