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The centripetal force acting on a satellite in orbit

krok68 [10]

The centripetal force (of gravity) on a satellite in orbit is an
unbalanced force (because there's no equal force pulling
the satellite away from Earth), changes the direction of the
satellite (into a closed orbit instead of a straight line), and
always acts toward the center of whatever curve the satellite
happens to be on at the moment.

8 0

3 years ago

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Dimension of radius of sphere

Readme [11.4K]

Answer:

The dimension is L

Explanation:

Dimension analysis is a method of representing quantities majorly with respect to some fundamental quantities of mass (M), length (L), time (T).

A sphere has a definite volume which relates to its radius by:

V = $\frac{4}{3}$ $\pi$ $r^{3}$

In this equation $\pi$ is a dimensionless quantity, and the unit of v is $m^{3}$ .

But, metre is a measure of length, thus it has a dimension of L.

So that,

$m^{3}$ ≅ $L^{3}$

Then,

$L^{3}$ = $r^{3}$

Find the cube root of both sides to have,

r = L

Therefore, the dimension of the radius of a sphere is L.

5 0

3 years ago

Water moves through a constricted pipe in steady, ideal flow. At the

Irina-Kira [14]

A) Speed in the lower section: 0.638 m/s

B) Speed in the higher section: 2.55 m/s

C) Volume flow rate: $1.8\cdot 10^{-3} m^3/s$

Explanation:

A)

To solve the problem, we can use Bernoulli's equation, which states that

$p_1 + \rho g h_1 + \frac{1}{2}\rho v_1^2 = p_2 + \rho g h_2 + \frac{1}{2}\rho v_2^2$

where

$p_1=1.75\cdot 10^4 Pa$ is the pressure in the lower section of the tube

$h_1 = 0$ is the heigth of the lower section

$\rho=1000 kg/m^3$ is the density of water

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

$v_1$ is the speed of the water in the lower pipe

$p_2$ is the pressure in the higher section

$h_2 = 0.250 m$ is the height in the higher pipe

$v_2$ is hte speed in the higher section

We can re-write the equation as

$v_1^2-v_2^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}$ (1)

Also we can use the continuity equation, which state that the volume flow rate is constant:

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-section of the lower pipe, with

$r_1 = 3.00 cm =0.03 m$ is the radius of the lower pipe (half the diameter)

$A_2 = \pi r_2^2$ is the cross-section of the higher pipe, with

$r_2 = 1.50 cm = 0.015 m$ (radius of the higher pipe)

So we get

$r_1^2 v_1 = r_2^2 v_2$

And so

$v_2 = \frac{r_1^2}{r_2^2}v_1$ (2)

Substituting into (1), we find the speed in the lower section:

$v_1^2-(\frac{r_1^2}{r_2^2})^2v_1^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}\\v_1=\sqrt{\frac{2(p_2-p_1+\rho g h_2)}{\rho(1-\frac{r_1^4}{r_2^4})}}=0.638 m/s$

B)

Now we can use equation (2) to find the speed in the lower section:

$v_2 = \frac{r_1^2}{r_2^2}v_1$

Substituting

v1 = 0.775 m/s

And the values of the radii, we find:

$v_2=\frac{0.03^2}{0.015^2}(0.638)=2.55 m/s$

C)

The volume flow rate of the water passing through the pipe is given by

$V=Av$

where

A is the cross-sectional area

v is the speed of the water

We can take any point along the pipe since the volume flow rate is constant, so

$r_1=0.03 cm$

$v_1=0.638 m/s$

Therefore, the volume flow rate is

$V=\pi r_1^2 v_1 = \pi (0.03)^2 (0.638)=1.8\cdot 10^{-3} m^3/s$

Learn more about pressure in a liquid:

brainly.com/question/9805263

#LearnwithBrainly

0 0

2 years ago

As light from a star spreads out and weakens, do gaps form between the photons?

Sati [7]

Answer:

There are no gaps in space between the photons as they travel.

Yes, you can form the shadow of a fire

Explanation:

8 0

2 years ago

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A bowling ball accidentally falls out of the cargo bay of an airliner as it flies along in a horizontal direction. As seen from

valina [46]

The path the bowling ball would most closely follow after leaving the airplane is horizontal direction.

<h3>Path of the bowling ball</h3>

Based on the law of inertia, which is the reluctance of an object to stop moving once in motion or start moving when it is at rest.

The bowling ball will maintain the path of the airline in the first few seconds of fall, after which it will change its path to vertical direction.

Thus, the path the bowling ball would most closely follow after leaving the airplane is horizontal direction.

Learn more about horizontal direction here: brainly.com/question/2534565

#SPJ1

3 0

2 years ago