PLEASE HELP!!!!!!!!!!!!!!!!!!!

O

Sergio [31]

see

below

Explanation:

refractive index = speed of light in vacuum / speed of light in medium

light travels at a speed of 3.0 x 10^8 m/s in vacuum

refractive index = 3.0 x 10^8 / 2.0 x 10^8

refractive index = 1.5

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4 0

3 years ago

For a point charge, how does the potential vary with distance from the point charge, r?

mr_godi [17]

For a point charge, how does the potential vary with distance from the point charge, r?

a constant

b. r.

c. 1/r.

d. $1/r^2$ .

e. $r^2$ .

Answer:

The correct option is C

Explanation:

Generally for a point charge the electric potential is mathematically represented as

$V = \frac{k Q }{r }$

Here we can deduce that the electric potential varies inversely with the distance i.e

$V \ \alpha \ \frac{1}{r}$

So

3 0

3 years ago

Please help me with this

AfilCa [17]

Answer:

20 N exerts no torque about the pivot.

14 N exerts a counterclockwise torque of 14 * .3 = .42 N-m

6 exerts a clockwise torque of 6 * .7 = .42 N-m

The meter stick will not turn because there is no net torque on the meter stick.

3 0

2 years ago

A spherical balloon is made from a material whose mass is 3.00 kg. The thickness of the material is negligible compared to the 1

vesna_86 [32]

To solve the problem it is necessary to apply the definition of Newton's second Law and the definition of density.

Density means the relationship between volume and mass:

$\rho = \frac{m}{V}$

While Newton's second law expresses that force is given by

F = ma

Where,

m = mass

a= acceleration (gravity at this case)

In the case of the given data we have to,

$m_b = 3Kg$

$r = 1.5m\\V = \frac{4}{3}\pi r^3 \\V = \frac{4}{3} \pi 1.5^3\\V = 14.13m^3$

In equilibrium, the entire system is equal to zero, therefore

$\sum F = 0$

$F_g +F_h-F_b = 0$

Where,

$F_g =$ Weight of balloon

$F_h =$ Weight of helium gas

$F_b =$ Bouyant force

Then we have,

$mg+V\rho g -V\rho_a g = 0$

$\rho = \rho_0-\frac{m}{V}$

Replacing the values we have that

$\rho = 1.19kg/m^3 -\frac{3Kg}{14.13m^3}$

$\rho = 0.978kg/m^3$

Now by ideal gas law we have that

$PV=nRT$

$P\frac{\rho}{m} = nRT$

$P = \rho \frac{n}{m}RT$

But the relation \frac{n}{m} is equal to the inverse of molar mass, that is

$P = \frac{\rho}{M_0} RT$

$P = \frac{0.978kg/m^3}{0.04kg/mol}*8.314J/K.Mol * 305K$

$P = 619995.7Pa$

Therefore the pressure of the helium gas assuming it is ideal is 0.61Mpa

5 0

3 years ago

The string constrains the rotational and translational motion of the falling cylinder, given that it doesn't slip. what is the r

Sindrei [870]

<span>Answer: First we need to find the acceleration. torque on cylinder Ď„ = T * r where T is the string tension; T = m(g - a) where a is the acceleration of the cylinder. Then Ď„ = m(g - a)r But also Ď„ = IÎ±. For a solid cylinder, I = Â˝mrÂ˛, and if the string doesn't slip, then Î± = a / r, so Ď„ = Â˝mrÂ˛ * a/r = Â˝mra. Since Ď„ = Ď„, we have m(g - a)r = Â˝mra â†’ m, r cancel, leaving g - a = Â˝a g = 3a/2 a = 2g/3 where g, of course, is gravitational acceleration. We know that v(t) = a*t, so for our cylinder v(t) = 2gt / 3 â—„ linear velocity and Ď‰ = v(t) / r = 2gt / 3r â—„ angular velocity</span>

3 0

4 years ago

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