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3 years ago

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How efficient is a pulley system if it enables you to lift a 600.0 Newton engine 0.600 meters if you exerted 35.7 Newtons of for

ce while pulling 11.43 meters of rope?

1 answer:

muminat3 years ago

4 0

Answer:

η = 0.882 = 88.2 %

Explanation:

The efficiency of the pulley system can be given as follows:

$\eta = \frac{W_{out}}{W_{In}}\\\\$

where,

η = efficiency of pulley system = ?

W_out = Output Work = (600 N)(0.6 m) = 360 J

W_in = Input Work = (35.7 N)(11.43 m) = 408.051 J

Therefore,

$\eta = \frac{360\ J}{408.051\ J}$

<u>η = 0.882 = 88.2 %</u>

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Answer

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Which of the following intermolecular forces explains why iodine (I2) is a solid at room temperature?

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3 years ago

Use the information from the graph to answer the question.

galina1969 [7]

The displacement of the object as determined from the velocity-time graph is 562.5 m.

<h3>What is a velocity-time graph?</h3>

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1 year ago

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3 years ago

A particle with a charge of +4.20nC is in a uniform electric field E⃗ directed to the left. It is released from rest and moves t

Morgarella [4.7K]

Answer:

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Explanation:

Given that,

Charge = 4.20 nC

Distance = 6.00 cm

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The particle start from rest.

So, the initial kinetic energy i zero.

(A). We need to calculate the work by the electric force

Using formula of work done

$W = \Delta K.E$

$W=K.E_{f}-K.E_{i}$

Put the value into the formula

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$W=1.50\times10^{-6}\ J$

The work done is $1.50\times10^{-6}\ J$ .

(B). We need to calculate the potential of the starting point with respect to the endpoint

We know that.

Change in potential energy = change in kinetic energy

$\Delta P.E=\Delta K.E$

So, $U = 1.50\times10^{-6}$

Using formula of potential

$V=\dfrac{U}{q}$

Put the value into the formula

$V=\dfrac{1.50\times10^{-6}}{4.20\times10^{-9}}$

$V=357.14\ V$

The potential of the starting point with respect to the endpoint is 357.14 V.

(C). We need to calculate the magnitude of E

Using formula of work done

$W=F\times r$ ....(I)

Using formula of force

$F=qE$

Put the value in the equation (I)

$W=qE\times r$

$E=\dfrac{W}{q\times r}$

Put the value into the formula

$E=\dfrac{1.50\times10^{-6}}{4.20\times10^{-9}\times6.00\times10^{-2}}$

$E=5952.38\ N/C$

The magnitude of E is 5952.38 N/C.

Hence, This is the required solution.

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3 years ago