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serg [7]
3 years ago
14

A positive point charge q is placed at the center of an uncharged metal sphere insulated from the ground. The outside of the sph

ere is then grounded as shown. Then the ground wire is removed. A is the inner surface and B is the outer surface. Which statement is correct

Physics
1 answer:
kenny6666 [7]3 years ago
4 0

Explanation:

the missing figure in the Question has been put in the attachment.

Then from the figure we can observe that

the center of the sphere is positive, therefore, negative charge will be  induced at A.

As B is grounded there will not be any charge on B

Hence the answer is A is negative and B is charge less.

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V = 387 V

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describe at least two ways in which parents are able to influence the drinking habits of their children
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A bakery measured the mass of a whole cake as 0.870 kg. A customer bought one slice of the cake and measured the mass of the sli
Brrunno [24]

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0.7549kg

Explanation:

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4 years ago
What's the difference between Uniform magnetic field and Non-uniform magnetic field?
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4 0
3 years ago
A physical pendulum consists of a meter stick that is pivoted at a small hole drilled through the stick a distance d from the 50
PSYCHO15rus [73]

Answer:

(a). The value of d is 0.056 cm and 1.496 cm.

(b). The time period is 1.35 sec.

Explanation:

Given that,

Length = 50.00 cm

Time period = 2.50 s

Time period of pendulum is defined as the time for one complete cycle.

The period depends on the length of the pendulum.

Using formula of time period

T=2\pi\sqrt{\dfrac{I}{mgh}}

Where, I = moment of inertia

We need to calculate the value of d

Using parallel theorem of moment of inertia

I=I_{cm}+md^2

For a meter stick mass m , the rotational inertia about it's center of mass

I_{cm}-\dfrac{mL^2}{12}

Where, L = 1 m

Put the value into the formula of time period

T=2\pi\sqrt{\dfrac{\dfrac{mL^2}{12}+md^2}{mgd}}

T=2\pi\sqrt{\dfrac{L^2}{12gd}+\dfrac{d}{g}}

T^2=4\pi^2(\dfrac{L^2}{12gd}+\dfrac{d}{g})

Multiplying both sides by d

tex]T^2d=4\pi^2(\dfrac{L^2}{12g}+\dfrac{d^2}{g})[/tex]

(\dfrac{4\pi^2}{g})d^2-T^2d+\dfrac{\pi^2L^2}{3g}=0

Put the value of T, L and g into the formula

4.028d^2-6.25d+0.336=0

d = 0.056\ m, 1.496\ m

The value of d is 0.056 cm and 1.496 cm.

(b). Given that,

L = 50-5 = 45 cm

We need to calculate the time period

Using formula of period

T=2\pi\sqrt{\dfrac{l}{g}}

Put the value into the formula

T=2\pi\sqrt{\dfrac{45\times10^{-2}}{9.8}}

T=1.35\ sec

Hence, (a). The value of d is 0.056 cm and 1.496 cm.

(b). The time period is 1.35 sec.

7 0
3 years ago
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