1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Marizza181 [45]
3 years ago
9

Radiation from the Sun reaching Earth (just outside the atmosphere) has an intensity of 1.39 kW/m2. (a) Assuming that Earth (and

its atmosphere) behaves like a flat disk perpendicular to the Sun's rays and that all the incident energy is absorbed, calculate the force on Earth due to radiation pressure. (b) For comparison, calculate the force due to the Sun's gravitational attraction. Assume that the speed of light and Earth radius are 2.998 × 108 m/s and 6.37 thousand km respectively.
Physics
2 answers:
Pachacha [2.7K]3 years ago
8 0

Answer:

(a) 5.91 × 10⁸ N; (b) 3.542 × 10²² N

Explanation:

Data:

  I = 1.39 kW/m² = 1.39 × 10³ N·m⁻³s⁻¹

 c = 2.998 × 10⁸ m/s  

 r = 6370 km = 6.37 × 10⁶ m

G  =  6.674 ×10⁻¹¹ N⋅m²kg⁻²

M = 1.989 × 10³⁰ kg  = mass of Sun

m = 5.972 × 10²⁴ kg = mass of Earth

d = 1.496 × 10¹¹ m     = distance from Earth to Sun

Calculations:  

(a) Force exerted by the radiation pressure

(i) Radiation pressure

All the incident radiation is absorbed, so

\begin{array}{rcl}p & = & \dfrac{I}{c}\\\\& = & \dfrac{\text{1.39 $\times$ 10$^{3}$ N$\cdot$ m$^{-3}$s$^{-1}$}}{2.998 \times 10^{8}\text{ m$\cdot$ s}^{-1}}\\\\& = & 4.636 \times 10^{-6}\text{ N$\cdot$m}^{-2}\\\end{array}

(ii) Area of Earth's disc

\begin{array}{rcl}A & = & \pi r^{2}\\& = & \pi \times (6.37 \times 10^{6} \text{ m})^{2}\\& = & 1.275 \times 10^{14} \text{ m}^{2}\\\end{array}

(iii) Radiation force

\begin{array}{rcl}p & = & \dfrac{F}{A}\\\\F & = & pA\\ & = & 4.636 \times 10^{-6}\text{ N$\cdot$m}^{-2} \times 1.275 \times 10^{14} \text{ m}^{2}\\ & = & 5.91 \times 10^{8}\text{ N}\\\end{array}

(b) Gravitational Force

\begin{array}{rcl}F & = & \dfrac{GMm}{r^{2}}\\\\& = & \dfrac{\text{6.674 $\times$ 10$^{-11}$ N $\cdot$ m$^{2}$kg$^{-2}\times$ 1.989 $\times$ 10$^{30}$ kg $\times$ 5.972 $\times$ 10$^{24}$ kg}}{(\text{1.496 $\times$ 10$^{11}$ m})^{2}}\\\\& = & 3.542 \times 10^{22} \text{ N}\\\end{array}

Zigmanuir [339]3 years ago
7 0

Answer:

F=5.8\times 10^{8}\ N

F=35.57\times 10^{21}\ N

Explanation:

Given that

Intensity I

I= 1.39\ KW/m^2

Speed\ of \ light = 2.99\times 10^8\ m/s

Radius of earth,R = 6370 Km

We know that surface area of earth, A

A=\pi R^2

A=\pi (6370\times 10^3)^2\ m^2

A=1.27\times 10^{14}\ m^2

As we know that pressure due to intensity given as

P=\dfrac{I}{V}

V =Velocity of light

V=3\times 10^8\ m/s

P=\dfrac{1.39\times 1000}{=3\times 10^8}

P=4.6\times 10^{-6}\ Pa

We know that force F

F = P .A

F=4.6\times 10^{-6}\times 1.27\times 10^{14}\ N

F=5.8\times 10^{8}\ N

b)Gravitational force F

F=G\dfrac{m.M}{r^2}

M = mass\ of\ sun = 2\times 10^{30} kg\\m = mass\ of\ earth = 6\times 10^{24}kg

r =1.5\times 10^{11}\ m

G =6.67\times 10^{-11}Nm^2/kg^2

So F

F=6.67\times 10^{-11}\times \dfrac{2\times 10^{30}\times 6\times 10^{24}kg}{1.5\times 10^{11}}

F=35.57\times 10^{21}\ N

You might be interested in
The magnetic field is 0.0 2M from the wire is 0.1 T what is the magnitude of the magnetic field 0.0 1M from the same wire
Kisachek [45]

Answer:

0.2 T

Explanation:

Magnetic field is inversely proportional to the distance from wire since the distance is halved therefore magnetic field will be doubled.

4 0
3 years ago
Weight is proportional to but not equal to mass. In which of the following situations would a person show an increase in weight
meriva

Answer: c living in a camber in an under water habitat

Explanation:

4 0
2 years ago
A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is free to rotate on a frictionless, vertical axle. a
USPshnik [31]
Refer to the diagram shown below.

Let I = the moment of inertia of the wheel.
α = 0.81 rad/s², the angular acceleration
r = 0.33 m, the radius of the weel
F = 260 N, the applied tangential force

The applied torque is
T = F*r
   = (260 N)*(0.33 m)
   = 85.8 N-m

By definition,
T = I*α

Therefore,
I = T/α
  = (85.8 N-m)/(0.81 rad/s²)
  = 105.93 kg-m²

Answer: 105.93 kg-m²

6 0
3 years ago
A piece of bismuth with a mass of 4.06 g 4.06 g gains 423 J 423 J of heat. If the specific heat of bismuth is 0.123 J / ( g ° C
Sholpan [36]

Answer: 846°C

Explanation:

The quantity of Heat Energy (Q) required to heat bismuth depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Given that:

Q = 423 joules

Mass of bismuth = 4.06g

C = 0.123 J/(g°C)

Φ = ?

Then, Q = MCΦ

423 J = 4.06g x 0.123 J/(g°C) x Φ

423 J = 0.5J/°C x Φ

Φ = (423J/ 0.5g°C)

Φ = 846°C

Thus, the change in temperature of the sample is 846°C

4 0
3 years ago
18)
Nina [5.8K]
Number 18 is C, i thin

5 0
3 years ago
Read 2 more answers
Other questions:
  • In which activity is no work done?
    15·2 answers
  • The following are examples of physical properties except
    11·2 answers
  • A school bus transporting students to school is an example of
    14·2 answers
  • The 68-kg crate is stationary when the force P is applied. Determine the resulting acceleration of the crate if (a) P = 0, (b) P
    5·1 answer
  • You are standing 1 meter from a squawking parrot. If you move to a distance three meters away, the sound
    12·2 answers
  • In which position would an object have a greater amount of potential energy
    6·2 answers
  • The topsoil layer has the greatest concentration of organic matter. The subsoil generally lacks organic matter, but it does rece
    6·1 answer
  • The note "Middle C" is known to have a frequency of 261.6 Hz. What would
    10·1 answer
  • when energy from battery was added to water, were the gases produced by this made of the same particles as those produced from h
    13·1 answer
  • Question 9 of 34
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!