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1 year ago

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two charges experience exert a force of 1n on each other when they are 1m apart. what force will these charges experience if the

y are placed 2m apart?

1 answer:

Kisachek [45]1 year ago

8 0

The force experienced by the charges when they are placed 2 m apart will be 0.25 N.

The force between two charges is given by Coulomb’s Law.

According to this law, the force between two charges varies inversely as the square of the distance between the charges and is directly proportional to the product of the magnitude of the two charges.

The above statement is represented by the following equation,

F= K Q1 x Q2/r^2

Here, F = Force between the two charges

K = Constant

Q1 and Q2 = Magnitude of the two charges

r = Separation between the two charges

According to this equation, the force is inversely proportional to the square of the distance between two charges.

Initially, when the two charges are 1 m apart, the force is 1 N.

When the distance between the two charges is doubled,

then according to Coulomb’s Law, the force should decrease by 4 times.

Hence, if the charges are placed 2 meters apart, the force becomes 1/4= 0.25 N.

To know more about "Coulomb's Law", refer to the following link:

brainly.com/question/9261306?referrer=searchResults

#SPJ4

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Answer:

Part 1)

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Part 3)

$\tau_3 = 1.4 N m$

Part 4)

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Explanation:

As we know that the block A is placed at distance

d = 50 cm from the hinge at 70 cm mark

So torque due to weight of A is given as

$\tau_1 = 5 \times (0.50) = 2.5 N m$

the block B is placed at distance

d = 30 cm from the hinge at 70 cm mark

So torque due to weight of B is given as

$\tau_2 = 14 \times (0.30) = 4.2 N m$

Now torque due to weight of the scale is given as

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$\tau_3 = 1.4 N m$

now torque on left side of scale is given as

$\tau_{left} = \tau_1 + \tau_3$

$\tau_{left} = 2.5 + 1.4 = 3.9 N m$

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Since torque on right side is more so here it will turn and slip over it

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3 years ago

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2 years ago

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Answer:

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$\rho_A\times g \times h_1 + \rho_G \times g \times h_2 = \rho_G \times g \times h_3$

Simplifying:

$\rho_A\times h_1 = \rho_G \times (h_3 - h_2) (1)$

On the other hand:

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Rearranging:

$h_1 - \Delta h = h_3 - h_2 (2)$

Replacing (2) in (1):

$\rho_A\times h_1 = \rho_G \times (h_1 - \Delta h)$

Rearranging:

$\frac{h_1 \times (\rho_A - \rho_G)}{- \rho_G} = \Delta h$

Data:

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3 years ago

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Explanation:

The given data is as follows.

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Angular velocity of automobile = w = $\frac{V}{R}$

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In the given case, angular velocity remains same for inner and outer wheel but there is change in linear velocity of inner wheel and outer wheel.

Now, we assume that

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and, u' = linear velocity of outer wheel.

Formula for angular velocity of inner wheel w = ,

Formula for angular velocity of outer wheel w =

Now, for inner wheels

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u = $V \times \frac{(R - d)}{R}$

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Since, velocity of wheel is u it will cover distance u in unit time(min)

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Now, rotation per minute of inner wheel is calculated as follows.

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3 years ago

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Answer:

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