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sleet_krkn [62]
3 years ago
7

A lever has an effort arm that is 8 meters long and the resistance arm that is 1.5 meters long, how much effort is needed to lif

t 209 newton weight?

Physics
2 answers:
sesenic [268]3 years ago
8 0

Answer: The effort force will be 8190.2 N.

Explanation:

Given that,

Effort arm = 8 m

Resistance arm = 1.5 m

Weight = 209 N

We know that,

Effort force multiply by effort arm=Weight multiply by resistance arm

F_{e}\times 8=209\times 1.5

F_{e}=8190.2 N

Hence, The effort force will be 8190.2 N.

ruslelena [56]3 years ago
6 0
We are given with two measurements of the arm and an input weight. To answer this problem,we need to balance the forces and use the lengths of the arms. 
209 N * 8 m = x * 1.5 m
x = 1114.67 N

it takes 1114.67 N to lift the input weight
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slamgirl [31]

Answer:

2.86×10⁻¹⁸ seconds

Explanation:

Applying,

P = VI................ Equation 1

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I = P/V................ Equation 2

From the question,

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I = 0.414/1.50

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Also,

Q = It............... Equation 3

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make t the subject of the equation

t = Q/I.................. Equation 4

From the question,

4.931020 electrons has a charge of (4.931020×1.6020×10⁻¹⁹) coulombs

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2 years ago
A 0.750 kg block is attached to a spring with spring constant 13.0 N/m . While the block is sitting at rest, a student hits it w
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To solve this problem we will apply the concepts related to energy conservation. From this conservation we will find the magnitude of the amplitude. Later for the second part, we will need to find the period, from which it will be possible to obtain the speed of the body.

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KE = PE

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Here,

m = Mass

v = Velocity

k = Spring constant

A = Amplitude

Rearranging to find the Amplitude we have,

A = \sqrt{\frac{mv^2}{k}}

Replacing,

A = \sqrt{\frac{(0.750)(31*10^{-2})^2}{13}}

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Replacing,

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We have all the values, then replacing,

v = \frac{2\pi}{1.509}\sqrt{(0.0744)^2-(0.750(0.0744))^2}

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