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2 years ago

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Two point charges are separated by 6.4 cm . The attractive force between them is 10 N . Suppose that the charges attracting each

other have equal magnitude. Rearrange Coulomb's law and find the magnitude of each charge. Express your answer to two significant figures and include the appropriate units.

1 answer:

LenaWriter [7]2 years ago

7 0

Answer:

Two point charges are separated by 6.4 cm . The attractive force between them is 10 N .

units.

Explanation:

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a)

Since the jack is 100% efficient, all the input work is converted into output work. So we can write:

$W_{in} = W_{out}\\F_{in} d_{in} = F_{out} d_{out}$

where:

$F_{in}$ is the input force applied on the jack

$d_{in}$ is the arm of the input force

$F_{out}$ is the output force applied on the jack

$d_{out}$ is the arm of the output force

Here we have:

$F_{out}=\frac{mg}{2}= \frac{(1200 kg)(9.8 m/s^2)}{2}=5880 N$ is the output force (half the weight of the car)

$d_{in} = 40.0 cm = 0.40 m$ is the arm of the input force

$d_{out} = 5.0 mm = 0.005 m$ is the arm of the output force

Solving for $F_{in}$ , we find the force that must be applied in input to lift the car:

$F_{in} = \frac{F_{out}d_{out}}{d_{in}}=\frac{(5880)(0.005)}{0.40}=73.5 N$

b)

The efficiency of the jack is given by the ratio between the output work and the input work:

$\eta = \frac{W_{out}}{W_{in}}=\frac{F_{out}d_{out}}{F_{in}d_{in}}$

where we have:

$F_{out}=5880 N$ is the output force (half the weight of the car)

$d_{in} = 40.0 cm = 0.40 m$ is the arm of the input force

$d_{out} = 5.0 mm = 0.005 m$ is the arm of the output force

Here we are told that the input force this time is

$F_{in}=90.0N$

Substituting into the equation, we find the new efficiency of the jack:

$\eta = \frac{(5880)(0.005)}{(90.0)(0.40)}=0.82$

Which means an efficiency of 82%.

Learn more about levers:

brainly.com/question/5352966

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