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Ludmilka [50]
3 years ago
15

A person is able to pull a rope with a force of 700 N. What is the minimum number of fixed and moveable pulleys that the person

will require to lift up a 27,800 N elephant? Assume the pulleys themselves are weightless and the system has 100% efficiency.
A. 10 fixed, 10 moveable
B. 15 fixed, 15 moveable
C. 20 fixed, 20 moveable
D. 30 fixed, 30 moveable
Physics
2 answers:
Sav [38]3 years ago
8 0

<u>Answer: </u>

C option is  correct

20 movable and 20 fixed pulleys required

<u>Explanation: </u>

To lift a 27,800 N elephant with a 700 N drive,  

27800/700=39.7  

the individual requires a mechanical favorable position of 39.7 which is equivalent to 40. A solitary movable pulley gives a mechanical favorable position of 2, two portable pulleys yield a mechanical preferred standpoint of 4, and a framework with three movable pulleys results in a mechanical favorable position of 6. In light of this example, 20 movable pulleys and 20 fixed are required to give a mechanical preferred standpoint of 40

Katen [24]3 years ago
4 0

Answer:

C. 20fixed, 20moveables

Explanation:

Efficiency of a machine = MA/VR × 100%

MA is the mechanical advantage

VR is the velocity ratio of the pulley

MA = Load/Effort

MA = 27,800N/700N

MA = 39.71

The velocity ratio formula depends on the type of machine. Since the machine in question is a block and tackle system, the velocity ratio will be the number of pulleys in the system.

Using the efficiency formula to get the VR

100 = 39.71/VR × 100

39.71/VR = 1

VR = 39.71

VR = 40(approximately)

This means that the minimum number of pulleys that the person will require to lift the load is 40pulleys i.e 20fixed, 20moveables

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II Force on a tennis ball. The record speed for a tennis ball that is served is 73.14 m/s. During a serve, the ball typically st
AveGali [126]

Answer:

F=248.5W N

Explanation:

Newton's 2nd Law tells us that F=ma. We will use their averages always. The average acceleration the tennis ball experimented is, by definition:

a=\frac{\Delta x}{\Delta t}=\frac{v-v_0}{t-t_0}

Since we start counting at 0s and the ball departs from rest, this is just a=\frac{v}{t}

So we can write:

F=ma=\frac{mv}{t}=\frac{gmv}{gt}

Where in the last step we have just multiplied and divided by g, the acceleration of gravity. This allows us to introduce the weight of the ball W since W=gm, so we have:

F=\frac{Wv}{gt}=\frac{v}{gt}W

Substituting our values:

F=\frac{(73.14m/s)}{(9.81m/s^2)(30\times10^{-3}s)}W=248.5W N

Where the average force exerted has been written it terms of the tennis ball's weight W.

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3 years ago
A 0.50-kg mass attached to the end of a string swings in a vertical circle (radius 2.0 m). When the mass is at the highest point
il63 [147K]

Answer:

31.1 N

Explanation:

m = mass attached to string = 0.50 kg

r = radius of the vertical circle = 2.0 m

v = speed of the mass at the highest point = 12 m/s

T = force of the string on the mass attached.

At the highest point, force equation is given as

T + mg =\frac{mv^{2}}{r}

Inserting the values

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3 years ago
A vector → A has a magnitude of 56.0 m and points in a direction 30.0° below the negative x axis. A second vector, → B , has a m
MissTica

Answer:

  • The magnitude of the vector \vec{C} is 107.76 m

Explanation:

To find the components of the vectors we can use:

\vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )

where | \vec{A} | is the magnitude of the vector, and θ is the angle over the positive x axis.

The negative x axis is displaced 180 ° over the positive x axis, so, we can take:

\vec{A} = 56.0 \ m \ ( \ cos( 180 \° + 30 \°) \ , \ sin (180 \° + 30 \°) \ )

\vec{A} = 56.0 \ m \ ( \ cos( 210 \°) \ , \ sin (210 \°) \ )

\vec{A} = ( \ -48.497 \ m \ , \ - 28 \ m \ )

\vec{B} = 82.0 \ m \ ( \ cos( 180 \° - 49 \°) \ , \ sin (180 \° - 49 \°) \ )

\vec{B} = 82.0 \ m \ ( \ cos( 131 \°) \ , \ sin (131 \°) \ )

\vec{B} = ( \ -53.797 \ m \ , \ 61.886\ m \ )

Now, we can perform vector addition. Taking two vectors, the vector addition is performed:

(a_x,a_y) + (b_x,b_y) = (a_x+b_x,a_y+b_y)

So, for our vectors:

\vec{C} = ( \ -48.497 \ m \ , \ - 28 \ m \ ) + ( \ -53.797 \ m \ ,  ) = ( \ -48.497 \ m \ -53.797 \ m , \ - 28 \ m \ + \ 61.886\ m \ )

\vec{C} = ( \ - 102.294 \ m , \ 33.886 m \ )

To find the magnitude of this vector, we can use the Pythagorean Theorem

|\vec{C}| = \sqrt{C_x^2 + C_y^2}

|\vec{C}| = \sqrt{(- 102.294 \ m)^2 + (\ 33.886 m \)^2}

|\vec{C}| =107.76 m

And this is the magnitude we are looking for.

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