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2 years ago

Two satellites, A and B are in different circular orbits

Physics

1 answer:

jek_recluse [69]2 years ago

6 0

Answer:

The ratio of the orbital time periods of A and B is $\frac{1}{2}$

Solution:

As per the question:

The orbit of the two satellites is circular

Also,

Orbital speed of A is 2 times the orbital speed of B

$v_{oA} = 2v_{oB}$ (1)

Now, we know that the orbital speed of a satellite for circular orbits is given by:

$v_{o} = \farc{2\piR}{T}$

where

R = Radius of the orbit

Now,

For satellite A:

$v_{oA} = \farc{2\piR}{T_{a}}$

Using eqn (1):

$2v_{oB} = \farc{2\piR}{T_{a}}$ (2)

For satellite B:

$v_{oB} = \farc{2\piR}{T_{b}}$ (3)

Now, comparing eqn (2) and eqn (3):

$\frac{T_{a}}{T_{b}} = \farc{1}{2}$

You might be interested in

A juggler is practicing with one ball. It takes 2.4 seconds for the ball to leave her hand and return to her hand. How fast did

nikklg [1K]

The kinematics for the vertical launch we can enter the initial velocity is 11.76 m / s

Given parameters

Time t = 2.4 s

To find

Initial velocity

Kinematics is the part of physics that establishes the relationships between the position, velocity, and acceleration of bodies.

In this case we have a vertical launch

y = y₀ + v₀ t - ½ g t²

Where y and y₀ are the final and initial positions, respectively, v₀ the initial velocity, g the acceleration of gravity (g = 9.8 m / s²) and t the time

With the ball in hand, its position is zero

0 = 0 + v₀ t - ½ g t²

v₀ t - ½ g t² = 0

v₀ = ½ g t

Let's calculate

v₀ = ½ 9.8 2.4

v₀ = 11.76 m / s

In conclusion using kinematics for the vertical launch we can enter the initial velocity is 11.76 m / s

Learn more about vertical launch kinematics here:

brainly.com/question/15068914

5 0

1 year ago

A single insulated duct flow experiment using air operating at steady-state is performed in a lab. One measurement location (Sta

weqwewe [10]

Answer:

a) -0.0934 kJ/kg. K

b) The direction of flow is from right to left.

Explanation:

A free flow diagram of the horizontal insulated duct is as shown below.

NOW,

Let assume that the direction of flow is from left to right and consider the following relation for the entropy rate balance equation for a control volume as:

$\frac{\sigma_{cv}}{m}= (s_2-s_1) \geq 0 \ \ \ -------> \ \ \ 1$

Now; if the value for this relation is greater than zero; then we conclude that our assumption is correct.

If the value is less than zero; then we conclude that the assumption is wrong.

Then, the flow is said to be in the opposite direction

Formula for the change in specific entropy can be calculated as:

$s_2-s_1 = s^0(T_2) - s^0(T_1)-R \ In ( \frac{P_2}{P-1}) \ \ \ -------> \ \ \ 2$

where;

$s_1, s_2 , s^0(T_2), s^0(T1)$ are specific entropies

R = universal gas constant

$P_1$ = pressure at location 1

$P_2$ = pressure at location 2

We obtain the specific properties of air at temperature at $T_1$ = (67°C + 273)K = 340 K from the table A-22 ( Ideal gas properties of air)

$s^0(T1)$ = 1.8279 kJ/kg.K

We also obtain the specific properties of air at temperature $T_2$ = 22°C + 273) K = 295 K

From the table A- 22

$s^0(T_2)$ = 1.68515 kJ/kg . K

R = $\frac{8.314 kJ}{28.97 kg.K}$

$P_1 =$ 0.95 bar

$P_2 =$ 0.8 bar

Now replacing our values into equation (2) from above; we have;

$s_2-s_1 = s^0(T_2) -s^0(T_1)-R \ In (\frac{P_2}{P_1} )$

$s_2-s_1 = 1.68515 -1.8279-\frac{8.314}{28.97} \ In (\frac{0.8}{0.95} )$

$s_2-s_1 = 1.68515 -1.8279+ 0.0493$

$s_2-s_1 =-0.0934 \ kJ/kg.K$

Equating our result to equation (1)

$s_2-s_1 \geq 0\\-0.0934 \leq 0$

Therefore , our assumption is wrong and the direction of flow is said to be from right to left.

