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dlinn [17]
3 years ago
11

Ocean waves are traveling to the east at 3.2 m/s with a distance of 19 m between crests. (a) With what frequency do the waves hi

t the front of a boat when the boat is at anchor? Hz (b) With what frequency do the waves hit the front of a boat when the boat is moving westward at 1.5 m/s?
Physics
1 answer:
rjkz [21]3 years ago
8 0

Answer:

(a) Frequency will be 0.168 Hz

(b) Frequency will be 0.0789 Hz

Explanation:

(a) We have given velocity of the ocean waves in east direction v = 3.2 m /sec

Distance between the crest \Delta x=\lambda =19m

We know that v=\lambda f

So 3.2=19\times  f

f=0.168Hz

(b)Velocity of the ocean waves in westward direction v = 1.5 m /sec

Distance between the crest \Delta x=\lambda =19m

We know that v=\lambda f

So 1.5=19\times  f

f=0.0789Hz

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A horizontal force of 750 N is needed to overcome the force of static friction between a level floor and a 250-kg crate. What is
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Answer:

The acceleration of the crate is 1.8 m/s² so the answer is a.

Explanation:

The very first thing you must do when solving this problem is to draw a free body diagram. (The body diagram is attached to this answer)

So once we got the free body diagram, we can analyze it and build our sum of forces in the x and y directions. Notice that according to the diagram, there are 4 forces to this problem, Normal (N), Weight (W), kinetic friction (fk) and the 750N force.

As one may see in the free body diagram, two of the forces are vertical forces: N and W, so we can use them to build a sum of forces:

Starting with the sum of forces in the y-direction, we get:

ΣF_{y}=0

We set the sum equal to zero because there is no movement in the y-direction, so the system is in vertical equilibrium.

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N-W=0

when solving for N we get that:

N=W

where W is found by multiplying the mass of the crate by the acceleration of gravity:

N=250kg*9.8m/s²

N=2450N

Once we found the normal force, we can use it to find the kinetic friction which is given by the following formula:

f_{k}=Nμ

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So we get that the kinetic friction is:

f_{k}=2450N*0.12

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f_{k}=294

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ΣF_{x}=ma

In this case the sum is equal to mass times acceleration because the crate is moving horizontally due to the action of a force, so it will have an acceleration.

so the sum of forces look like this:

750N-f_{k}=ma

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750N-294N=(250kg)a

when solving for a we get:

a=\frac{759N-294N}{250kg}\\ \\a=1.8m/s^{2}

so the crate's acceleration is 1.82m/s².

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