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Lynna [10]
2 years ago
15

Someone plans to float a small, totally absorbing sphere 0.500 m above an isotropic point source of light,so that the upward rad

iation force from the light matches the downward gravitational force on the sphere. The sphere’s density is 19.0 g/cm3, and its radius is 2.00 mm. (a) What power would be required of the light source?
Physics
1 answer:
mote1985 [20]2 years ago
4 0

Answer:

468449163762.0812 W

Explanation:

m = Mass = \rhoV

V = Volume =\dfrac{4}{3}\pi r^3

r = Distance of sphere from isotropic point source of light = 0.5 m

R = Radius of sphere = 2 mm

\rho = Density = 19 g/cm³

c = Speed of light = 3\times 10^8\ m/s

A = Area = \pi R^2

I = Intensity = \dfrac{P}{4\pi r^2}

g = Acceleration due to gravity = 9.81 m/s²

Force due to radiation is given by

F=\dfrac{IA}{c}\\\Rightarrow F=\dfrac{\dfrac{P}{4\pi r^2}{\pi R^2}}{c}\\\Rightarrow F=\dfrac{PR^2}{4r^2c}

According to the question

F=mg\\\Rightarrow \dfrac{PR^2}{4r^2c}=\rho \dfrac{4}{3}\pi R^3g\\\Rightarrow P=\dfrac{16r^2\rho c\pi Rg}{3}\\\Rightarrow P=\dfrac{16\times 0.002\times 19000\times \pi\times 0.5^2\times 9.81\times 3\times 10^8}{3}\\\Rightarrow P=468449163762.0812\ W

The power required of the light source is 468449163762.0812 W

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Someone who wants to sell you a Superball claims that it will bounce to a height greater than the height from which it is droppe
sergeinik [125]

Answer:

No

Explanation:

Unless there are other external forces, this will never be true. Because according to energy conservation, potential energy will be converted to kinetic energy as the ball falls down (so it loses height and gain speed). And vice versa, kinetic to potential when it bounces back. So the potential energy after must be the same (or smaller if losing heat to external environment), so it can only get the the same height or less, but not more.

7 0
2 years ago
The velocity of a an object in linear motion changes from +25 meters per second to +15 meters per second in 2.0 seconds.
kodGreya [7K]

Answer:

-5.0 m/s^2

Explanation:

Missing question:

What is the object's acceleration?

Solution:

The acceleration of an object is given by

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time taken for the change in velocity

For the object in this problem,

u = +25 m/s

v = +15 m/s

t = 2.0 s

Substituting,

a=\frac{15-25}{2}=-5.0 m/s^2

And the acceleration is negative because its direction is opposite to that of the velocity.

5 0
2 years ago
At a certain instant the current flowing through a 5.0-H inductor is 3.0 A. If the energy in the inductor at this instant is inc
lys-0071 [83]

Answer:

The current is changing at the rate of 0.20 A/s

Explanation:

Given;

inductance of the inductor, L = 5.0-H

current in the inductor, I = 3.0 A

Energy stored in the inductor at the given instant, E = 3.0 J/s

The energy stored in inductor is given as;

E = ¹/₂LI²

E = ¹/₂(5)(3)²

E = 22.5 J/s

This energy is increased by 3.0 J/s

E = 22.5 J/s + 3.0 J/s = 25.5 J/s

Determine the new current at this given energy;

25.5 = ¹/₂LI²

25.5 = ¹/₂(5)(I²)

25.5 = 2.5I²

I² = 25.5 / 2.5

I² = 10.2

I = √10.2

I = 3.194 A/s

The rate at which the current is changing is the difference between the final current and the initial current in the inductor.

= 3.194 A/s - 3.0 A/s

= 0.194 A/s

≅0.20 A/s

Therefore, the current is changing at the rate of 0.20 A/s.

5 0
2 years ago
Evaporation of sweat requires energy and thus take excess heat away from the body. Some of the water that you drink may eventual
kotegsom [21]

Answer:

The amount of heat required is H_t =  1.37 *10^{6} \ J

Explanation:

From the question we are told that

The mass of water is m_w  =  20 \ ounce = 20 * 28.3495 = 5.7 *10^2 g

The temperature of the water before drinking is T_w  =  3.8 ^oC

The temperature of the body is T_b  =  36.6^oC

Generally the amount of heat required to move the water from its former temperature to the body temperature is

H=  m_w  *  c_w * \Delta T

Here c_w is the specific heat of water with value c_w = 4.18 J/g^oC

So

H=   5.7 *10^2 * 4.18 * (36.6 - 3.8)

=> H= 7.8 *10^{4} \  J

Generally the no of mole of sweat present mass of water is

n = \frac{m_w}{Z_s}

Here Z_w is the molar mass of sweat with value

Z_w =  18.015 g/mol

=> n = \frac{5.7 *10^2}{18.015}

=> n = 31.6 \  moles

Generally the heat required to vaporize the number of moles of the sweat is mathematically represented as

H_v  =  n  *  L_v

Here L_v is the latent heat of vaporization with value L_v  = 7 *10^{3} J/mol

=> H_v  =  31.6 * 7 *10^{3}

=> H_v  = 1.29 *10^{6} \  J

Generally the overall amount of heat energy required is

H_t =  H +  H_v

=> H_t =  7.8 *10^{4} +  1.29 *10^{6}

=> H_t =  1.37 *10^{6} \ J

4 0
2 years ago
A future use of space stations may be to provide hospitals for severely burned persons. it is very painful for a badly burned pe
natta225 [31]
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5 0
2 years ago
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