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IgorC [24]
3 years ago
5

To hoist himself into a tree, a 72.0-kg man ties one end of a nylon rope around his waist and throws the other end over a branch

of the tree. he then pulls downward on the free end of the rope with a force of 371 n. neglect any friction between the rope and the branch, and determine the manâs upward acceleration. use g =9.80 m/sec2.
Physics
1 answer:
taurus [48]3 years ago
4 0
<span>0.506 m/s^2 Given the arrangement of rope and the man, he's effectively suspended by a rope from a frictionless pulley. So he's pulling the rope with a force of 371 n causing him to be pulled upwards by his arms. Additionally, the rope goes up to the pulley and back down to the man causing an additional 371 n of force pulling him upwards. So the total force being used to lift the man is 742 n. Additionally, the man is being pulled downwards by gravity at 9.80 m/s^2. And since he masses 72.0 kg, the downward force in newtons is 72.0 kg * 9.80 m/s^2 = 705.6 n downward. So the total force being exerted on the man is 742 n - 705.6 n = 36.4 n To calculate his acceleration, simply divide the number of newtons applied by his mass. 36.4 kg m/s^2 / 72.0 kg = 0.505556 m/s^2 And round to 3 significant figures. 0.505556 m/s^2 = 0.506 m/s^2</span>
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sammy [17]

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Ohm's Law states that the voltage across an electric conductor is directly proportional to the current(I) passing through it provided the resistant is constant.

So;

V ∝ I

V = IR  

where

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The objective of this question want us to determine: How did the current change for each test provided that Avery uses a 1.5-volt battery, then she uses a 3-volt battery and lastly she uses a 9-volt battery, given that the resistance is constant through out the whole process.

In the first experiment;

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In the second experiment;

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In the third experiment;

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Therefore, we can conclude that the current in each experiment increases with increase in the voltage. Similarly, the association between resistance and the current in a circuit shows that increase in the resistance shows a reduction in the current, vice versa.

Learn more about Ohm's Law here:

brainly.com/question/14296509

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Pepsi [2]

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3 years ago
What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates
34kurt

The given question is incomplete. The complete question is as follows.

A parallel-plate capacitor has capacitance C_{0} = 8.50 pF when there is air between the plates. The separation between the plates is 1.00 mm.

What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed 3.00 \times 10^{4} V/m?

Explanation:

It is known that relation between electric field and the voltage is as follows.

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Now,  

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or,           Q = C \times Ed

Therefore, substitute the values into the above formula as follows.

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Hence, we can conclude that the maximum magnitude of charge that can be placed on each given plate is 2.55 \times 10^{-10} C.

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