As thermal energy is added to a sample of water, the kinetic energy of its

use the general formulas for gravitational force and centripetal force to derive the relationship between speed and orbital radi

MrRissso [65]

Answer:

$v = \sqrt{\frac{GM}{r}}$

Explanation:

The universal law of gravitation is defined as:

$F = G\frac{Mm}{r^{2}}$ (1)

Where G is the gravitational constant, M and m are the masses of the two objects and r is the distance between them.

The centripetal force can be found by means of Newton's second law:

$F = ma$ (2)

Since it is a circular motion, the acceleration can be defined as:

$a = \frac{v^{2}}{r}$ (3)

Where v is the velocity and r is the orbital radius.

Replacing equation (3) in equation (2) it is gotten:

$F = m\frac{v^{2}}{r}$ (4)

Hence,

$m\frac{v^{2}}{r} = G\frac{Mm}{r^{2}}$

Then, v can be isolated:

$mv^{2} = G\frac{Mmr}{r^{2}}$

$mv^{2} = G\frac{Mm}{r}$

$v^{2} = G\frac{Mm}{mr}$

$v^{2} = \frac{GM}{r}$

$v = \sqrt{\frac{GM}{r}}$

So the relationship between speed and orbital radius is given by the expression $v = \sqrt{\frac{GM}{r}}$

8 0

3 years ago

. Find the buoyant force exerted on a ball with radius of 2 meters, when the ball is entirely immersed in the water.

almond37 [142]

Explanation:

Buoyancy force is equal to the weight of the displaced fluid:

B = ρVg

where ρ is the density of the fluid,

V is the volume of the displaced fluid,

and g is the acceleration due to gravity.

The fluid is water, so ρ = 1000 kg/m³.

The volume displaced is that of a sphere with radius 2 m:

V = 4/3 π r³

V = 4/3 π (2 m)³

V ≈ 33.5 m³

The buoyancy force is therefore:

B = (1000 kg/m³) (33.5 m³) (9.8 m/s²)

B ≈ 328,400 N

Round as needed.

8 0

3 years ago

(b) The figure below shows a model train, travelling at speed , approaching a

Andrew [12]

Answer:

0.62 m/s

Explanation:

Given that a model train, travelling at speed , approaching a buffer model train buffer spring

The train, of mass 2.5 kg, is stopped by compressing a spring in the buffer.

After the train has stopped, the energy stored in the spring is 0.48 J.

Calculate the initial speed v of the train

Solution

The total energy stored in the spring will be equal to the kinetic energy of the train.

That is,

1/2fe = 1/2mv^2

Substitutes the spring energy, and mass into the formula

0.48 = 1/2 × 2.5 × V^2

2.5V^2 = 0.96

V^2 = 0.96 / 2.5

V^2 = 0.384

V = sqrt ( 0.384 )

V = 0.62 m/s

Therefore, the initial velocity of the train is 0.62 metres per second.

3 0

3 years ago

Ayden and Steven are playing catch with a football. Ayden throws the football at a velocity of 15 m/s with an angle of 60° above

Pachacha [2.7K]

1) 1.33 s

2) 8.6 m

3) 19.9 m

Explanation:

1)

The motion of the football is a projectile motion, which consists of two independent motions:

- A uniform motion (=constant velocity) along the horizontal direction

- A uniformly accelerated motion (=constant acceleration) along the vertical direction

The hang time of the football is the time it takes for the football to reach its maximum height. It is given by:

$t=\frac{u_y}{g}$

where

$u_y = u sin \theta$ is the initial vertical velocity of the ball, where

u = 15 m/s is the initial velocity

$\theta=60^{\circ}$ is the angle of projection of the ball

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

Substituting, we find the hang time:

$t=\frac{u sin \theta}{g}=\frac{(15)(sin 60^{\circ})}{9.8}=1.33 s$

2)

The motion of the ball along the vertical direction is a uniformly accelerated motion, so we can use the suvat equation:

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

where:

s is the vertical displacement

$u_y = u sin \theta$ is the initial vertical velocity

t is the time

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

In this part, we want to find the maximum height, so the height reached by the ball when the time is

t = 1.33 s

Therefore, by substituting the values in the equation, we can find the maximum height:

$s=usin \theta t-\frac{1}{2}gt^2=(15)(sin 60^{\circ})(1.33)-\frac{1}{2}(9.8)(1.33)^2=8.6 m$

3)

Here we want to find the horizontal range covered by the ball during its flight.

The horizontal range for a projectile is given by the equation

$d=\frac{u^2 sin(2\theta)}{g}$

where

u is the initial velocity

$\theta$ is the angle of projection

g is the acceleration due to gravity

For the ball in this problem we have:

u = 15 m/s

$\theta=60^{\circ}$

$g=9.8 m/s^2$

Substituting into the equation, we find the horizontal distance covered by the ball:

$d=\frac{(15)^2sin(2\cdot 60^{\circ})}{9.8}=19.9 m$

6 0

4 years ago

You are going sledding with your friends, sliding down a snowy hill. Friction can't be ignored. Riding solo on your sled, you ha

Musya8 [376]

If you let a friend ride with you, increasing the mass does increase the net force on the system, but it also increases the inertia. a=Fnetm. Since both the net force and mass are increased they still cancel, leaving the acceleration the same.

<h3>Further explanation </h3>

Inertia is the tendancy of a moving object to keep moving in a straight line or of any object to resist a change in motion.

According to the picture below, we choose x-direction parallel to the inclined surface. Our positive direction is to the right. Hence, we will calculate the acceleration when riding solo first:

$\Sigma F_x = mg sin \theta - F_k = m a_x\\mg sin \theta - F_k = m a_x\\a_x = \frac{mg sin \theta - F_k }{m} ... (1)\\$