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3 years ago

A positively charged light metal ball is suspended between two oppositely charged metal plates on

Physics

1 answer:

Ipatiy [6.2K]3 years ago

8 0

Answer:

The positively charged ball moves between both charged plates till the plates and the ball all become neutral.

Check Explanation for more.

Explanation:

Let the ball be in square brackets, and the plates in normal brackets.

(+) [+] (-)

From the law that like charges repel and unlike charges attract.

The positive ball would go first to the negatively charged plate. After which, the ball would hold more negative charges overall than before.

Because the ball is now more negatively charged, it then travels towards the positive plate. In the same manner, the ball would transfer negative electrons to the positive plate.

So, when leaving the positive plate, the ball would be more positive and be drawn towards the negative plate once more. In doing so, it would make the negative plate more positive.

Then, the ball again holds more negative electrons and is drawn towards the positive plate once more.

This back and forth process continues until the once-positive and once-negative plates become neutral, that is, they are discharged.

The ball hanging on the insulated thread becomes neutral too at this point.

Hope this Helps!!!

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by scale drawing, find the resultant of vectors 70N inclined at100N and direction of the resultant 100N

Thepotemich [5.8K]

Answer: 70n=

100n=

Explanation:

4 0

3 years ago

What property of light is suggested by the formation of shadows

Mandarinka [93]

Answer:

The reflection and rectilinear propagation of light helps in the formation of shadows and also tells light doesn't penetrate opaque materials.

3 0

3 years ago

A tank has the shape of an inverted circular cone with height 16m and base radius 3m. The tank is filled with water to a height

rewona [7]

Answer:

$W=17085KJ$

Explanation:

From the question we are told that:

Height $H=16m$

Radius $R=3$

Height of water $H_w=9m$

Gravity $g=9.8m/s$

Density of water $\rho=1000kg/m^3$

Generally the equation for Volume of water is mathematically given by

$dv=\pi*r^2dy$

$dv=\frac{\piR^2}{H^2}(H-y)^2dy$

Where

y is a random height taken to define dv

Generally the equation for Work done to pump water is mathematically given by

$dw=(pdv)g (H-y)$

Substituting dv

$dw=(p(=\frac{\piR^2}{H^2}(H-y)^2dy))g (H-y)$

$dw=\frac{\rho*g*R^2}{H^2}(H-y)^3dy$

Therefore

$W=\int dw$

$W=\int(\frac{\rho*g*R^2}{H^2}(H-y)^3)dy$

$W=\rho*g*R^2}{H^2}\int((H-y)^3)dy)$

$W=\frac{1000*9.8*3.142*3^2}{9^2}[((9-y)^3)}^9_0$

$W=3420.84*0.25[2401-65536]$

$W=17084965.5J$

$W=17085KJ$

4 0

3 years ago

The angular velocity of a flywheel obeys the equa tion w(1) A Br2, where t is in seconds and A and B are con stants having numer

makkiz [27]

Answer:

$A \to rad/s$

$B \to rad/s^3$

Explanation:

$\omega_z(t)=A + Bt^2$

Required

The units of A and B

From the question, we understand that:

$\omega_z(t) \to rad/s$

This implies that each of $A$ and $Bt^2$ will have the same unit as $\omega_z(t)$

So, we have:

$A \to rad/s$

$Bt^2 \to rad/s$

The unit of t is (s); So, the expression becomes

$B * s^2 \to rad/s$

Divide both sides by $s^2$

$B \to \frac{rad/s}{s^2}$

$B \to rad/s^3$

5 0

3 years ago

How do i find stretch? The problem in questioning has already given me the elastic energy and k-value, but I have no idea how to

finlep [7]

Answer:

Stretch can be obtained using the Elastic potential energy formula.

The expression to find the stretch (x) is $x=\sqrt{\frac{2\times EPE}{k}}$

Explanation:

Given:

Elastic potential energy (EPE) of the spring mass system and the spring constant (k) are given.

To find: Elongation in the spring (x).

We can find the elongation or stretch of the spring using the formula for Elastic Potential Energy (EPE).

The formula to find EPE is given as:

$EPE=\frac{1}{2}kx^2$

Rewriting the above expression in terms of 'x', we get:

$x=\sqrt{\frac{2\times EPE}{k}}$

Example:

If EPE = 100 J and spring constant, k = 2 N/m.

Elongation or stretch is given as:

$x=\sqrt{\frac{2\times EPE}{k}}\\\\x=\sqrt{\frac{2\times 100}{2}}\\\\x=\sqrt{100}=10\ m$

Therefore, the stretch in the spring is 10 m.

So, stretch in the spring can be calculated using the formula for Elastic Potential Energy.

6 0

3 years ago