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LuckyWell [14K]
2 years ago
12

How can you tell that an electroscope has an electrical charge? can you tell from an electroscope alone what kind of charge it h

as?

Physics
1 answer:
tester [92]2 years ago
8 0

Answer:

An electroscope is a device which used to detect charge on an object. It consists of conducting sphere at top with a metal rod and metal leaves. This is enclosed in a glass jar. A charged object is brought near the top of the electroscope. this causes electrons to transfer to or from the charged object.

If the electroscope has charge, the leaves repel each other and move away from each other. This is because charge conducts from the metallic sphere through rod to both leaves. Having same charge, the leaves repel each other and open up. No change occurs in neutral state.

We cannot tell from an electroscope alone what kind of charge it has because the leaves have like charges and hence repel each other whether it is positive or negative charge. Thus, we cannot differentiate the charge.

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A boy throws a ball vertically up it returns the ground after 10 seconds find the maximum height reached by the ball
Akimi4 [234]

Answer:

Approximately 122.625\; {\rm m} (assuming that g = 9.81\; {\rm m\cdot s^{-2}}, the ball was launched from ground level, and that the drag on the ball is negligible.)

Explanation:

Let v_{0} denote the velocity at which the ball was thrown upward.

If the drag (air friction) on the ball is negligible, the ball would land with a velocity of exactly (-v_{0}). The velocity of the ball would be changed from v to (-v_{0})\! (such that \Delta v = (-v_{0}) - v_{0} = (-2\, v_{0})) within t = 10\; {\rm s}.

Also because the drag on the ball is negligible, the acceleration of the ball would be a = -g = -9.81\; {\rm m\cdot s^{-2}}. Thus:

\Delta v = a\, t = 10\; {\rm s} \times (-9.81\; {\rm m\cdot s^{-2}}) = -98.1\; {\rm m\cdot s^{-1}}.

Since \Delta v = (-2\, v_{0}):

-2\, v_{0} = \Delta v = -98.1\; {\rm m\cdot s^{-1}.

\begin{aligned}v_{0} &= \frac{-98.1\; {\rm m\cdot s^{-1}}}{-2}= 49.05\; {\rm m \cdot s^{-1}}\end{aligned}.

The ball reaches maximum height when its velocity is v_{1} = 0\; {\rm m\cdot s^{-1}}. Apply the SUVAT equation x = ({v_{1}}^{2} - {v_{0}}^{2}) / (2\, a) to find the displacement x between the original position (ground level, where v_{0} = 49.05\; {\rm m\cdot s^{-1}}) and the max-height position of the ball (where v_{1} = 0\; {\rm m\cdot s^{-1}}.)

\begin{aligned}x &= \frac{(0\; {\rm m\cdot s^{-1}})^{2} - (49.05\; {\rm m\cdot s^{-1}})^{2}}{2 \times (-9.81\; {\rm m\cdot s^{-2}})} \\ &\approx 122.625\; {\rm m\cdot s^{-1}}\end{aligned}.

7 0
2 years ago
what can you say about the motion of an object if its speed time graph is a straight line parallel to the time axis ? Also draw
MaRussiya [10]

Answer:

It says about the motion and the graph of the object is stationary, basically travelling at the same speed at any time of the graph. It will never change.

Explanation:

To draw a diagram:

1. Draw an object and represent the speed as stationary and constant at any time.

5 0
2 years ago
Two forces, one four times as large as the other, pull in the same direction on a 10kg mass and impart to it an acceleration of
notsponge [240]

Answer:

The acceleration of the mass is 2 meters per square second.

Explanation:

By Newton's second law, we know that force (F), measured in newtons, is the product of mass (m), measured in kilograms, and net acceleration (a), measured in meters per square second. That is:

F = m\cdot a (1)

The initial force applied in the mass is:

F = (10\,kg)\cdot \left(2.5\,\frac{m}{s^{2}} \right)

F = 25\,N

In addition, we know that force is directly proportional to acceleration. If the smaller force is removed, then the initial force is reduced to \frac{4}{5} of the initial force. The acceleration of the mass is:

\frac{25\,N}{20\,N} = \frac{2.5\,\frac{m}{s^{2}} }{a}

a = 2\,\frac{m}{s^{2}}

The acceleration of the mass is 2 meters per square second.

4 0
3 years ago
What is the speed of sound in air with temperature of 355.8 k​
nikklg [1K]

Answer:

Explanation:

20.05 √Tk = 20.05 √355.8 = 378.196... ≈ 378 m/s

7 0
3 years ago
What is the total distance, side to side, that the top of the building moves during such an oscillation? The New England Merchan
kramer
I think the question should be the below:

<span>What is the total distance, side to side, that the top of the building moves during such an oscillation?
</span>
Answer is the below:

 <span>Acceleration .. a = (-) ω² x </span>
<span>(ω = equivalent ang. vel. = 2π.f) (x = displacement from equilibrium position) </span>

<span>x (max) = a(max) /ω² </span>

<span>x = (0.015 x 9.8m/s²) / (2π.f)² .. .. (0.147) / (2π*0.22)² .. .. ►x(max) = 0.077m .. (7.70cm)</span>
5 0
3 years ago
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