We therefore conclude that the direction of flow is from right to left.

3 0

2 years ago

Verify that the SI unit of impulse is the same as the SI unit of momentum.

lys-0071 [83]

Maybe this will help you out:

Momentum is calculate by the formula:

$P = mv$

Where:

P = momentum

m = mass

v = velocity

The SI unit:

$mass = kg\\ velocity = \dfrac{m}{s}$

So the unit of momentum would be:

$kg.\dfrac{m}{s}$

Impulse is defined as the change in momentum or how much force changes momentum. It can be calculate with the formula:

I = FΔt

where:

I = impulse

F = Force

Δt = change in time

The SI unit:

F = Newtons (N) or $kg.\dfrac{m}{s^{2} }$

t = Seconds (s)

So the unit of impulse would be derived this way:

I = FΔt

I = $kg.\dfrac{m}{s^{2} }$ x $s$

$\dfrac{kg.m.s}{s^{2}} = \dfrac{kg.m.s}{s.s}$

You can then cancel out one s each from the numerator and denominator and you'll be left with:

$kg.\dfrac{m}{s}$

So then:

Momentum: Impulse

$kg.\dfrac{m}{s}$ $kg.\dfrac{m}{s}$

4 0

2 years ago

How would a spinning disk's kinetic energy change if its moment of inertia was five times larger but its angular speed was five

Keith_Richards [23]

Answer:

The kinetic energy of a spinning disk will be reduced to a tenth of its initial kinetic energy if its moment of inertia is made five times larger, but its angular speed is made five times smaller.

Explanation:

Let us first consider the initial characteristics of the angular motion of the disk

moment of inertia = $I$

angular speed = ω

For the second case, we consider the characteristics to now be

moment of inertia = $5I$ (five times larger)

angular speed = ω/5 (five times smaller)

Recall that the kinetic energy of a spinning body is given as

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

therefore,

for the first case, the K.E. is given as

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

and for the second case, the K.E. is given as

$KE = \frac{1}{2}(5I)(\frac{w}{5} )^{2} = \frac{5}{50}Iw^{2}$

$KE = \frac{1}{10}Iw^{2}$

this is one-tenth the kinetic energy before its spinning characteristics were changed.

This implies that the kinetic energy of the spinning disk will be reduced to a tenth of its initial kinetic energy if its moment of inertia is made five times larger, but its angular speed is made five times smaller.

6 0

2 years ago

Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.57 A out of the jun

AlekseyPX

Answer:

a. 1.56 × 10¹⁸ electrons per second

b. The electrons in wire 3 flow into the junction.

Explanation:

Here is the complete question

Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.65 A out of the junction. (a) How many electrons per second move past a point in wire 3? (b) In which direction do the electrons move in wire 3 -- into or out of the junction?

Solution

(a) How many electrons per second move past a point in wire 3?

Using Kirchhoff's current law, at the junction, i₁ + i₂ + i₃ = 0 where i₁ = current in wire 1 = 0.40 A, i₂ = current in wire 2 = 0.65 A and i₃ = = current in wire 3,

So, i₃ = -(i₁ + i₂)

taking current flowing into the junction as positive and those leaving as negative, i₁ = + 0.40 A and i₂ = -0.65 A

So, i₃ = -(i₁ + i₂)

i₃ = -(0.40 A + (-0.65 A))

i₃ = -(0.40 A - 0.65 A)

i₃ = -(-0.25 A)

i₃ = 0.25 A

Since i₃ = 0.25 C/s and we have e = 1.602 × 10⁻¹⁹ C per electron, then the number of electrons flowing in wire 3 per second is i₃/e = 0.25 C/s ÷ 1.602 × 10⁻¹⁹ C per electron = 0.1561 × 10¹⁹ electrons per second = 1.561 × 10¹⁸ electrons per second ≅ 1.56 × 10¹⁸ electrons per second

(b) In which direction do the electrons move -- into or out of the junction?

Given that i₃ = + 0.25 A and that positive flows into the junction, thus, the electrons in wire 3 flow into the junction.

8 0

2 years ago