Calculate the work done on a Pressure / Volume (PV) isothermal expansion where 400 Pa and 0.08 volume in m3?

If the star Sirius emits 23 times more energy than the Sun, why does the Sun appear brighter in the sky?

Ganezh [65]

Answer:

As b ∝ (L/r²) and

the distance of the sun from the earth is 0.00001581 light years

and

the distance of the Sirius from the earth is 8.6 light years

hence,

the Sun appear brighter in the sky

Explanation:

The brightness (b) is directly proportional to the Luminosity of the star (L) and inversely proportional to the square of the distance between the star and the observer (r).

thus, mathematically,

b ∝ (L/r²)

now,

given

L for sirius is 23 times more than the sun i.e 23L

now,

the distance of the sun from the earth is 0.00001581 light years

and

the distance of the Sirius from the earth is 8.6 light years

thus,

using the the relation between conclude that the value of brightness for the Sirius comes very very low as compared to the value for brightness for the Sun.

hence, the sun appears brighter

5 0

3 years ago

What are beats? A. periodic fluctuations in the velocity of sound waves B. periodic fluctuations in the wavelength of sound wave

777dan777 [17]

The answer is

C. periodic fluctuations in the intensity if sound waves.

3 0

3 years ago

Read 2 more answers

The cabinet is mounted on coasters and has a mass of 45 kg. The casters are locked to prevent the tires from rotating. The coeff

stira [4]

Answer:

the force P required for impending motion is 132.3 N

the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m

Explanation:

Given that:

mass of the cabinet m = 45 kg

coefficient of static friction μ = 0.30

A free flow body diagram illustrating what the question represents is attached in the file below;

The given condition from the question let us realize that ; the casters are locked to prevent the tires from rotating.

Thus; considering the forces along the vertical axis ; we have :

$\sum f_y =0$

The upward force and the downward force is :

$N_A+N_B = mg$

where;

$\mathbf { N_A \ and \ N_B}$ are the normal contact force at center point A and B respectively .

$N_A+N_B = 45*9.8$

$N_A+N_B = 441$ ------- equation (1)

Considering the forces on the horizontal axis:

$\sum f_x = 0$

$F_A +F_B = P$

where ;

$\mathbf{ F_A \ and \ F_B }$ are the static friction at center point A and B respectively.

which can be written also as:

$\mu_s N_A + \mu_s N_B = P$

$\mu_s( N_A + N_B) = P$

replacing our value from equation (1)

P = 0.30 ( 441)

P = 132.3 N

Thus; the force P required for impending motion is 132.3 N

b) Since the horizontal distance between the casters A and B is 480 mm; Then half the distance = 480 mm/2 = 240 mm = 0.24 cm

the largest value of "h" allowed for the cabinet is not to tip over is calculated by determining the limiting condition of the unbalanced torque whose effect is canceled by the normal reaction at $N_A$ and it is shifted to $N_B$ :