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mezya [45]
3 years ago
11

A stubborn, 120 kg mule sits down and refuses to move. To drag the mule to the barn, the exasperated farmer ties a rope around t

he mule and pulls with his maximum force of 800 N. The coefficients of friction between the mule and the ground are μs=0.8 and μk=0.5. Is the farmer able to move the mule?
Physics
1 answer:
Helen [10]3 years ago
8 0

Answer:

No

Explanation:

In order for the farmer to be able to put the mule in motion, the force he applies must be larger than the maximum value of the force of static friction exerted by the ground on the mule.

The maximum value of the force of static friction on the mule is given by:

F_f=\mu_s mg

where:

\mu_s=0.8 is the coefficient of static friction

m = 120 kg is the mass of the mule

g=9.8 m/s^2 is the acceleration due to gravity

Substituting, we find:

F_f=(0.8)(120)(9.8)=941 N

Here in this problem, the force applied by the farmer through the rope is

F = 800 N

We see that this force is smaller than the value of the maximum force of friction: therefore, the farmer will not be able to move the mule.

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Answer:

b)

Explanation:

By convention, the electric field lines (which are tangent to the direction of the electric field at a given point) always begin at positive charges, and finish at negative charges.

This is a consequence of the convention that states that the electric field has the direction of the trajectory of a positive test charge when released from rest in an electric field.

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The answer is B.
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Erica (38 kg ) and danny (43 kg ) are bouncing on a trampoline. just as erica reaches the high point of her bounce, danny is mov
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<span>2.5 m/s going upward.
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3 years ago
if a ball is thrown straight up into the air with an initial velocity of 8080 ft/s, its height in feet after tt second is given
yanalaym [24]

The average velocity for the time period beginning when t=1 and lasting

(i) 0.01 seconds = 63.84 ft/s

(ii) 0.001 seconds = 63.984 ft/s

Given that a ball is thrown with an initial velocity = 80 ft/s

Let 'y' be the height in feet after 't' seconds.

Given,  y=80t-16t^2 gives the height in 't' seconds.

Average velocity = Rate of change of distance

                             = Change in distance/Change in time.

The initial time can be taken as 0 s.

When t =1 s, y = 80 - 16 = 64 ft

(1)  t = 0.01 s

    y = 80 x 0.01 - 16 x 0.01 x 0.01 = 0.7984 ft

    Average velocity = (64 - 0.7984) / (1 -0.01) = 63.84 ft/s

(2) t = 0.001 s

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The question is incomplete. Find out the complete question below:

If a ball is thrown straight up into the air with an initial velocity of 80 ft/s, it height in feet after t second is given by  y=80t-16t^2 .Find the average velocity for the time period beginning when t=1 and lasting

(i) 0.01 seconds

(ii) 0.001 seconds

Learn more about average velocity at brainly.com/question/6504879

#SPJ4

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m = \frac{(0.4)^2X (0.305)^2 X1.602X 10^-^1^9}{2X 2000}\\\\

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