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Kay [80]
3 years ago
14

A 10 cm diameter pulley is used to lift a bucket of cement weighing 400 N. How much force must be applied to the rope to lift th

e bucket at the other end?​
Physics
1 answer:
Alina [70]3 years ago
3 0

hi brainly user! ૮₍ ˃ ⤙ ˂ ₎ა

⊱┈────────────────────────┈⊰

\large \bold {ANSWER}

Considering that the pulley is fixed, the force applied should be equal to the weight of the object - of 400N.

\large \bold {EXPLANATION}

Pulleys or pulleys are mechanical tools used to assist in the movement of objects and bodies. There are two types of pulleys: fixed and movable. While the fixed pulley changes the direction of force, the moving pulley helps to decrease the force needed to move the object or body in question.

As the statement only tells us a pulley, we must consider that it is fixed, <u>because generally when it is mobile, this information is highlighted in the question</u>.

In this way, a fixed pulley only changes the direction of the applied force. Thus, the force must have the same magnitude as the weight of the object to be moved. If the bucket weighs 400N, the force applied to the pulley must be 400N.

<u>Therefore, having a fixed pulley, the force applied must be equal to the weight of the object, and will be 400N.</u>

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eimsori [14]

Answer:

Its position after 4 seconds is 62 meters.

Explanation:

It is given that,

The acceleration of the particle is given by equation :

a(t)=5+4t-2t^2

Also, a=\dfrac{dv}{dt}

v=\int\limits {a.dt}

v=\int\limits {(5+4t-2t^2).dt}

v=5t+2t^2-\dfrac{2}{3}t^3+c

At t = 0, v(0)=3\ m/s. So, c = 3

v=5t+2t^2-\dfrac{2}{3}t^3+3

Also, v=\dfrac{ds}{dt}, s is the position

s=\int\limits {v.dt}

s=\int\limits {(5t+2t^2-\dfrac{2}{3}t^3+3).dt}

s=\dfrac{5}{2}t^2+\dfrac{2}{3}t^3-\dfrac{t^4}{6}+3t+c'

At t = 0, s(0)=10\ m. So, c' = 10

s=\dfrac{5}{2}t^2+\dfrac{2}{3}t^3-\dfrac{t^4}{6}+3t+10

At t = 4 s

s=\dfrac{5}{2}(4)^2+\dfrac{2}{3}(4)^3-\dfrac{(4)^4}{6}+3(4)+10

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5 0
3 years ago
Water is circulating through a closed system of pipes in a two floor apartment. On the first floor, the water has a gauge pressu
anastassius [24]

Answer:

The value of gauge pressure at outlet = -38557.224 pascal

Explanation:

Apply Bernoulli' s Equation

\frac{P_{1}}{9810} + \frac{V_{1} ^{2}}{19.62} + h_{1} = \frac{P_{2}}{9810} + \frac{V_{2} ^{2}}{19.62} + h_{2} --------------(1)

Where

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V_{1} = velocity at inlet =  2.4 \frac{m}{sec}

P_{2} = Gauge pressure at outlet = we have to calculate

V_{2} = velocity at outlet = 3.5 \frac{m}{sec}

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Put all the values in equation (1) we get,

⇒ \frac{3.70105}{9810} + \frac{2.4 ^{2}}{19.62} = \frac{P_{2}}{9810} + \frac{3.5 ^{2}}{19.62} + 3.6

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