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2 years ago

22. State any three features of the electroscope.

1 answer:

7 0

-1- was created in the 1600 by william gilbert
-2-When the charge is positive, electrons in the metal of the electroscope are attracted to the charge and move upward out of the leaves. This results in the leaves to have a temporary positive charge and because like charges repel, the leaves separate. When the charge is removed, the electrons return to their original positions and the leaves relax
3-

An electroscope is made up of a metal detector knob on top which is connected to a pair of metal leaves hanging from the bottom of the connecting rod. When no charge is present the metals leaves hang loosely downward. But, when an object with a charge is brought near an electroscope, one of the two things can happen.

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A person carries a plank of wood 1.6 m long with one hand pushing down on it at one end with a force F1 and the other hand holdi

Elis [28]

Answer: $115.52\ N$

Explanation:

Given

Length of plank is 1.6 m

Force $F_1$ is applied on the left side of plank

Force $F_2$ is applied 43 cm from the left end O.

Mass of the plank is $m=13.7\ kg$

for equilibrium

Net torque must be zero. Taking torque about left side of the plank

$\Rightarrow mg\times 0.8-F_2\times 0.43=0\\\\\Rightarrow F_2=\dfrac{13.7\times 9.8\times 0.8}{0.43}\\\\\Rightarrow F_2=249.78\ N$

Net vertical force must be zero on the plank

$\Rightarrow F_1+W-F_2=0\\\Rightarrow F_1=F_2-W\\\Rightarrow F_1=249.78-13.7\times 9.8\\\Rightarrow F_1=115.52\ N$

8 0

2 years ago

A weight of 30.0 N is suspended from a spring that has a force constant of 220 N/m. The system is undamped and is subjected to a

Nimfa-mama [501]

Answer:

$F_0 = 393 N$

Explanation:

As we know that amplitude of forced oscillation is given as

$A = \frac{F_0}{ m(\omega^2 - \omega_0^2)}$

here we know that natural frequency of the oscillation is given as

$\omega_0 = \sqrt{\frac{k}{m}}$

here mass of the object is given as

$m = \frac{W}{g}$

$\omega_0 = \sqrt{\frac{220}{\frac{30}{9.81}}}$

$\omega_0 = 8.48 rad/s$

angular frequency of applied force is given as

$\omega = 2\pi f$

$\omega = 2\pi(10.5) = 65.97 rad/s$

now we have

$0.03 = \frac{F_0}{3.06(65.97^2 - 8.48^2)}$

$F_0 = 393 N$

6 0

3 years ago

Where does current flows maximum in? series connection or parallel connection?

notsponge [240]

Current flow depends on other things in addition to the circuit configuration. If the SAME voltage is applied to some arrangement of the SAME components, the greatest current will occur when they are all in parallel.

3 0

3 years ago

A skater slides across the ice with an initial velocity of 5.0 m/s. She slows 10 points

zvonat [6]

Explanation:

Given that,

The initial velocity of a skater is, u = 5 m/s

She slows to a velocity of 2 m/s over a distance of 20 m.

We can find the acceleration of skater. It is equal to the rate of change of velocity. So, it can be calculated using third equation of motion as follows :

$v^2-u^2=2as$