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Flauer [41]
3 years ago
8

Need help ??? Please

Physics
1 answer:
Yanka [14]3 years ago
8 0
Is there information in the previous question which relates to this one?
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Radiation from the Sun reaching Earth (just outside the atmosphere) has an intensity of 1.39 kW/m2. (a) Assuming that Earth (and
Zigmanuir [339]

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

7 0
3 years ago
Read 2 more answers
A wave with a frequency of 60 Hz is traveling along a string whose linear mass density is 230 g/m and whose tension is 65 N. If
matrenka [14]

To develop this problem we will use the concepts related to Speed in a string that is governed by Tension (T) and linear density (µ)

V = \sqrt{\frac{T}{\mu}}

Our values are given as:

f = 60Hz\\\mu = 230 g/m = 0.230kg/m\\T = 65N\\P = 75w

Replacing we have that the velocity is

V = \sqrt{\frac{T}{\mu}}

V = \sqrt{\frac{65}{0.230}}

V = 16.81m/s

From the theory of wave propagation the average power wave is given as

P =\frac{1}{2} \mu \omega^2 A^2 V

Where,

A = Amplitude

\omega = 2\pi f \rightarrow Angular velocity

A^2 = \frac{2P}{\mu \omega^2 V}

A^2 = \frac{2P}{\mu (2\pi f)^2 V}

Replacing,

A^2 = \sqrt{\frac{2(75)}{(0.230)(2\pi 60)^2(16.81)}}

A = 0.0165m

Therefore the amplitude of the wave should be 0.0165m

8 0
3 years ago
What is the name for the area of dust and debris which orbited the young sun and eventually became planets and moons?
alexandr402 [8]
I believe the answer is C. protoplanetary disc. Let me know if this helps
5 0
3 years ago
Read 2 more answers
A Viking ship roller coaster at the fair has a mass of 36,000 kg. If at its peak it reaches a height of 20 m off the ground, how
adell [148]

Answer:

7056 kJ

Explanation:

Given that,

Mass of a ship roller coaster is 36,000 kg.

It reaches a height of 20 m off the ground

We need to find the gravitational potential energy does it have. The formula for the gravitational potential energy ios given by :

E = mgh

g is acceleration due to gravity

E = 36,000 kg × 9.8 m/s² × 20 m

= 7056000 J

or

E = 7056 kJ

So, it will have 7056 kJ of gravitational potential energy.

8 0
3 years ago
A 0.150 kg mass is attached to a spring with k = 18.9 N/m. At the equilibrium position, it moves 2.39 m/s. How much mechanical e
Nastasia [14]

Answer:0.428

Explanation: got it right on Acellus

5 0
2 years ago
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