Ask question

Ask question

All categories

4 years ago

7

Two small metal spheres are 25 cm apaft.The spheres have equal amount of negative charge and repel each other with a force of 0.

036 N. What is the magnitude of the charge on each sphere? (a) 2x10-3 C (h) 3 x 10-4C

1 answer:

Mars2501 [29]4 years ago

8 0

Answer:

$0.5\times 10^{-6}C$

Explanation:

According to coulombs law force between two charges is given by $F=\frac{1}{4\pi \epsilon _0}\frac{Q_1Q_2}{R^2}$ here R is the distance between both the charges which is given as 25 cm

We have given force F =0.036 N

So $0.036=\frac{1}{4\pi \times 8.85\times 10^{-12}}\frac{Q^2}{(0.25^2)}$ As $\epsilon _0$ is constant which value is $8.85\times 10^{-12}$

$Q^2=0.250\times 10^{-12}$

$Q=0.5\times 10^{-6}C$

You might be interested in

If a pendulum clock keeps perfect time at the base of a mountain, will it also keep perfect time when it is moved to the tip of

vazorg [7]

Answer:

No

Explanation:

The period of a pendulum is given by

$T=2\pi \sqrt{\frac{L}{g}}$

where

L is the length of the pendulum

g is the acceleration due to gravity

We see that the period of the pendulum depends on the value of g. However, the value of the gravitational acceleration is different at different locations on Earth. In particular, at the top of the mountain the value of g is slightly lower than the value of g at the base of the mountain; in fact, g is given by

$g=\frac{GM}{r^2}$

where

G is the gravitational constant

M is the Earth's mass

r is the distance from the Earth's center

so since r is greater at the top of the mountain, g is lower, and therefore the period of the pendulum will be slightly longer.

8 0

3 years ago

A 0.700-kg particle has a speed of 1.90 m/s at point circled A and kinetic energy of 7.20 J at point circled B. (a) What is its

Savatey [412]

Answer:

a). $E_{kA}=1.2635 J$

b). $V_{B}=4.535\frac{m}{s}$

c). Δ $E_{t}=8.4635 J$

Explanation:

ΔE=kinetic energy

a).

$E_{kA}=\frac{1}{2}*m*v_{A} ^{2} \\ v_{A}=1.9 \frac{m}{s}\\ m=0.70kg\\E_{kA}=\frac{1}{2}*0.70kg*(1.9 \frac{m}{s})^{2} \\E_{kA}=1.2635 J$

b).

$E_{kB}=\frac{1}{2}*m*v_{B} ^{2}$

$V_{B}^{2}=\frac{E_{kB}*2}{m} \\V_{B}=\sqrt{\frac{E_{kB}*2}{m}} \\V_{B}=\sqrt{\frac{7.2J*2}{0.70kg}} \\V_{B}=4.53 \frac{m}{s}$

c).

net work= EkA+EkB

$E_{t}=E_{kA}+ E_{kB}\\E_{t}=1.2635J+7.2J\\E_{t}=8.4635J$

3 0

3 years ago

The concrete sections of a certain superhighway are designed to have a length of 23.0 m. The sections are poured and cured at 10

loris [4]

Answer:

0.88 cm

Explanation:

Initial length at 10°c, L = 23 m

Rise of temperature, Δt = 42 - 10 = 32°c

Coefficient of linear expansion of concrete, α = 12 x 10^(-6) per°c

Minimum spacing to be left = α L Δt

$=12\times10^{-6}\times23\times32$

$=8.832\times10^{-3}m$

= 8.832 x 10^{-1} cm

= 0.88 cm

Note: I have a chosen a general value of 12 x 10^-6 per deg c for coefficient of expansion of concrete. However, please refer to the value given in your text book and substitute it for an accurate answer.

8 0

3 years ago

The Electric Field 9.0m from a +25μC charge is 

salantis [7]

The electric field intensity generated by a single point charge is given by

$E(r)=k\frac{q}{r^2}$

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, the charge is $q=25 \mu C=25 \cdot 10^{-6} C$ and we are asked to calculate the field at distance $r=9.0 m$ , so the electric field is

$E(9.0 m)=(8.99 \cdot 10^9 Nm^2C^{-2})\frac{25 \cdot 10^{-6} C}{(9 m)^2}=2775 V/m$

5 0

3 years ago

An object in motion will remain in motion and an object at rest will remain at rest until a greater force interrupts it. Explain

Fittoniya [83]

Answer:

A object, lets say a cup. This cup will never, ever move unless something or someone disturbs it. If something touches or hits this cup the cup will move. But, until the cup gets touched, nothing will EVER make it move.

Explanation:

<em>I hope this helps!!</em>

4 0

3 years ago