All categories

3 years ago

Express the kinetic energy K in terms of the potential energy U. K=GMm/2R

Physics

1 answer:

max2010maxim [7]3 years ago

7 0

Answer:

K = -½U

Explanation:

From Newton's law of gravitation, the formula for gravitational potential energy is;

U = -GMm/R

Where,

G is gravitational constant

M and m are the two masses exerting the forces

R is the distance between the two objects

Now, in the question, we are given that kinetic energy is;

K = GMm/2R

Re-rranging, we have;

K = ½(GMm/R)

Comparing the equation of kinetic energy to that of potential energy, we can derive that gravitational kinetic energy can be expressed in terms of potential energy as;

K = -½U

You might be interested in

How can a cyclist minimize friction as he or she rides? A. by increasing the weight of the bike B. by increasing the tread on hi

jek_recluse [69]

B I believe is the answer!

Hope this helps and have a great day!!!

6 0

3 years ago

Read 2 more answers

A brave but inadequate rugby player is being pushed backward by an opposing player who is exerting a force of 800 N on him. The

uysha [10]

Answer:

f = 692 N

Explanation:

given data:

f =800N

a =1.2 m s^{2}

m= 90 kg

from newton's second law

net force $F_{net} =\sum F = F_1 +F_2 +..... = ma$

therefore we have from above equation $F_{net} = F -f = ma$

ma =F - f

putting all value to get force of friction

1.2*90 = 800 - f

f = 692 N

8 0

3 years ago

3. A football is kicked with a speed of 35 m/s at an angle of 40°.

jarptica [38.1K]

a) 22.5 m/s

The initial vertical velocity is given by:

$u_y = u sin \theta$

where

u = 35 m/s is the initial speed

$\theta=40^{\circ}$ is the angle of projection of the ball

Substituting into the equation, we find

$u_y = (35)(sin 40)=22.5 m/s$

b) 26.8 m/s

The initial horizontal velocity is given by:

$u_x = u cos \theta$

where

u = 35 m/s is the initial speed

$\theta=40^{\circ}$ is the angle of projection of the ball

Substituting into the equation, we find

$u_x = (35)(cos 40)=26.8 m/s$

c) 2.30 s

The time it takes for the ball to reach the maximum heigth can be found by considering the vertical motion only. This is a uniformly accelerated motion (free-fall), so we can use the suvat equation

$v_y = u_y + at$

where

$v_y$ is the vertical velocity at time t

$u_y = 22.5 m/s$

$a=g=-9.8 m/s^2$ is the acceleration of gravity (negative because it is downward)

At the maximum height, the vertical velocity becomes zero, $v_y =0$ ; substituting, we find the time t at which this happens:

$0=u_y + gt\\t=-\frac{u_y}{g}=-\frac{22.5}{-9.8}=2.30 s$

d) 25.8 m

The maximum height can also be found by considering the vertical motion only. We can use the following suvat equation:

$s=u_y t + \frac{1}{2}gt^2$

where

s is the vertical displacement at time t

$u_y = 22.5 m/s$

$g=-9.8 m/s^2$

Substituting t = 2.30 s, we find the displacement at maximum height, so the maximum height:

$s=(22.5)(2.30)+\frac{1}{2}(-9.8)(2.30)^2=25.8 m$

e) 123.3 m

In order to find how far does the ball lands, we have to consider the horizontal motion.

First of all, the time it takes for the ball to go back to the ground is twice the time needed for reaching the maximum height:

$t=2(2.30 s)=4.60 s$

Then, we consider the horizontal motion. There is no acceleration along this direction, so the horizontal velocity is constant:

$v_x = 26.8 m/s$

Therefore, the horizontal distance travelled during the whole motion is

$d=v_x t = (26.8)(4.60)=123.3 m$

So, the ball lands 123.3 m far from the initial point.

4 0

3 years ago

The answer would be ?

Cerrena [4.2K]

Answer:

D. demand; increased

Explanation:

Demand is how much people want it.

4 0

3 years ago

At what height is an object that weighs 490 newtons if its gravitational potential energy is 4900 J?

Sergio [31]

Answer:

The height at which the object is moved is 10 meters.

Explanation:

Given that,

Force acting on the object, W = F = 490 N

The gravitational potential energy, P = 4900 J

We need to find the height at which the object is moved. We know that the gravitational potential energy is possessed due to its position. It is given by :

$P=mgh\\\\h=\dfrac{P}{mg}\\\\h=\dfrac{4900\ J}{490\ N}\\\\h=10\ m$

So, the height at which the object is moved is 10 meters. Hence, this is the required solution.

5 0

3 years ago