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alexira [117]
3 years ago
6

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.
Physics
1 answer:
Rufina [12.5K]3 years ago
6 0
This assumption is correct. The emissivity of a substance is the ratio of the energy radiated from a body to the energy radiated from a black body (perfect emitter). Because these stars reflect negligible radiation compared to the amount they emit.
You might be interested in
Par 1/2
BaLLatris [955]

part 1

mass = ρ x V

mass = 1739 kg/m³ x 3.8 km³ = 6608.2 kg

PE (potential energy)= mgh

PE = 6608.2 kg x 9.81 x 403

PE = 2.61 x 10⁷ J

part 2

megaton of TNT (Mt) =4.2 x 10¹⁵ J

convert PE to Mt:

2.61 x 10⁷ J : 4.2 x 10¹⁵ J = 6.21 x 10⁻⁹ Mt

4 0
3 years ago
A 30-cm long string, with one end clamped and the other free to move transversely, is vibrating in its second harmonic. The wave
Ann [662]

Answer:

\lambda = 40 cm

Explanation:

given data

string length = 30 cm

solution

we take here equation of length that is

L = n \times \frac{1}{4} \lambda     ...............1

so

total length will be here

L = \frac{\lambda}{2} +  \frac{\lambda}{4}\\

L = \frac{3 \lambda }{4}

so \lambda  will be

\lambda = \frac{4L}{3}\\\lambda = \frac{4\times 30}{3}

\lambda = 40 cm

5 0
3 years ago
When an ice cube is placed in a glass of warm water, how are the signs and values q for the ice and the warm water related, assu
Valentin [98]

Answer:

q(ice) = -q(warm water)

Explanation:

Ice removes energy if ice is put in warm water and ice retains the very same amount of energy. Thus the water temperature heat sign is negative, and the ice heat sign is positive.

3 0
3 years ago
The light rays in the illustration below do not properly focus at the focal point. this problem occurs with?
tankabanditka [31]

Answer:

The problem occurs with all spherical mirrors.

Spherical mirrors are practical up to about inches in diameter.

Reflecting telescopes use spherical mirrors for apertures up to about 4 ".

Larger aperture telescopes use parabolic mirrors to obtain sharp focus.

7 0
3 years ago
A force of 1.00 x 10^2 pounds acts at an angle of 60.0° to the x-axis. (The force is accurate to 3 significant figures.) What ar
tamaranim1 [39]

Answer:

Fx= 50.0 Pounds : Components of the force along the x-axis

Fy= 86.6 Pounds : Component of the force along the y-axis

Explanation:

Conceptual Analysis

To find the components (Fx, Fy) of the total force (F), we apply the trigonometric concepts for a right triangle, where the perpendicular sides of the triangle are the components (Fx, Fy) of the force (F), the hypotenuse (h) is the magnitude of the total force F and β is the angle that forms the horizontal component with the hypotenuse.

Formulas

cos β : x/h  :    x: side adjacent to the β angle  h: hypotenuse  (1)

sin β = y/h  :    y: side opposite to the β angle  h: hypotenuse  (2)

Known Data

Known data

F= 1.00 * 10² pounds  = 100 pounds :  magnitude of total force

β =  60.0° to the x-axis. : Angle that forms the force with the x-axis

Problem Development

We apply the formula 1 to calculate horizontal component (Fx)

cos β :Fx/F

Fx= F cosβ  = 100*cos 60° = 50.0 Pounds

We apply the formula 2 to calculate vertical component (Fy)

sin β = Fy/F

Fy= F sinβ = 100*sin 60° = 86.6 Pounds

6 0
4 years ago
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