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RideAnS [48]
3 years ago
5

A rock slides down from A to B along the inside surface of a frictionless hemispherical bowl. As the rock slides, mechanical ene

rgy is conserved (ignoring air resistance) because: (Mark all that apply.) Group of answer choices the bowl is hemispherical in shape. the normal force on the rock is balanced by the centrifugal force. the normal force on the rock is balanced by the centripetal force. the normal force on the rock acts perpendicular to the bowl's surface. the rock's acceleration is perpendicular to the bowl's surface. NONE of the other choices.
Physics
1 answer:
DIA [1.3K]3 years ago
6 0

Answer:

the normal force on the rock acts perpendicular to the bowl's surface.

Explanation:

As we know that Normal force is the reaction force of two contact surfaces which always act perpendicular to the contact surfaces

Here we know that the rock is moving inside the bowl

So Normal force on the rock must perpendicular to the surface of the bowl which always passes through the center of the bowl.

Since the rock is moving in vertical plane so it must have two acceleration

1) Tangential acceleration which will increase the magnitude of the speed along the tangential path

2) Centripetal acceleration which will change the direction of the rock

So here only correct option will be

the normal force on the rock acts perpendicular to the bowl's surface.

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Answer:

The electric field at origin is 3600 N/C

Solution:

As per the question:

Charge density of rod 1, \lambda = 1\ nC = 1\times 10^{- 9}\ C

Charge density of rod 2, \lambda = - 1\ nC = - 1\times 10^{- 9}\ C

Now,

To calculate the electric field at origin:

We know that the electric field due to a long rod is given by:

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\vec{E} = \frac{2K\lambda }{R}                  (1)

where

K = electrostatic constant = \frac{1}{4\pi \epsilon_{o} R}

R = Distance

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In case, the charge is positive, the electric field is away from the rod and towards it if the charge is negative.

At x = - 1 cm = - 0.01 m:

Using eqn (1):

\vec{E} = \frac{2\times 9\times 10^{9}\times 1\times 10^{- 9}}{0.01} = 1800\ N/C

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\vec{E_{net}} = 1800 + 1800 = 3600\ N/C

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