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2 years ago

7

two charged particles repel each other with a force of 10N when they are 30cm apart if they are repel each other by a force of 4

N what would be the separation distance

1 answer:

KonstantinChe [14]2 years ago

5 0

Answer:

The pattern between electrostatic force and distance can be further characterized as an inverse square relationship

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How do I find the speed of sound when given Frequency and displacement?

IceJOKER [234]

Answer: Taking into account sound is a wave, we can use the information of the displacement (generally given as a graph) to find the wavelength and frequency, then we can calculate the speed with the formula of the speed of a wave.

Explanation:

If we have the displacement graph of the sound wave, we can find its amplitude, its wavelength and period (which is the inverse of frequency).

Now, if we additionally have the frequency as data, we can use the equation of the speed of a wave:

$s=\lambda f$

Where:

$s$ is the speed of the sound wave

$\lambda$ is the wavelength

$f$ is the frequency

3 0

3 years ago

A brass washer has an outside diameter of 4.50 cm with a hole of diameter 1.25 cm and is 1.50 mm thick. The density of brass is

-Dominant- [34]

We will find the mass from
mass = density x volume
We are told the density and must find the volume from the dimensions given
the volume of the washer will be the area x thickness (remembering to convert all measurements to meters)
if the washer had no hole, its area would be pi (0.0225m)^2 (remember to convert to meters and to use radius)
the area of the hole is pi(0.00625m)^2
so the area of the washer is pi[(0.0225m)^2 - (0.00625m)^2] = 1.5x10^-3 m
the volume of the washer is 1.5x10^-3 m x 1.5x10^-3 m = 2.25x10^-6 m^3 (the thickness of the washer is 1.5 mm = 1.5x10^-3m)
thus, the mass of the washer = 8598kg/m^3 x 2.25x10^-6m^3 = 0.0189kg = 18.9 grams

6 0

3 years ago

Coherent light with wavelength 599 nm passes through two very narrow slits with separation of 20 μm, and the interference patter

goblinko [34]

Answer:

134.77 mm

Explanation:

Wave length of light λ = 599 x 10⁻⁹ m

Slit separation d = 20 x 10⁻⁶ m

Screen distance D = 3 m

Distance of second dark fringe from centre

= 1.5 x λ D / d

Putting the values given above

distance = $\frac{1.5\times599\times10^{-9}\times 3}{20\times10^{-6}}$

= 134.77 x 10⁻³ m

= 134.77 mm.

7 0

4 years ago

A 52.0-kg sandbag falls off a rooftop that is 22.0 m above the ground. The collision between the sandbag and the ground lasts fo

sergiy2304 [10]

Answer:

F = 7,916,955.0N

Explanation:

According to newtons second law

Force = mass * acceleration

Given

mass = 52.0kg

distance S = 22.0m

time t = 17.0 ms = 0.017s

We need to get the acceleration first using the formula;

S = ut+ 1/2at²

22 = 0 + 1/2 a(0.017²)

22 = 0.0001445a

a = 22/0.0001445

a = 152,249.13m/s²

The magnitude of the average force exerted will be;

F = ma

F = 52 * 152,249.13

F = 7,916,955.0N

4 0

3 years ago

Two identical small metal spheres with q1 > 0 and |q1| > |q2| attract each other with a force of magnitude 72.1 mN when se

Brrunno [24]

1) $+2.19\mu C$

The electrostatic force between two charges is given by

$F=k\frac{q_1 q_2}{r^2}$ (1)

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of

$\frac{Q}{2}$

where Q is the total charge between the two spheres.

So we can actually rewrite the force as

$F=k\frac{(\frac{Q}{2})^2}{r^2}$

And since we know that

r = 1.41 m (distance between the spheres)

$F= 21.63 mN = 0.02163 N$

(the sign is positive since the charges repel each other)

We can solve the equation for Q:

$Q=2\sqrt{\frac{Fr^2}{k}}=2\sqrt{\frac{(0.02163)(1.41)^2}{8.98755\cdot 10^9}}}=4.37\cdot 10^{-6} C$

So, the final charge on the sphere on the right is

$\frac{Q}{2}=\frac{4.37\cdot 10^{-6} C}{2}=2.19\cdot 10^{-6}C=+2.19\mu C$

2) $q_1 = +6.70 \mu C$

Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that

$F = -72.1 mN = -0.0721 N$ (we put a negative sign since the force is attractive, which means that the charges have opposite signs)

r = 1.41 m is the separation between the charges

And also,

$q_2 = Q-q_1$

So we can rewrite eq.(1) as

$F=k \frac{q_1 (Q-q_1)}{r^2}$

Solving for q1,

$Fr^2=k (q_1 Q-q_1^2})\\kq_1^2 -kQ q_1 +Fr^2 = 0$

Since $Q=4.37\cdot 10^{-6} C$ , we can substituting all numbers into the equation:

$8.98755\cdot 10^9 q_1^2 -3.93\cdot 10^4 q_1 -0.141 = 0$

which gives two solutions:

$q_1 = 6.70\cdot 10^{-6} C\\q_2 = -2.34\cdot 10^{-6} C$

Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

$q_1 = +6.70 \mu C$

8 0

4 years ago