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shutvik [7]
2 years ago
8

The force shown in the attached figure is the net eastward force acting on a ball. The force starts rising at t = 0.012 s, falls

back to zero at t = 0.062 s, and reaches a maximum force of 35 N at the peak. Determine with an error no bigger than 25% (high or low) the magnitude of the impulse (in N-s) delivered to the ball. Hint: Do not use J = FΔt. Look at the figure. Find the area of a nearly equally sized triangle.

Physics
1 answer:
kifflom [539]2 years ago
6 0
Impulse = Integral of F(t) dt from 0.012s to 0.062 s

Given that you do not know the function F(t) you have to make an approximation.

The integral is the area under the curve.

The problem suggest you to approximate the area to a triangle.

In this triangle the base is the time: 0.062 s - 0.012 s = 0.050 s

The height is the peak force: 35 N.

Then, the area is [1/2] (0.05s) (35N) = 0.875 N*s

Answer> 0.875 N*s
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Answer:

1.056 x 10⁷ lb-ft

Explanation:

v = Speed of the bike = 20 mph

t = time of travel = 2 h

d = distance traveled by cyclist

Distance traveled by cyclist is given as

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d = (20) (2)

d = 40 miles

We know that, 1 mile = 5280 ft

d = 40 (5280) ft

d = 211200 ft

F = force applied by cyclist = 50 lb

W = work done by cyclist

Work done by cyclist is given as

W = F d

W = (50) (211200)

W = 1.056 x 10⁷ lb-ft

5 0
2 years ago
On a snowy day, when the coefficient of friction μs between a car’s tires and the road is 0.50, the maximum speed that the car c
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Answer:

28.1 mph

Explanation:

The force of friction acting on the car provides the centripetal force that keeps the car in circular motion around the curve, so we can write:

F=\mu mg = m \frac{v^2}{r} (1)

where

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

v is the maximum speed of the car

r is the radius of the trajectory

On the snowy day,

\mu=0.50\\v = 20 mph = 8.9 m/s

So the radius of the curve is

r=\frac{v^2}{\mu g}=\frac{(8.9)^2}{(0.50)(9.8)}=16.1 m

Now we can use this value and re-arrange again the eq. (1) to find the maximum speed of the car on a sunny day, when \mu=1.0. We find:

v=\sqrt{\mu g r}=\sqrt{(1.0)(9.8)(8.9)}=12.6 m/s=28.1 mph

7 0
2 years ago
Read 2 more answers
10. A girl pulls a wagon along a level path for a distance of 44 m. The handle of
Rina8888 [55]

The work done on the wagon is 3549 J

Explanation:

The work done by a force when moving an object is given by

W=Fd cos \theta

where :

F is the magnitude of the force

d is the displacement

\theta is the angle between the direction of the force and of the displacement

In this problem we have the following data:

F = 87 N is the magnitude of the force

d = 44 m is the displacement of the wagon

\theta=22^{\circ} is the angle between the direction of the force and the displacement

Substituting, we find the work done

W=(87)(44)(cos 22^{\circ})=3549 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

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2 years ago
Derive an algebraic equation for the vertical force that the bench exerts on the book at the lowest point of the circular path i
fiasKO [112]

Answer:

The algebraic equation is:

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Explanation:

Given information:

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vb = tangential speed

R = radius of the path

Question: Derive an algebraic equation for the vertical force, Fv = ?

To derive the equation, we need to draw a force diagram for this case, please, see the attached diagram. As you can see, there are three types of forces acting on the system. Two up and one of the weight acting down. Therefore, the algebraic equation is as follows:

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2 years ago
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Answer:

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Based on the aforementioned, let's analyze the statements in order of importance

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c) false. The amount of vaporized material on the moon is large

Therefore, the order of importance must be

         b e a f c

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