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Advocard [28]
3 years ago
9

A person standing on the surface of the earth has a weight of 650 N. What is its weight at a distance of two earth radii from th

e surface? If the person is at a distance of two earth radii from the surface, he/she is at a distance of three earth radii from the earth’s center. Since we have an inverse square law, three times the distance means 1/9th of the weight.
Physics
1 answer:
snow_lady [41]3 years ago
4 0

Answer:

72.22 N

Explanation:

F = weight

m = mass of body

M = mass of earth

R = radius of earth

G = universal constant of gravitation

F_1= 650 N

F_1 = GMm/R^2

two earth radius above the surface of the earth:

F_2=  GMm/(3R)^2=  GMm/9R^2= F_1/9= 650/9

=72.22 N

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A car increased its velocity to 62m/s in 10s starting from rest. Calculate the distance it covers during this time?
Roman55 [17]

Answer:

distance= velocity ×time

distance= 62×10

distance=620m

hope it helps you mate please mark me as brainliast

6 0
3 years ago
Read 2 more answers
he fan blades on a jet engine make one thousand revolutions in a time of 54.9 ms. What is the angular frequency of the blades?
Gnesinka [82]

So, the angular frequency of the blades approximately <u>36.43π rad/s</u>.

<h3>Introduction</h3>

Hi ! Here I will discuss about the angular frequency or what is also often called the angular velocity because it has the same unit dimensions. <u>Angular frequency occurs, when an object vibrates (either moving harmoniously / oscillating or moving in a circle)</u>. Angular frequency can be roughly interpreted as the magnitude of the change in angle (in units of rad) per unit time. So, based on this understanding, the angular frequency can be calculated using the equation :

\boxed{\sf{\bold{\omega = \frac{\theta}{t}}}}

With the following condition :

  • \sf{\omega} = angular frequency (rad/s)
  • \sf{\theta} = change of angle value (rad)
  • t = interval of the time (s)

<h3>Problem Solving</h3>

We know that :

  • \sf{\theta} = change of angle value = 1,000 revolution = 1,000 × 2π rad = 2,000π rad/s >> Remember 1 rev = 2π rad/s.
  • t = interval of the time = 54.9 s.

What was asked :

  • \sf{\omega} = angular frequency = ... rad/s

Step by step :

\sf{\omega = \frac{\theta}{t}}

\sf{\omega = \frac{2,000 \pi}{54.9}}

\boxed{\sf{\omega \approx 36.43 \pi \: rad/s}}

<h3>Conclusion :</h3>

So, the angular frequency of the blades approximately 36.43π rad/s.

8 0
2 years ago
Gasoline burns in the cylinder of an automobile engine. During the combustion reaction, the production of gas forces the piston
serg [7]

Answer:

\Delta U = 1640 J

Explanation:

As we know by first law of thermodynamics that for ideal gas system we have

Heat given = change in internal energy + Work done

so here we will have

Heat given to the system = 2.2 kJ

Q = 2200 J

also we know that work done by the system is given as

W = 560 J

so we have

\Delta U = Q - W

\Delta U = 2200 - 560

\Delta U = 1640 J

6 0
3 years ago
To pull an old stump out of the ground, you and a friend tie two ropes to the stump. You pull on it with a force of 500 N to the
zhannawk [14.2K]

Answer:

C. less than 950 N.

Explanation:

Given that

Force in north direction F₁ = 500 N

Force in the northwest F₂ = 450 N

Lets take resultant force R

The angle between force = θ

θ = 45°

The resultant force R

R=\sqrt{F_1^2+F_2^2+2F_1F_2cos\theta}

R=\sqrt{500^2+450^2+2\times 450\times 500\times cos\theta}

R= 877.89 N

Therefore resultant force is less than 950 N.

C. less than 950 N

Note- When these two force will act in the same direction then the resultant force will be 950 N.

8 0
3 years ago
Question 14 (1 point)
Karo-lina-s [1.5K]

Answer:

car travel

precipitation

O temperature

Explanation:

Jet streams which is the ability of the object to move at a high speed due to its power is common among some given set of objects. Some are powered by the objects fuel while others are entirely different.

The above given options are actually affected by the jet streams.

5 0
3 years ago
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