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Kay [80]
3 years ago
15

An upright clothes washer spins clothes around 30 times in 12 seconds. Its radius is 0.20m. What is the tangential velocity of t

he clothes in the washer?

Physics
2 answers:
ankoles [38]3 years ago
4 0

Explanation:

Below is an attachment containing the solution.

ziro4ka [17]3 years ago
4 0

Answer:

<em>3.142 m/sec</em>

Explanation:

Tangential velocity is the velocity measured along the path of a rotating body, which is tangent forms a tangent with the path.

The tangential velocity of the washer can be obtained using equation 1:

V = ω x r

where is the tangential velocity

ω is the angular velocity

r is the radius = 0.20 m.

Calculating for the angular velocity

ω = \frac{Number of revolutions}{time}

given the number of revolution as 30 and time 12 secs our angular velocity can be obtained as;

ω = \frac{30}{12 secs}

ω  = 2.5 rev/secs

angular velocity is measured in rad/sec so we have to convert our answer;

1 rev/sec = 2π rad/rev

2.5 rev/secs = 2π rad/rev x 2.5 re/secs = 15.71 rad/sec

the answer will be substituted into the expression in equation 1 to obtain the tangential velocity;

V = 15.71 rad/sec x 0.20 m

V = 3.142 m/sec

Therefore the tangential velocity of the washer is <em>3.142 m/sec</em>

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1) A boy drags a wooden crate with a mass of 20 kg, a distance of 12 m, across a rough level floor at a constant speed of 1.5 m/
mojhsa [17]

Answer: a) 49.560 and 21.13 b) i) 50 N, ii) 196 N iii) 196 N iv) 47.685 N

c) i) 594.72 ii) 0 iii) 0 iv) 0

d) 594.72

Explanation: question a)

The force is inclined at an angle of 25° to the horizontal

The horizontal component of force = 50 cos 25° = 49.560 N

The vertical component of force = 50 sin 30°= 21.130N

Question b)

i) according to the question applied force is 50 N

ii) if g = 9.8m/s², w=mg where m = mass of object = 20kg hence weight = 20* 9.8 = 196 N

iii) the normal force is the force the floor exerts on the body as a result of the weight of the object.

Normal reaction R = W = mg, we already deduced that w = mg, hence R = 196 N.

iv) according to newton's laws of motion

F - Fr = ma

F = applied force = horizontal component of force = 49.560 N.

We need to get the acceleration (a) by using Newton laws of motion before we can be able to compute the frictional force..

The body started from rest hence initial velocity u = 0

Final velocity v = 1.5m/s distance covered (s) = 12m

v ² = u² + 2as

But u = 0

v² = 2as

1.5² = 2(a) * 12

2.25 = 24a

a = 2.25/24 = 0.09735m/s²

From F - Fr = ma

49.560 - Fr = 20 * 0.09735

49.560 - Fr = 1.875

Fr = 49.560 - 1.875

Fr = 47.685 N

Question c)

i) The applied force = 49.560 N, distance covered = 12m

Work done = force * distance

Work done = 49.560 * 12

Work done = 594.72 J

ii) the weight of the object does not make the object move a distance, hence work done = 0 ( since distance covered is 0)

iii) the normal force is the same thing as the weight and they did not cover any distance hence work done is zero.

iv) the frictional force does not cover any distance, hence work done is zero.

Question d)

The total work done = work done by applied force + work done by weight + work done by normal reaction + work done by frictional force.

Total work done = 594.72 + 0 + 0 + 0 = 594.72 J

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a linear function has the same y-intercept as x + 4y equals 16 and it's graph contains the point (4,5). Find the slope of the li
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Answer:  \bold{\text{Slope (m)}=\dfrac{1}{4}}

<u>Explanation:</u>

A linear equation is of the form: y = mx + b   where

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