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2 years ago

Can anyone help me out with conservation of energy

Physics

1 answer:

Keith_Richards [23]2 years ago

3 0

Answer:

Whenever energy gets transformed, the total

energy remains unchanged. This is the law of

conservation of energy. According to this law,

energy can only be converted from one form

to another; it can neither be created nor

destroyed. The total energy before and after

the transformation remains the same.

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If earth had no atmosphere, would a falling object ever reach terminal velocity?

stepan [7]

No, because terminal velocity is when the acceleration of the Earth’s gravity is balanced by the air resistance of the atmosphere.

5 0

3 years ago

1) On the way to the moon, the Apollo astro-

kramer

(1) You must find the point of equilibrium between the two forces,

G * MTms / (R−x)^2 = G * MLms / x^2
MT / (R-x)^2 = ML / x^2

So,

x = R * sqrt(ML * MT) - ML / (MT - ML)
R = is the distance between Earth and Moon.

The result should be,
x = 3.83 * 10^7m
from the center of the Moon, and

R - x = 3.46*10^8 m
from the center of the Earth.

(2) As the distance from the center of the Earth is the number we found before,
d = R - x = 3.46*10^8m
The acceleration at this point is
g = G * MT / d^2
g = 3.33*10^-3 m/s^2

6 0

2 years ago

If 34.7 g of O2 reacts with iron to form 79.34 g of iron oxide, how much iron was used in the reaction?

zhuklara [117]

Answer: B. 44.64 g

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

Mass of reactants = mass of iron + mass of oxygen = mass of iron + 34.7 g

Mass of product = mass of iron oxide = 79.34 g

As Mass of reactants = Mass of product

mass of iron + 34.7 g = 79.34 g

mass of iron = 44.64 g

Thus 44.64 g of iron was used in the reaction

6 0

3 years ago

A 1300-N crate rests on the floor. How much work is required to move it at constant speed (a)

kherson [118]

a) The work done is 920 J

b) The work done is 5200 J

Explanation:

In this first part of the problem, the crate is moved horizontally at constant speed.

The work required in this case is given by

$W=Fd cos \theta$

where

F is the magnitude of the force applied

d is the displacement of the crate

$\theta$ is the angle between the direction of the force and of the displacement

Here the crate is moved at constant speed: this means that the acceleration of the crate is zero, and so according to Newton's second law, the net force on the crate is zero: this means that the force applied, F, must be equal to the force of friction (but in opposite direction), so

F = 230 N

The displacement is

d = 4.0 m

And the angle is $\theta=0^{\circ}$ , since the force is applied horizontally. Therefore, the work done is

$W=(230)(4.0)(cos 0^{\circ})=920 J$

In this case, the crate is moved vertically. The force that must be applied to lift the crate must be equal to the weight of the crate (in order to move it a constant speed), therefore

F = W = 1300 N

The displacement this time is again

d = 4.0 m

And the angle is $\theta=0^{\circ}$ , since the force is applied vertically, and the crate is moved also vertically. Therefore, the work done on the crate this time is

$W=(1300)(4.0)(cos 0^{\circ})=5200 J$