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iris [78.8K]
3 years ago
12

Which of Newton’s laws, when combined algebraically with his law of universal gravitation, can be used to calculate Earth’s mass

?
A) first
B) second
C) third
Physics
1 answer:
andriy [413]3 years ago
4 0

B) second law of motion which states that F=ma

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A railroad car is pulled through the distance of 960 m by a train that did 578 kJ of work during this pull.
Wittaler [7]

Answer:

<h2>602.08 N</h2>

Explanation:

The force supplied by the train can be found by using the formula

f =   \frac{w}{d}  \\

w is the workdone

d is the distance

From the question we have

f =  \frac{578000}{960}  \\  = 602.083333...

We have the final answer as

<h3>602.08 N</h3>

Hope this helps you

7 0
3 years ago
"My distance to the center of the earth is about 4000 miles when I am on the surface. If I go to a height of 8000 miles above th
lord [1]

Given,

Distance from the surface to the center of the earth, d=4000 miles

Distance from the center to you at a height of 8000 miles, a= 8000+4000=12000 miles

The gravitational force acting on a person at the surface is equal to his weight.

From Newton's Universal Law of Gravitation, the gravitational force is

F=\frac{G\times M\times m}{r^2}

Where G is the gravitational constant, M is the mass of the earth, m is the mass of the object/person, r is the distance between the center of the earth and the object/person

At the surface, this force is equal to the weight of the person, W=mg

i.e.

F_s=\frac{G\times M\times m}{d^2}=W

On substituting the of d,

W=\frac{\text{GMm}}{4000^2}

At a height of 8000 miles from the surface, the gravitational force is equal to,

F_a=\frac{GMm}{12000^2}

On dividing the above two equations,

\frac{F_a}{W}=\frac{4000^2^{}}{12000^2}=\frac{1}{9}

Therefore,

F_a=\frac{1}{9}W

Therefore at a height of 8000 miles above the surface of the earth, the force of gravity becomes 1/9 time your weight.

5 0
1 year ago
Does anyone taste the difference between left and right twix?
4vir4ik [10]

I personally don't (I can't speak for others tho) but I say right twix is better for the memes

3 0
3 years ago
.A hard rubber ball, released at chest height, falls to the pavement and bounces back to nearly the same height. When it is in c
ohaa [14]

Answer:

 a = 1.1 10⁵ m / s²

Explanation:

This is a momentum exercise, where we use the relationship between momentum and momentum

          I = ∫ F dt = Δp

= p_f - p₀

as they indicate that the ball bounces at the same height, we can assume that the moment when it reaches the ground is equal to the moment when it bounces, but in the opposite direction

        F t = 2 (m v)

therefore the average force is

         F = 2 m v / t

where in general the mass of the ball unknown, the velocity of the ball can be calculated using the conservation of energy

starting point. Done the ball is released with zero initial velocity

        Em₀ = U = mgh

final point. Upon reaching the ground, just before the deformation begins

        Em_f = K = ½ m v²

energy is conserved in this system

        Em₀ = Em_f

        m g h = ½ m v²

        v = √ (2gh)

This is the velocity of the body when it reaches the ground, so the force remains

        F = 2m √(2gh)   /t

where the height of the person's chest is known and the time that the impact with the floor lasts must be estimated in general is of the order of milli seconds

knowing this force let's use Newton's second law

          F = m a

          a = F / m

 

          a = 2 √(2gh) / t

We can estimate the order of magnitude of this acceleration, assuming the person's chest height of h = 1.5 m and a collision time of t = 1 10⁻³ s

         a = 2 √ (2 9.8 1.5) / 10⁻³

         a = 1.1 10⁵ m / s²

6 0
3 years ago
How much heat energy is required to raise the temperature of 5 kilograms of coal from 20°C to 220°C? A. 314 J B. 6,573 J C. 1,31
Vika [28.1K]
Just took the test and the answer is <span>C. 1,314,718.

</span>
3 0
3 years ago
Read 2 more answers
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