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3 years ago

Planet A and planet B are in circular orbits around a distant star.

Physics

1 answer:

evablogger [386]3 years ago

8 0

Answer:

$\frac{v_A}{v_B}=0.402$

Explanation:

The speed that planets must have in order for their orbit to be stable, is given by:

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

Here v It is called orbital speed, G is the gravitational constant, M is the mass of the star and r is the radius of the orbit. In this case we have:

$r_A=6.2r_B$

So, the ratio of their speed is:

$\frac{v_A}{v_B}=\frac{\sqrt{\frac{GM}{r_A}}}{\sqrt{\frac{GM}{r_B}}}\\\frac{v_A}{v_B}=\sqrt{\frac{r_B}{r_A}}\\\frac{v_A}{v_B}=\sqrt{\frac{r_B}{6.2r_B}}\\\frac{v_A}{v_B}=\sqrt{\frac{1}{6.2}}\\\frac{v_A}{v_B}=0.402$

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A tennis player smashes a ball of mass m horizontally at a vertical wall. The ball rebounds at the same speed v with which it st

Morgarella [4.7K]

Answer:

B.0

Explanation:

Change in momentum: This is defined as the product of mass and change in velocity of a body. or it can be defined as the product of force and time of a body. The fundamental unit of change in momentum is kg.m/s

Change in momentum = M(V-U)......................... Equation 1

where M = mass of the ball, V = final velocity of the ball, U = initial velocity of the ball.

Let: M = m kg and V = U = v m/s

Substituting these values into equation 1

Change in momentum = m(v-v)

Change in momentum = m(0)

Change in momentum = 0 kg.m/s

Therefore the momentum of the ball has not changed.

The right option is B.0

5 0

3 years ago

A sample of monatomic ideal gas occupies 5.00 L at atmospheric pressure and 300 K (point A). It is warmed at constant volume to

leonid [27]

Answer:

(a) 0.203 moles

(b) 900 K

(d) 15 L

(e) A → B, W = 0, Q = Eint = 1,518.91596 J

B → C, W = Q ≈ 1668.69974 J Eint = 0 J

C → A, Q = -2,531.5266 J, W = -1,013.25 J, Eint = -1,518.91596 J

(g) ∑Q = 656.089 J, ∑W = 655.449 J, ∑Eint = 0 J

Explanation:

At point A

The volume of the gas, V₁ = 5.00 L

The pressure of the gas, P₁ = 1 atm

The temperature of the gas, T₁ = 300 K

At point B

The volume of the gas, V₂ = V₁ = 5.00 L

The pressure of the gas, P₂ = 3.00 atm

The temperature of the gas, T₂ = Not given

At point C

The volume of the gas, V₃ = Not given

The pressure of the gas, P₃ = 1 atm

The temperature of the gas, T₂ = T₃ = 300 K

(a) The ideal gas equation is given as follows;

P·V = n·R·T

Where;

P = The pressure of the gas

V = The volume of the gas

n = The number of moles present

R = The universal gas constant = 0.08205 L·atm·mol⁻¹·K⁻¹

n = PV/(R·T)

∴ The number of moles, n = 1 × 5/(0.08205 × 300) ≈ 0.203 moles

The number of moles in the sample, n ≈ 0.203 moles

(b) The process from points A to B is a constant volume process, therefore, we have, by Gay-Lussac's law;

P₁/T₁ = P₂/T₂

∴ T₂ = P₂·T₁/P₁

From which we get;

T₂ = 3.0 atm. × 300 K/(1.00 atm.) = 900 K

The temperature at point B, T₂ = 900 K

(d) For a constant temperature process, according to Boyle's law, we have;

P₂·V₂ = P₃·V₃

V₃ = P₂·V₂/P₃

∴ V₃ = 3.00 atm. × 5.00 L/(1.00 atm.) = 15 L

The volume at point C, V₃ = 15 L

(e) The process A → B, which is a constant volume process, can be carried out in a vessel with a fixed volume

The process B → C, which is a constant temperature process, can be carried out in an insulated adjustable vessel

The process C → A, which is a constant pressure process, can be carried out in an adjustable vessel with a fixed amount of force applied to the piston

(f) For A → B, W = 0,

Q = Eint = n·cv·(T₂ - T₁)

Cv for monoatomic gas = 3/2·R

∴ Q = 0.203 moles × 3/2×0.08205 L·atm·mol⁻¹·K⁻¹×(900 K - 300 K) = 1,518.91596 J

Q = Eint = 1,518.91596 J

For B → C, we have a constant temperature process

Q = n·R·T₂·㏑(V₃/V₂)

∴ Q = 0.203 moles × 0.08205 L·atm/(mol·K) × 900 K × ln(15 L/5.00 L) ≈ 1668.69974 J

Eint = 0

Q = W ≈ 1668.69974 J

For C → A, we have a constant pressure process

Q = n·Cp·(T₁ - T₃)

∴ Q = 0.203 moles × (5/2) × 0.08205 L·atm/(mol·K) × (300 K - 900 K) = -2,531.5266 J

Q = -2,531.5266 J

W = P·(V₂ - V₁)

∴ W = 1.00 atm × (5.00 L - 15.00 L) = -1,013.25 J

W = -1,013.25 J

Eint = n·Cv·(T₁ - T₃)

Eint = 0.203 moles × (3/2) × 0.08205 L·atm/(mol·K) × (300 K - 900 K) = -1,518.91596 J

Eint = -1,518.91596 J

(g) ∑Q = 1,518.91596 J + 1668.69974 J - 2,531.5266 J = 656.089 J

∑W = 0 + 1668.69974 J -1,013.25 J = 655.449 J

∑Eint = 1,518.91596 J + 0 -1,518.91596 J = 0 J

5 0

3 years ago

In physics grade 10 number 8 in the pic how do we do it thanks in advance

Rom4ik [11]

Hope this helps you.

5 0

3 years ago

un litro de un gas es calentado a presión constante desde 20°C hasta 60°C que volumen final ocupará dicho gas?

WINSTONCH [101]

Answer:

Final volume, V2 = 3 Litres

Explanation:

Given the following data;

Initial volume, V1 = 1 litre

Initial temperature, T1 = 20°C

Final temperature, T2 = 60°C

To find the final volume, we would use Charles' law;

Charles states that when the pressure of an ideal gas is kept constant, the volume of the gas is directly proportional to the absolute temperature of the gas.

Mathematically, Charles is given by;

V1/T1 = V2/T2

Making V2 as the subject formula, we have;

V1T2 = V2T1

V2 = (V1T2)/T1

Substituting into the formula, we have;

V2 = (1 * 60)/20

V2 = 60/20

Final volume, V2 = 3 Litres

4 0

3 years ago

A common pickup for an electric guitar consists of a coil of wire around a small permanent magnet, as described in Figure 25.5.

Orlov [11]

Guitar pickups are tiny coils with magnets inside them and they work via the principal of electromagnetic inductance. The metal strings vibrate within the magnetic field of the pickup which generates an electrical current. Since nylon is not a metal, it will not cause fluctuations in that magnetic field. Hope this answer helps.

6 0

3 years ago