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Zarrin [17]
3 years ago
13

This problem explores the behavior of charge on conductors. We take as an example a long conducting rod suspended by insulating

strings. Assume that the rod is initially electrically neutral. For convenience we will refer to the left end of the rod as end A, and the right end of the rod as end B. In the answer options for this problem, "strongly attracted/repelled" means "attracted/repelled with a force of magnitude similar to that which would exist between two charged balls.A small metal ball is given a negative charge, then brought near (i.e., within about 1/10 the length of the rod) to end A of the rod. What happens to end A of the rod when the ball approaches it closely this first time?
Physics
1 answer:
Olegator [25]3 years ago
3 0

Answer:

rod end A is strongly attracted towards the balls

rod end B is weakly repelled by the ball as it is at a greater distance

Explanation:

When the ball with a negative charge approaches the A end of the neutral bar, the charge of the same sign will repel and as they move they move to the left end, leaving the rod with a positive charge at the A end and a negative charge of equal value at end B.

Therefore rod end A is strongly attracted towards the balls and

rod end B is weakly repelled by the ball as it is at a greater distance

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Two point sources, vibrating in phase, produce an interferencepattern in a ripple tank. If the frequency is increased by 20%, th
Vera_Pavlovna [14]

Answer:

option A

Explanation:

given,

frequency is increased by 20%

we know,

\dfrac{x_n}{L}d = (n-\dfrac{1}{2})\times \lambda...........(1)

where

x_n is the perpendicular distance between the point the interference pattern is obtained,

L is the distance between the center of the two point sources

and λ is  the wavelength of light.

If the frequency is increased by 20%, then the number of nodal lines is increased by 20%.

From equation (1),we observe that the frequency is directly proportional to the number of nodal lines.

Hence, the correct answer is option A

8 0
3 years ago
A model airplane with a mass of 0.741kg is tethered by a wire so that it flies in a circle 30.9 m in radius. The airplane engine
liq [111]

(a) 24.6 Nm

The torque produced by the net thrust about the center of the circle is given by:

\tau = F r

where

F is the magnitude of the thrust

r is the radius of the wire

Here we have

F = 0.795 N

r = 30.9 m

Therefore, the torque produced is

\tau = (0.795 N)(30.9 m)=24.6 N m

(b) 0.035 rad/s^2

The equivalent of Newton's second law for a rotational motion is

\tau = I \alpha

where

\tau is the torque

I is the moment of inertia

\alpha is the angular acceleration

If we consider the airplane as a point mass with mass m = 0.741 kg, then its moment of inertia is

I=mr^2 = (0.741 kg)(30.9 m)^2=707.5 kg m^2

And so we can solve the previous equation to find the angular acceleration:

\alpha = \frac{\tau}{I}=\frac{24.6 Nm}{707.5 kg m^2}=0.035 rad/s^2

(c) 1.08 m/s^2

The linear acceleration (tangential acceleration) in a rotational motion is given by

a=\alpha r

where in this problem we have

\alpha = 0.035 rad/s^2 is the angular acceleration

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Substituting the values, we find

a=(0.035 rad/s^2)(30.9 m)=1.08 m/s^2

4 0
3 years ago
A particle initially located at the origin has an acceleration of a=3jm/s^2 and an initial velocity of Vi=5im/s.
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Given the particle's acceleration is

\vec a(t) = \left(3\dfrac{\rm m}{\mathrm s^2}\right)\vec\jmath

with initial velocity

\vec v(0) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath

and starting at the origin, so that

\vec r(0) = \vec 0

you can compute the velocity and position functions by applying the fundamental theorem of calculus:

\vec v(t) = \vec v(0) + \displaystyle \int_0^t \vec a(u)\,\mathrm du

\vec r(t) = \vec r(0) + \displaystyle \int_0^t \vec v(u)\,\mathrm du

We have

• velocity at time <em>t</em> :

\vec v(t) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \displaystyle \int_0^t \left(3\dfrac{\rm m}{\mathrm s^2}\right)\,\vec\jmath\,\mathrm du \\\\ \vec v(t) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \left(3\dfrac{\rm m}{\mathrm s^2}\right)t\,\vec\jmath \\\\ \boxed{\vec v(t) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \left(3\dfrac{\rm m}{\mathrm s^2}\right)t\,\vec\jmath}

• position at time <em>t</em> :

\vec r(t) = \displaystyle \int_0^t \left(\left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \left(3\dfrac{\rm m}{\mathrm s^2}\right)u\,\vec\jmath\right) \,\mathrm du \\\\ \boxed{\vec r(t) = \left(5\dfrac{\rm m}{\rm s}\right)t\,\vec\imath + \frac12 \left(3\frac{\rm m}{\mathrm s^2}\right)t^2\,\vec\jmath}

8 0
3 years ago
A sample of radium-226 will decay to 1/4 of its original amount after 3200 years. What is the half-life of radium-226?
Alenkinab [10]

Answer:1600

Explanation:Just took the test

5 0
3 years ago
A cartoon plane with four engines can accelerate at 8.9 m/s2 when one engine is running. What is the acceleration of the plane i
egoroff_w [7]

Answer:

The acceleration of the plane is 35.6 meters per square second.

Explanation:

By Newton's Laws of Motion, we know that force (F), in newtons, is directly proportional to acceleration (a), in meters per square second. That is:

F \propto a

If four engines are running simultaneously and each produce the same force, then we find that acceleration of the plane is:

a_{T} = 4\cdot \left(8.9\,\frac{m}{s^{2}} \right)

a_{T} = 35.6\,\frac{m}{s^{2}}

The acceleration of the plane is 35.6 meters per square second.

5 0
3 years ago
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