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Alex787 [66]
3 years ago
15

The hot glowing surfaces of stars emit energy in the form of electromagnetic radiation. It is a good approximation to assume tha

t the emissivity eee is equal to 1 for these surfaces.
Required:
a. Find the radius RRigel of the star Rigel, the bright blue star in the constellation Orion that radiates energy at a rate of 2.7 x 10^31 W and has a surface temperature of 11,000 K.
b. Find the radius RProcyonB of the star Procyon B, which radiates energy at a rate of 2.1 x 10^23 W and has a surface temperature of 10,000 K. Assume both stars are spherical. Use σ=5.67 x 10−8^ W/m^2*K^4 for the Stefan-Boltzmann constant.
Physics
1 answer:
belka [17]3 years ago
8 0

Given that,

Energy H=2.7\times10^{31}\ W

Surface temperature = 11000 K

Emissivity e =1

(a). We need to calculate the radius of the star

Using formula of energy

H=Ae\sigma T^4

A=\dfrac{H}{e\sigma T^4}

4\pi R^2=\dfrac{H}{e\sigma T^4}

R^2=\dfrac{H}{e\sigma T^4\times4\pi}

Put the value into the formula

R=\sqrt{\dfrac{2.7\times10^{31}}{1\times5.67\times10^{-8}\times(11000)^4\times 4\pi}}

R=5.0\times10^{10}\ m

(b). Given that,

Radiates energy H=2.1\times10^{23}\ W

Temperature T = 10000 K

We need to calculate the radius of the star

Using formula of radius

R^2=\dfrac{H}{e\sigma T^4\times4\pi}

Put the value into the formula

R=\sqrt{\dfrac{2.1\times10^{23}}{1\times5.67\times10^{-8}\times(10000)^4\times4\pi}}

R=5.42\times10^{6}\ m

Hence, (a). The radius of the star is 5.0\times10^{10}\ m

(b). The radius of the star is 5.42\times10^{6}\ m

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The angle at which a wave strikes a surface is ___ the angle at which the wave is reflected.
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Answer:

a)equal to

Explanation:

properties of a wave

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5 0
3 years ago
A 26.0 g ball is fired horizontally with initial speed v0 toward a 110 g ball that is hanging motionless from a 1.10 m -long str
mart [117]

Answer:

7.3 ms^{-1}

Explanation:

Consider the motion of the ball attached to string.

In triangle ABD

Cos50 = \frac{AB}{AD} \\Cos50 = \frac{AB}{L}\\AB = L Cos50

height gained by the ball is given as

h = BC = AC - AD \\h = L - L Cos50\\h = 1.10 - 1.10 Cos50\\h = 0.393 m

M  = mass of the ball attached to string = 110 g

V = speed of the ball attached to string just after collision

Using conservation of energy

Potential energy gained = Kinetic energy lost

Mgh = (0.5) M V^{2} \\V = sqrt(2gh)\\V = sqrt(2(9.8)(0.393))\\V = 2.8 ms^{-1}

Consider the collision between the two balls

m  = mass of the ball fired = 26 g

v_{o} = initial velocity of ball fired before collision = ?

v_{f} = final velocity of ball fired after collision = ?

using conservation of momentum

m v_{o} = MV + m v_{f}\\26 v_{o} = (110)(2.8) + 26 v_{f}\\v_{f} = v_{o} - 11.85

Using conservation of kinetic energy

m v_{o}^{2} = MV^{2} + m v_{f}^{2} \\26 v_{o}^{2} = 110 (2.8)^{2} + 26 (v_{o} - 11.85)^{2} \\v_{o} = 7.3 ms^{-1}

3 0
3 years ago
A pitcher throws an overhand fastball from an approximate height of 2.65 m and at an angle of 2.5° below horizontal. The catcher
rodikova [14]

Answer:

The initial velocity of the pitch is approximately 36.5 m/s

Explanation:

The given parameters of the thrown fastball are;

The height at which the pitcher throws the fastball, h₁ = 2.65 m

The angle direction in which the ball is thrown, θ = 2.5° below the horizontal

The height above the ground the catcher catches the ball, h₂ = 1.02 m

The distance between the pitcher's mound and the home plate = 18.5 m

Let 'u' represent the initial velocity of the pitch

From h = u_y·t + 1/2·g·t², we have;

u_y = The vertical velocity = u·sin(θ) = u·sin(2.5°)

h = 2.65 m - 1.02 m = 1.63 m

uₓ·t = u·cos(θ) = u·cos(2.5°) × t = 18.5 m

∴ t = 18.5 m/(u·cos(2.5°))

∴ h = u_y·t + 1/2·g·t² =  (u·sin(2.5°))×(18.5/(u·cos(2.5°))) + 1/2·g·t²

1.63 = 8.5·tan(2.5°) + 1/2 × 9.8 × t²

t² = (1.63 - 8.5·tan(2.5°))/(1/2 × 9.8) = 0.25691469087

t = √(0.25691469087) ≈ 0.50686752763

t ≈ 0.50686752763 seconds

u = 18.5 m/(t·cos(2.5°)) = 18.5 m/(0.50686752763 s × cos(2.5°)) = 36.5334603 m/s ≈ 36.5 m/s

The initial velocity of the pitch = u ≈ 36.5 m/s.

3 0
3 years ago
A sealed test tube traps 25.0 cm3 of air at a pressure of 1.00 atm and temperature of 18°C. The test tube’s stopper has a diamet
puteri [66]

Answer:

180° C

Explanation:

First we start by finding the area of the stopper.

A = πd²/4, where d = 1.5 cm = 0.015 m

A = 3.142 * 0.015² * ¼

A = 1.767*10^-4 m²

Next we find the force on the stopper

F = (P - P•)A, where

F = 10 N

P = pressure inside the tube,

P• = 1 atm

10 = (P - 101325) * 1.767*10^-4

P - 101325 = 10/1.767*10^-4

P - 101325 = 56593

P = 56593 + 101325

P = 157918 Pascal

Now, remember, in an ideal gas,

P1V1/T1 = P2V2/T2, where V is constant, then we have

P1/T1 = P2/T2, and when we substitute the values, we have

101325/(273 + 18) = 157918/ T2

101325/291 = 157918/ T2

T2 = (157918 * 291)/101325

T2 = 453 K

T2 = 453 - 273 = 180° C

3 0
4 years ago
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