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seropon [69]
2 years ago
8

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 e is equal to 1 for these surfaces. Part A Find the radius RRigel of the star Rigel, the bright blue star in the constellation Orion that radiates energy at a rate of 2.7×1031W and has a surface temperature of 11,000 K. Assume that the star is spherical. Use σ=5.67×10−8W/m2⋅K4 for the Stefan-Boltzmann constant and express your answer numerically in meters to two significant figures. RRigel = 5.1×1010 m SubmitHintsMy AnswersGive UpReview Part Correct This is over 50 times the size of our own sun and about a third of the orbital radius of the earth around the sun. Rigel is an example of a supergiant star. Part B Find the radius RProcyonB of the star Procyon B, which radiates energy at a rate of 2.1×1023W and has a surface temperature of 10,000 K. Assume that the star is spherical. Use σ=5.67×10−8W/m2⋅K4 for the Stefan-Boltzmann constant and express your answer numerically in meters to two significant figures. RProcyonB = 1.61•1011 m
Physics
1 answer:
valkas [14]2 years ago
4 0

Answer:

A) 5.1*10^10m B) 5.4*10^6m

Explanation:

Using the surface radiation formula of P (energy per second in Watt) = emissivity constant * surface area * Stefan-boltzmann constant*Temperature in kelvin^4*

2.7*10^31 = 1* 5.67*10^-8*A*11000^4

Making A subject of the formula = 2.7*10^31/ (5.67*10^-8*1.46*10^16) = 0.3261*10^23m^2

Since the shape is a sphere, the surface area = 4πR^2(radius of the Rigel)

R = √ (0.3261*10^23/ 4*π) = 5.1 * 10^ 10m

B) repeating the same step

2.1 *10^23 = 1*A*5.67*10^-8*10000^4 where A is the surface area of the Procyon

Make A subject of the formula

A = 2.1*10^23/(5.67*10^-8*10^16)

A = 0.37*10^15

The star assumed to be a sphere;

A = 4πR^2 where R is the radius of Procyon

R = √(0.37*10^15/4π) = 5.4*10^6m

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How does climate change lead to an increase in algal blooms? a. Decreased temperatures lead to an increase in phytoplankton grow
asambeis [7]

Answer;

B. Increased levels of carbon dioxide, a greenhouse gas, leads to increased phytoplankton growth.

Explanation;

-A combination of warm water, high nutrient levels, and adequate sunlight may cause a harmful algae bloom. These blooms may damage aquatic ecosystems by blocking sunlight and depleting oxygen that other organisms need to survive.

-Algae blooms have been increasing globally, and climate change may be playing a role in the increment. For instance, during the warm summer season or when water is warmer, some harmful types of algae to grow faster than other, more benign varieties.

-Additionally, the warmer surface water also prevents water from mixing vertically, allowing algae to grow thicker and faster.

4 0
2 years ago
Read 2 more answers
An electron starts from rest 3.00 cm from the center of a uniformly charged sphere of radius 2.00 cm. if the sphere carries a to
Sidana [21]
Answer:
velocity = 7.26 * 10^6 m/sec

Explanation:
The rule that is used to solve this problem is shown in the attached image.
The variables are as follows:
k = 8.99 * 10^9 Nm^2 / C^2
e is the electron charge = -1.6 * 10^-19 C
q is the charge given = 1 * 10^-9 C
m is the mass of the electron = 9.11 * 10^-31
r1 is the radius of starting point = 3 cm = 0.03 m
r2 is the radius of the sphere = 2 cm = 0.02 m
Substitute with the givens in the equation to get the value of the velocity

Hope this helps :)

7 0
2 years ago
The small piston of a hydraulic lift has a cross-sectional of 3 00 cm2 and its large piston has a cross-sectional area of 200 cm
Nesterboy [21]

Q: The small piston of a hydraulic lift has a cross-sectional of 3.00 cm2 and its large piston has a cross-sectional area of 200 cm2. What downward force of magnitude must be applied to the small piston for the lift to raise a load whose weight is Fg = 15.0 kN?

Answer:

225 N

Explanation:

From Pascal's principle,

F/A = f/a ...................... Equation 1

Where F = Force exerted on the larger piston, f = force applied to the smaller piston, A = cross sectional area of the larger piston, a = cross sectional area of the smaller piston.

Making f the subject of the equation,

f = F(a)/A ..................... Equation 2

Given: F = 15.0 kN = 15000 N, A = 200 cm², a = 3.00 cm².

Substituting into equation 2

f = 15000(3/200)

f = 225 N.

Hence the downward force that must be applied to small piston = 225 N

8 0
2 years ago
A farm hand does 972 J of work pulling an empty hay wagon along level ground with a force of 310 N [23° below the horizontal]. T
irina1246 [14]

Answer:

Option d)

Solution:

As per the question:

Work done by farm hand, W_{FH} = 972J

Force exerted, F' = 310 N

Angle, \theta = 23^{\circ}

Now,

The component of force acting horizontally is F'cos\theta

Also, we know that the work done is the dot or scalar product of force and the displacement in the direction of the force acting on an object.

Thus

W_{FH} = \vec{F'}.\vec{d}

972 = 310\times dcos23^{\circ}

d = 3.406 m = 3.4 m

3 0
2 years ago
A 4.0-m-diameter playground merry-go-round, with a moment of inertia of 350 kg⋅m2 is freely rotating with an angular velocity of
Flauer [41]

Answer:

v = 4.375\,\frac{m}{s}

Explanation:

The situation of the system Ryan - merry-go-round is modelled after the Principle of the Angular Momentum Conservation:

(350\,kg\cdot m^{2})\cdot (1.5\,\frac{rad}{s} ) - (2\,m)\cdot (60\,kg)\cdot v = 0\,kg\cdot \frac{m^{2}}{s}

The initial speed of Ryan is:

v = 4.375\,\frac{m}{s}

5 0
2 years ago
Read 2 more answers
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