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3 years ago

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Need help ??? Please

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