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Kaylis [27]
3 years ago
14

A yacht is pushed along at a constant velocity by the wind. The wind force is 180N at an angle of 25 degrees to the direction of

the of the motion of the yacht. By resolving the wind force parallel and perpendicular to the direction of motion, determine the component of the wind force in the direction of motion.

Physics
1 answer:
damaskus [11]3 years ago
4 0
The parallel component is given by
F=180cos(25)=163.14N
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The fixed hydraulic cylinder C imparts a constant upward velocity v = 2.2 m/s to the collar B, which slips freely on rod OA. Det
Olenka [21]

Answer:

so angular velocity is 7.13128 sec−1

Explanation:

velocity v = 2.2 m/s

displacement s = 220 mm = 0.220 m

distance d = 510 mm = 0.510 m

to find out

angular velocity

solution

we know that

angular velocity will be velocity ( v)  / (displacement²  +  distance²)   .....1

now put all these value in equation 1 and we get angular velocity i.e.

angular velocity =  velocity ( v)  / (displacement²  +  distance²)

angular velocity = 2.2  / (0.22²  +  0.51²)

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6 0
4 years ago
A man applies a force of 100 N to a rock for 60 seconds, but the rock does not move what is the amount of work done by the man o
Art [367]
600. I forgot the measurement. but 600 is correct
6 0
3 years ago
How was newton's laws used to solve problems on apollo 13 ship<br> Explain in two full paragraphs
MaRussiya [10]

Answer:

actually ships are made in newtons third law of motion.it states to every action there is equal and opposite reaction. curved is made in downwards to maintain upthrust and to made balance.

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3 0
3 years ago
A tank holds a 1.44-m thick layer of oil that floats on a 0.98-m thick layer of brine. Both liquids are clear and do not intermi
denpristay [2]

Answer:

Angle of ray makes with the vertical is 62.1 degree

Explanation:

As per the ray diagram we know that the angle of incidence on oil brine interface will be given as

tan\theta_i = \frac{0.7}{0.98}

\theta_i = 35.5^0

now by Snell'a law at that interface we have

\mu_1 sin\theta_i = \mu_2 sin \theta_r

now we will have

1.52  sin35.5 = 1.40 sin\theta_r

\theta_r = 39.12^0

now this is the angle of incidence for oil air interface

so now again by Snell's law we will have

\mu_2 sin\theta_i' = \mu_{air} sin\theta

1.40 sin39.12 = 1 sin\theta

\theta = 62.1^0

3 0
3 years ago
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