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Paladinen [302]
4 years ago
10

A binary star system in the constellation orion has an angular separation of 10-5 rad. if the wavelength of the light from the s

ystem is 500 nm, what is the smallest aperture (diameter) telescope that can just resolve two stars?
Physics
1 answer:
Dahasolnce [82]4 years ago
3 0
The answer is: Diameter = d = 6.1 cm

Explanation:
Angular separation between the two stars = Q = 10^-5 rad.
Wavelength =λ= 500 * 10^-9 m
The smallest diameter = d = ?

As Q = 1.22*(λ/ d)

From this
d = 1.22*(λ/Q) --- (1)

Plug in the value in (1):
(1) => d = 1.22*((500*10^-9)/10^-5))
d = 0.061 m
d = 6.1 cm

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What are the five basic postulates of kinetic-molecular theory?
lesantik [10]

Explanation:

The five basic postulate of kinetic molecular theory includes:

1) All gases consist of large amount and numbers of tiny particles that are far apart from each other and also relative to their size.

2) The collisions between gas particles and gas particles against container walls is refer to as  elastic collision.

3) All gas particles are in a continuous random and rapid motion. They possess kinetic energy which is energy of motion.

4) There are no attractive force between gas particles.

5) The temperature of a gas depends on the average kinetic energy of the gas particle.

8 0
3 years ago
Train cars are coupled together by being bumped into one another. Suppose two loaded cars are moving toward one another, the fir
tia_tia [17]

Answer:

7560 Joules

Explanation:

m_1 = Mass of first car = 1.5\times 10^5\ kg

m_2 = Mass of second car = 2\times 10^5\ kg

u_1 = Initial Velocity of first car = 0.3 m/s

u_2 = Initial Velocity of second car = -0.12 m/s

v = Velocity of combined mass

As linear momentum of the system is conserved

m_1u_1 + m_2u_2 =(m_1 + m_2)v\\\Rightarrow v=\frac{m_1u_1 + m_2u_2}{m_1 + m_2}\\\Rightarrow v=\frac{1.5\times 10^5\times 0.3 + 2\times 10^5\times -0.12}{1.5\times 10^5 + 2\times 10^5}\\\Rightarrow v=0.06\ m/s

Energy lost is

\Delta E=\Delta E_i-\Delta E_f\\\Rightarrow \Delta=\frac{1}{2}(m_1u_1^2 + m_2u_2^2-(m_1+m_2)v^2)\\\Rightarrow \Delta=\frac{1}{2}(1.5\times 10^5\times 0.3^2 + 2\times 10^5\times (-0.12)^2-(1.5\times 10^5 + 2\times 10^5)\times 0.06^2)\\\Rightarrow \Delta=7560\ J

The Energy lost in the collision is 7560 Joules

7 0
3 years ago
An Earth satellite is in a circular orbit at an altitude of 500 km.
Dafna1 [17]

Answer:

Speed Unchanged.

Explanation:

As work is defined as a product of force over a distance. If the distance in altitude is constant = 500km, there's 0 change in distance and force, no work would be done by the gravitational force.

Since potential energy of the satellite is unchanged, unless there's additional internal energy source, the kinetic energy would remain unchanged, so would its speed.

8 0
3 years ago
A rigid adiabatic container is divided into two parts containing n1 and n2 mole of ideal gases respectively, by a movable and th
kicyunya [14]

Answer:

Explanation:

Given

Pressure, Temperature, Volume of gases is

P_1, V_1, T_1 & P_2, V_2, T_2

Let P & T be the final Pressure and Temperature

as it is rigid adiabatic container  therefore Q=0 as heat loss by one gas is equal to heat gain by another gas

-Q=W+U_1----1

Q=-W+U_2-----2

where Q=heat loss or gain (- heat loss,+heat gain)

W=work done by gas

U_1 & U_2 change in internal Energy of gas

Thus from 1 & 2 we can say that

U_1+U_2=0

n_1c_v(T-T_1)+n_2c_v(T-T_2)=0

T(n_1+n_2)=n_1T_1+n_2T_2

T=\frac{n_1+T_1+n_2T_2}{n_1+n_2}

where n_1=\frac{P_1V_1}{RT_1}

n_2=\frac{P_2V_2}{RT_2}

T=\frac{\frac{P_1V_1}{RT_1}\times T_1+\frac{P_2V_2}{RT_2}\times T_2}{\frac{P_1V_1}{RT_1}+\frac{P_2V_2}{RT_2}}

T=\frac{P_1V_1+P_2V_2}{\frac{P_1V_1}{T_1}+\frac{P_2V_2}{T_2}}

and P=\frac{P_1V_1+P_2V_2}{V_1+V_2}

6 0
3 years ago
If an object is moving up, is it considered to have a positive or negative velocity?
grin007 [14]

Answer:

A positive velocity

Explanation:

7 0
3 years ago
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