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chubhunter [2.5K]

3 years ago

6

A 1-kg rock is suspended from the tip of a horizontal meterstick at the 0-cm mark so that the meterstick barely balances like a

seesaw when its fulcrum is at the 12.5-cm mark. From this information, the mass of the meterstick is _________.
A) 1/4 kg.B) 1/2 kg.C) 3/4 kg.D) 1 kg.E) none of the above

1 answer:

tigry1 [53]3 years ago

5 0

Explanation:

Given that,

Mass if the rock, m = 1 kg

It is suspended from the tip of a horizontal meter stick at the 0-cm mark so that the meter stick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark.

We need to find the mass of the meter stick. The force acting by the stone is

F = 1 × 9.8 = 9.8 N

Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.

$9.8\times 12.5=W\times (50-12.5)\\\\W=\dfrac{9.8\times 12.5}{37.5}$

W = 3.266 N

The mass of the meters stick is :

$m=\dfrac{W}{g}\\\\m=\dfrac{3.266}{9.81}\\\\m=0.333\ kg$

So, the mass of the meter stick is 0.333 kg.

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A flexible loop has a radius of 10.5 cm and it is in a magnetic field of B = 0.117 T. The loop is grasped at points A and B and

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Explanation:

Given that,

Radius = 10.5 cm

Magnetic field = 0.117 T

Time = 0.243 s

After stretched, area is zero

(I). We need to calculate the magnetic flux through the loop before stretched

Using formula of magnetic flux

$\phi=B\times A$

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Here, A = 0

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So, The magnetic flux through the loop after stretched is zero.

(III). We need to calculate the magnitude of the average induced electromotive force

Using formula of the induced electromotive force

$\epsilon=-\dfrac{d\phi}{dt}$

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3 years ago

An 800 V/m electric field is directed along the +x-axis. If the potential at x = 0 m is 2000 V, what is the potential at x = 2 m

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3 years ago