A wave front has the form of a

With "normal" gravity, we used a potential energy of mgh. Now with the gravity that is more accurate over longer distances we us

Artemon [7]

Answer:

A general solution is $\Delta U=mh\frac{GM}{r^{2}}\frac{r}{r+\Delta h}$ and a particualr case is mgh, it is just to distance around the radius Earth.

Explanation:

We can use a general equation of the potential energy to understand the particular and general case:

The potential energy is defined as $U=-\int F\cdot dx$ , we know that the gravitational force is $F=GmM/r^{2}$ , so we could find the potential energy taking the integral of F.

$U=-GmM/r$ (1)

We can find the particular case, just finding the gravitational potential energy difference:

$\Delta U=U_{f}-U_{i}$ . Here Uf is the potential evaluated in r+Δh and Ui is the potential evaluated in r.

Using (1) we can calculate ΔU.

$\Delta U=-\frac{GmM}{r+\Delta h}+\frac{GmM}{r}$

Simplifying and combining terms we have a simplified expression.

$\Delta U=mh\frac{GM}{r^{2}}\frac{r}{r+\Delta h}$ (2)

Let's call $g=\frac{GM}{r^{2}}$ . It is the acceleration due to gravity on the Earth's surface, if r is the radius of Earth and M is the mass of the Earth and we can write (2) as ΔU=mgh, but if we have distance grader than r we should use (2), otherwise, we could get incorrect values of potential energy.

I hope i hleps you!

3 0

4 years ago

The mirror used in search light is:

dolphi86 [110]

Answer:

B

Explanation:

6 0

3 years ago

For a freely falling object weighing 3 kg : A. what is the object's velocity 2 s after it's release. B. What is the kinetic ener

Fed [463]

A) 19.6 m/s (downward)

B) 576 J

C) 19.6 m

D) Velocity: not affected, kinetic energy: doubles, distance: not affected

Explanation:

A)

An object in free fall is acted upon one force only, which is the force of gravity.

Therefore, the motion of an object in free fall is a uniformly accelerated motion (constant acceleration). Therefore, we can find its velocity by applying the following suvat equation:

$v=u+at$

where:

v is the velocity at time t

u is the initial velocity

$a=g=9.8 m/s^2$ is the acceleration due to gravity

For the object in this problem, taking downward as positive direction, we have:

$u=0$ (the object starts from rest)

$a=9.8 m/s^2$

Therefore, the velocity after

t = 2 s

is:

$v=0+(9.8)(2)=19.6 m/s$ (downward)

B)

The kinetic energy of an object is the energy possessed by the object due to its motion.

It can be calculated using the equation:

$KE=\frac{1}{2}mv^2$

where

m is the mass of the object

v is the speed of the object

For the object in the problem, at t = 2 s, we have:

m = 3 kg (mass of the object)

v = 19.6 m/s (speed of the object)

Therefore, its kinetic energy is:

$KE=\frac{1}{2}(3)(19.6)^2=576 J$

C)

In order to find how far the object has fallen, we can use another suvat equation for uniformly accelerated motion:

$s=ut+\frac{1}{2}at^2$

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

For the object in free fall in this problem, we have:

u = 0 (it starts from rest)

$a=g=9.8 m/s^2$ (acceleration of gravity)

t = 2 s (time)

Therefore, the distance covered is

$s=0+\frac{1}{2}(9.8)(2)^2=19.6 m$

D)

Here the mass of the object has been doubled, so now it is

M = 6 kg

For part A) (final velocity of the object), we notice that the equation that we use to find the velocity does not depend at all on the mass of the object. This means that the value of the final velocity is not affected.

For part B) (kinetic energy), we notice that the kinetic energy depends on the mass, so in this case this value has changed.

The new kinetic energy is

$KE'=\frac{1}{2}Mv^2$

where

M = 6 kg is the new mass

v = 19.6 m/s is the speed

Substituting,

$KE'=\frac{1}{2}(6)(19.6)^2=1152 J$

And we see that this value is twice the value calculated in part A: so, the kinetic energy has doubled.

Finally, for part c) (distance covered), we see that its equation does not depend on the mass, therefore this value is not affected.

5 0

3 years ago

What type of machine would most likely be used to move buckets of the mineral pieces up to the surface of the Earth? a wedge a p

inessss [21]

The type of machine would be a pulley.

3 0

3 years ago

All waves must travel up and down. True False

const2013 [10]

Answer:

True all waves that go up must come down

I feel like the follow the rule of gravity:everything that goes up must come d5

8 0

3 years ago