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ivanzaharov [21]

2 years ago

11

Most stars are main-sequence stars, a group of stars for which size, mass, surface temperature, and radiated power are closely r

elated. The sun, for instance, is a yellow main sequence star with a surface temperature of 5800 K. For a main-sequence star whose mass M is more than twice that of the sun, the total radiated power, relative to the sun, is approximately P/Psun = 1.5(M/Msun)3.5. The star Regulus A is a bluish main-sequence star with mass 3.8Msun and radius 3.1Rsun. What is the surface temperature of Regulus A?

1 answer:

quester [9]2 years ago

4 0

Answer:

the surface temperature of Regulus A is 11724.13 K

Explanation:

Given the data in the question;

Sun's surface temperature T = 5800 K

total radiated power, relative to the sun is; P/P $_{sun$ = $($ 1.5(M/M $_{sun$ $)^{3.5$

The star Regulus A is a bluish main-sequence star with mass 3.8M $_{sun$ and radius 3.1R $_{sun$ .

First, we determine the value power emitted by the sun or sun as follows;

P = eσAT⁴

where P is the power, e is surface emissivity, σ is Stefan Boltzmann, A is area and T is temperature.

so, lets assume emissivity of star and sun is same;

let p be power related to star and p $_{sun$ be power related sun.

Ratio of power radiated by star and power radiated by sun;

P/P $_{sun$ = eσAT⁴ / eσA $_{sun$ T $_{sun$ ⁴

we know that AREA A = πR²

we input the formula for area

P/P $_{sun$ = eσ(πR²)T⁴ / eσ(π(R $_{sun$ )²)T $_{sun$ ⁴

such that we now have;

P/P $_{sun$ = R²T⁴ / R $_{sun$ ²T $_{sun$ ⁴

given that P/P $_{sun$ = $($ 1.5(M/M $_{sun$ $)^{3.5$ , we substitute

$($ 1.5(M/M $_{sun$ $)^{3.5$ = R²T⁴ / R $_{sun$ ²T $_{sun$ ⁴

we find temperature of the star T

T = 5800 × $[$ 1.5 $($ M/M $_{sun$ $)^{3.5$ (R $_{sun$ ²/R²) $]^{1/4$

Given that; mass M is 3.8M $_{sun$ and radius R is 3.1R $_{sun$ .

we substitute

T = 5800 × $[$ 1.5 $($ 3.8M $_{sun$ /M $_{sun$ $)^{3.5$ (R $_{sun$ ²/( 3.1R $_{sun$ )²) $]^{1/4$

T = 5800 × $[$ 1.5 $($ 3.8 $)^{3.5$ ( 1/( 3.1)²) $]^{1/4$

T = 5800 × $[$ 1.5( 106.9652 ) ( 1/(9.61) $]^{1/4$

T = 5800 × $[$ 16.69592 $]^{1/4$

T = 5800 × 2.02140152

T = 11724.13 K

Therefore, the surface temperature of Regulus A is 11724.13 K

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An object of mass 6 kg. is resting on a horizontal surface. A horizontal force

son4ous [18]

Answer:

a) The work done by the applied force is 1500 joules.

b) The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) 300 joules of energy are lost during motion.

Explanation:

a) Since the object has a constant mass, on which a constant horizontal force is exerted. The work done by the force ( $W$ ), measured in joules, is defined by the following expression:

$W = F\cdot \Delta x$ (1)

Where:

$F$ - Force, measured in newtons.

$\Delta x$ - Distance, measured in meters.

If we know that $F = 15\,N$ and $\Delta x = 100\,m$ , then the work done by the force exerted on the object is:

$W = (15\,N)\cdot (100\,m)$

$W = 1500\,J$

The work done by the applied force is 1500 joules.

b) At first we need to calculate the net acceleration of the object ( $a$ ), measured in meters per square second. By assuming a constant acceleration, we use the following kinematic formula:

$\Delta x = v_{o}\cdot t +\frac{1}{2}\cdot a\cdot t^{2}$ (2)

Where $v_{o}$ is the initial velocity of the object, measured in meters per second.

