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In-s [12.5K]
3 years ago
15

1. The gravitational force acting on a falling body and its weight is constant. But the law of universal gravitation tells us th

at the gravitational force on a body increases as it gets closer to Earth’s center. Is there a contradiction here?
Physics
1 answer:
vagabundo [1.1K]3 years ago
3 0
It's not so much a "contradiction" as an approximation. Newton's law of gravitation is an inverse square law whose range is large. It keeps people on the ground, and it keeps satellites in orbit and that's some thousands of km. The force on someone on the ground - their weight - is probably a lot larger than the centripetal force keeping a satellite in orbit (though I've not actually done a calculation to totally verify this). The distance a falling body - a coin, say - travels is very small, and over such a small distance gravity is assumed/approximated to be constant.
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A 60 kg skydiver is falling at a terminal velocity of 50 m/s.
marishachu [46]

Answer:

The gravitational force is definitely acting downwards towards the ground and this is equal to the weight of the skydiver.

the acceleration a = 7.8 m/s²

Explanation:

Given that :

the mass of the skydiver = 60 kg

Velocity = 50 m/s

Thus;  gravitational force is definitely acting downwards towards the ground and this is equal to the weight of the skydiver.

Also; the air resistance is acting upward and the resultant of both forces = mass×acceleration

So;

mg-R = ma

60(9.8) - 120 = 60(a)

588 -120 = 60a

468 = 60a

a = \frac{468}{60}

a = 7.8 m/s²

Hence, the acceleration a = 7.8 m/s²

5 0
3 years ago
A charge Q exerts a 1.2 N force on another charge q. If the distance between the charges is doubled, what is the magnitude of th
Anon25 [30]

Answer:

0.3 N

Explanation:

Electromagnetic force is F= Kq1q2/r^2, where r is the distance between charges. If r is doubled then the force will be 1/4F which is 0.3 N.

8 0
3 years ago
Consider an elevator carrying Kermit the frog weighing 4000.0 N is held 5.00 m above a spring with a force constant of
tigry1 [53]

Answer:

The maximum compression distance of the spring is 0.375 m

Explanation:

The given parameters are;

The weight of the elevator and the frog = 4,000.0 N

The location of the elevator above the spring = 5.00 m

The force constant of the spring, k = 8,000.0 N/m

The frictional force of the brakes = 1,000.0 N

The net force, F, of the elevator on the spring is F = 4,000.0 N - 1,000.0 N = 3,000.0 N

F = 3,000.0 N

The maximum compression distance of the spring, x is given as follows;

F = k × x

Substituting the known values gives;

3,000.0 N = 8,000.0 N/m × x

∴ x = (3,000.0 N)/(8,000.0 N/m) = 0.375 m

x = 0.375 m = 37.5 cm

The maximum compression distance of the spring, x = 0.375 m.

4 0
3 years ago
A conveyor belt at a recycling plant launches bottles and bottle caps into the air, so that an automatic image recognition devic
dem82 [27]

(a) The initial speed at which the bottles are launched is 4.27 m/s.

(b) The horizontal displacement at which the bottle land is 1.75 m.

<h3>Initial speed of the bottle</h3>

The initial speed of the bottle is calculated as follows;

T = \frac{2usin\theta}{g}

where;

  • T is time of flight
  • u is the initial speed

2usinθ = Tg

u = Tg/(2sinθ)

u = (0.5 x 9.8)/(2 x sin35)

u = 4.27 m/s

<h3>Horizontal displacement of the bottle</h3>

X = u²sin(2θ)/g

X = (4.27² x sin(70))/(9.8)

X = 1.75 m

Learn more about projectile here: brainly.com/question/12870645

#SPJ1

6 0
2 years ago
How would you eliminate the unit in this conversion? 5 mil to m
m_a_m_a [10]

Answer:

https://youtu.be/9iulv2QvKwo

3 0
3 years ago
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