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Vesna [10]
3 years ago
7

A projectile of mass m is fired straight upward from the surface of an airless planet of radius R and mass M with an initial spe

ed equal to the escape speed vesc (meaning the projectile will just barely escape the planet's gravity -- it will asymptotically approach infinite distance and zero speed.) What is the correct expression for the projectile's kinetic energy when it is a distance 9R from the planet's center (8R from the surface). Ignore the gravity of the Sun and other astronomical bodies. KE (at r = 9R) is:a. GMm/9Rb. GMm/8Rc. 1/2mvesc^2d. -GMm/8Re. None of these
Physics
1 answer:
motikmotik3 years ago
6 0

Answer:

K = G Mm / 9R

Explanation:

Expression for escape velocity V_e = \sqrt{\frac{2GM}{R} }

Kinetic energy at the surface = 1/2 m V_e ²

= 1/2 x m x 2GM/R

GMm/R

Potential energy at the surface

= - GMm/R

Total energy = 0

At height 9R ( 8R from the surface )

potential energy

= - G Mm / 9R

Kinetic energy = K

Total energy will be zero according to law of conservation of mechanical energy

so

K  - G Mm / 9R = 0

K = G Mm / 9R

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A ball of mass 0.160 kg is dropped from a height of 2.25 m. When it hits the ground it compresses 0.087 m.
Studentka2010 [4]

A) 6.64 m/s downward

B) 0.026 s

C) -40.9 N

Explanation:

A)

We can solve this problem by using the law of conservation of energy.

In fact, since the total mechanical energy of the ball must be conserved, this means that the initial gravitational potential energy of the ball before the fall is entirely converted into kinetic energy just before it reaches the floor.

So we can write:

PE=KE\\mgh = \frac{1}{2}mv^2

where

m = 0.160 kg is the mass of the ball

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

h = 2.25 m is the initial height of the ball

v is the final velocity of the ball before hitting the ground

Solving for v, we find:

v=\sqrt{2gh}=\sqrt{2(9.8)(2.25)}=6.64 m/s

And the direction of the velocity is downward.

B)

The motion of the ballduring the collision is a uniformly accelerated motion (= with constant acceleration), so the time of impact can be found by using a suvat equation:

s=(\frac{u+v}{2})t

where:

v is the final velocity

u is the initial velocity

s is the displacement of the ball during the impact

t is the time

Here we have:

u = 6.64 m/s is the velocity of the ball before the impact

v = 0 m/s is the final velocity after the impact (assuming it comes to a stop)

s = 0.087 m is the displacement, as the ball compresses by 0.087 m

Therefore, the time of the impact is:

t=\frac{2s}{u+v}=\frac{2(0.087)}{0+6.64}=0.026 s

C)

The force exerted by the floor on the ball can be found using the equation:

F=\frac{\Delta p}{t}

where

\Delta p is the change in momentum of the ball

t is the time of the impact

The change in momentum can be written as

\Delta p = m(v-u)

So the equation can be rewritten as

F=\frac{m(v-u)}{t}

Here we have:

m = 0.160 kg is the mass of the ball

v = 0 is the final velocity

u = 6.64 m/s is the initial velocity

t = 0.026 s is the time of impact

Substituting, we find the force:

F=\frac{(0.160)(0-6.64)}{0.026}=-40.9 N

And the sign indicates that the direction of the force is opposite to the direction of motion of the ball.

4 0
3 years ago
PART ONE
Lina20 [59]

Explanation:

Make a table, listing the x and y coordinates of each square's center of gravity and its mass.  Multiply the coordinates by the mass, add the results for each x and y, then divide by the total mass.

\left\begin{array}{ccccc}x&y&m&xm&ym\\\frac{a}{2} &\frac{a}{2} &10&5a&5a\\\frac{3a}{2}&\frac{a}{2}&70&105a&35a\\\frac{a}{2}&\frac{3a}{2}&80&40a&120a\\\frac{3a}{2}&\frac{3a}{2}&50&75a&75a\\&\sum&210&225a&235a\\&&Avg&\frac{15a}{14}&\frac{47a}{42}\end{array}\right

The x-coordinate of the center of gravity is 15/14 a.

The y-coordinate of the center of gravity is 47/42 a.

4 0
3 years ago
Question: A 0.50 m wave has a frequency of 686 Hz. What is the speed of the wave?
Dafna1 [17]
C. 340 
Frequency is the number of wavelengths per second and since the length is 0.5 you multiply 0.5*686 and get 343.
the question not allowing for one position thus the answer is c

7 0
3 years ago
If the change in velocity increases, what happens to the acceleration during the same time period?
4vir4ik [10]
When the velocity is increasing the acceleration increases too
4 0
3 years ago
Read 2 more answers
Is erosion a constructive or destructive force and why?
Andrei [34K]
Destructive force
It’s a force that slowly wears away mountains and other features on earths surface, such as erosion.
5 0
3 years ago
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