We clear the acceleration within the equation above:

$\frac{1}{2}\cdot a \cdot t^{2} = \Delta x-v_{o}\cdot t$

$a = \frac{2\cdot (\Delta x - v_{o}\cdot t)}{t^{2}}$

If we know that $\Delta x = 100\,m$ , $v_{o} = 0\,\frac{m}{s}$ and $t = 10\,s$ , then the net acceleration experimented by the object is:

$a = \frac{2\cdot \left[100\,m-\left(0\,\frac{m}{s} \right)\cdot (10\,s)\right]}{(10\,s)^{2}}$

$a = 2\,\frac{m}{s^{2}}$

By the 2nd Newton's Law, we construct the following equation of equilibrium under the consideration of a friction force acting against the motion of the object:

$\Sigma F = F - f = m\cdot a$ (3)

Where:

$F$ - External force exerted on the object, measured in newtons.

$f$ - Kinetic friction force, measured in newtons.

If we know that $F = 15\,N$ , $m = 6\,kg$ and $a = 2\,\frac{m}{s^{2}}$ , the kinetic friction force is:

$f = F-m\cdot a$

$f = 15\,N-(6\,kg)\cdot \left(2\,\frac{m}{s^{2}} \right)$

$f = 3\,N$

The work done by friction ( $W'$ ), measured in joules, is:

$W' = f\cdot \Delta x$ (4)

$W' = (3\,N) \cdot (100\,m)$

$W' = 300\,J$

And the net work experimented by the object is:

$\Delta W = 1500\,J - 300\,J$

$\Delta W = 1200\,J$

By the Work-Energy Theorem we understand that change in translational kinetic energy ( $\Delta K$ ), measured in joules, is equal to the change in net work. That is:

$\Delta K = \Delta W$ (5)

If we know that $\Delta W = 1200\,J$ , then the change in translational kinetic energy is:

$\Delta K = 1200\,J$

The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) The energy lost by the object is equal to the work done by the force of friction. Therefore, 300 joules of energy are lost during motion.

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3 years ago

Instruments in an airplane which is in level flight indicate that the velocity relative to the air (airspeed) is 180.00 km/h and

snow_lady [41]

Answer:

Explanation:

From the given information:

The coordinate axis is situated in the east and north direction.

So, the north will be the y-axis and the east will be the x-axis

Similarly, the velocity of the plane in regard to the air in the coordinate system will be $v_{P/A} = v( cos \theta \ i + sin \theta \ j)$

where:

$v_{P/A}$ = velocity of the plane in regard to the air

v = velocity

θ = angle of inclination of the plane with respect to the horizontal

replacing v = 180 km/ and θ = 20° in above equation, then:

The velocity of the airplane in the coordinate system as:

$v_{P} = v_o( cos \phi \ i + sin \phi \ j)$

where;

$v_p$ = velocity of the airplane

$v_o$ = velocity

∅ = angle of inclination with regard to the base axis;

Then; replacing $v_o$ = 150 km/h and ∅ = 30°

Therefore, the velocity of the plane in the system is :

$v_p = v_A + v_{P/A}$

$v_A= v_P -v_{P/A}$ --- (1)

$v_A=$ ( 150 cos 30° - 180 cos 20°)i + ( 150 sin 30° - 180sin 20°)j

$v_A=$ (-39.24 km/h)i + (13.44 km/h) j

The magnitude is:

$v_A= (-39.24 km/h)i + (13.44 km/h) j$

$|v_A|^2 = \sqrt{ (-39.24 km/h)^2+ (13.44 km/h)^2}$

$v_A$ = 41.48 km/h

The airplane is moving at an angle of the inverse tangent to the abscissa and ordinate.

The angle of motion is:

tan θ = 39.24/13.44

tan θ = 2.9

θ = $tan ^{-1} (2.9)$

θ = 70.97°

The angle of motion is 70.97° from west of north with a velocity of 41.48 km/h.

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