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konstantin123 [22]
3 years ago
5

Suppose the Sun appeared to you 900 times dimmer than it does now. How far away from the Sun would you be? (A) 1/9 AU (B) 3 AU

(C) 30 AU (D) 90 AUs (E) 900 AUs
Physics
1 answer:
Harrizon [31]3 years ago
5 0

Answer:

The correct answer is option 'c': 30 AUs

Explanation:

For a spherical wave front emitted by sun with total energy 'E' the energy density over the surface when it is at a distance 'r' from the sun is given by

e=\frac{E}{4\pi r^{2}}

This energy per unit area is sensed by observer as intensity of the sun.

Let the initial intensity of sun at a distance r_{1} be e_{1}

Thus if the sun becomes 900 times dimmer we have

e'=\frac{e_{1}}{900}\\\\\frac{E}{4\pi r_{2}^{2}}=\frac{1}{900}\times \frac{E}{4\pi r_{1}^{2}}\\\\\Rightarrow r_{2}^{2}={r_{1}^{2}}\times 900\\\\\therefore r_{2}={r_{1}}\times {30}

Thus the distance increases 30 times.

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Define couple and give 2 examples
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Answer:

Two equal and opposite parallel forces not acting along the same line, form a couple. A couple is always needed to produce the rotation.

For example, turning a key in a lock and turning a steering wheel.

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2 years ago
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An athlete needs to lose weight and decides to do it by “pumping iron.” (a) How many times must an 90.0 kg weight be lifted a di
Wittaler [7]

Answer:

230kg would be the best answer

Explanation:

7 0
3 years ago
Caleb rides his horse with a constant speed of 12 miles per hour. How far can she travel in 3 1/2 hours?
baherus [9]
12 × 3= 36
12 × 1/2 = 6
36+ 6= 42
6 0
3 years ago
Calculate the potential energy of a 20-kg sled at 40 meters
irakobra [83]

Answer: 7840N

Explanation:

Given that

Potential energy = ?

Mass of sled = 20-kg

Distance = 40 meters

Acceleration due to gravity = 9.8m/s^2

Recall that potential energy is the energy possessed by a body at rest

i.e potential energy = mass m x acceleration due to gravity g x distance h

P.E = mgh

P.E = 20kg x 9.8m/s^2 x 40m

P.E = 7840N

Thus, the potential energy of the sled is 7840N

5 0
3 years ago
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The volume of water in the Pacific Ocean is about 7.0 × 10 8 km 3 . The density of seawater is about 1030 kg/m3. (a) Determine t
Novay_Z [31]

To solve the problem it is necessary to consider the concepts related to Potential Energy and Kinetic Energy.

Potential Energy because of a planet would be given by the equation,

PE=\frac{GMm}{r}

Where,

G = Gravitational Universal Constant

M = Mass of Ocean

M = Mass of Moon

r = Radius

From the data given we can calculate the mass of the ocean water through the relationship of density and volume, then,

m = \rho V

m = (1030Kg/m^3)(7*10^8m^3)

m = 7.210*10^{11}Kg

It is necessary to define the two radii, when the ocean is far from the moon and when it is facing.

When it is far away, it will be the total diameter from the center of the earth to the center of the moon.

r_1 = 3.84*10^8 + 6.4*10^6 = 3.904*10^8m

When it's near, it will be the distance from the center of the earth to the center of the moon minus the radius,

r_2 = 3.84*10^8-6.4*10^6 - 3.776*10^8m

PART A) Potential energy when the ocean is at its furthest point to the moon,

PE_1 = \frac{GMm}{r_1}

PE_1 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.904*10^8}

PE_1 = 9.05*10^{15}J

PART B) Potential energy when the ocean is at its closest point to the moon

PE_2 = \frac{GMm}{r_2}

PE_2 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.776*10^8}

PE_2 = 9.361*10^{15}J

PART C) The maximum speed. This can be calculated through the conservation of energy, where,

\Delta KE = \Delta PE

\frac{1}{2}mv^2 = PE_2-PE_1

v=\sqrt{2(PE_2-PE_1)/m}

v = \sqrt{\frac{2*(9.361*10^{15}-9.05*10^{15})}{7.210*10^{11}}}

v = 29.4m/s

8 0
3 years ago
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