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schepotkina [342]
3 years ago
12

Is the star with the longest total lifetime Also the farthest from earth? Explain.

Physics
1 answer:
Stels [109]3 years ago
6 0
E) No. Ollie will shine for 30 Billion years but is only 10,000 LY from Earth.
F) No. Cosmo will shine for 3 Million years but is 10 Billion LY from Earth.
G) Yes. Ollie is only 10.000 LY away but will shine for 30 Billion years.
Ga) No. Stars such as Cosmo shine for 3 Million years.
Gb) If Cosmo was also 3 Million LY away we would see it now.
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A large fraction of the ultraviolet (UV) radiation coming from the sun is absorbed by the atmosphere. The main UV absorber in ou
Sergeeva-Olga [200]

Answer:

The wavelength of the radiation absorbed by ozone is 319.83 nm

Explanation:

Given;

frequency of absorbed ultraviolet (UV) radiation, f = 9.38×10¹⁴ Hz

speed of the absorbed ultraviolet (UV) radiation, equals speed of light, v = 3 x 10⁸ m/s

wavelength of the absorbed ultraviolet (UV) radiation, λ = ?

Apply wave equation for speed, frequency and wavelength;

v = fλ

λ = v / f

λ = (3 x 10⁸) / (9.38×10¹⁴)

λ = 3.1983 x 10⁻⁷ m

λ = 319.83 x  10⁻⁹ m

λ = 319.83 nm

Therefore, the wavelength of the radiation absorbed by ozone is 319.83 nm

5 0
3 years ago
The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by
Eduardwww [97]

Complete Question:

A 10 kg block is pulled across a horizontal surface by a rope that is oriented at 60° relative to the horizontal surface.

The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by the tension in the rope if the block when the block is 5 m away from its starting point? The coefficient of kinetic friction between the block and the floor is 0.2 and you may assume that the block starting at rest.

Answer:

Power = 54.07 W

Explanation:

Mass of the block = 10 kg

Angle made with the horizontal, θ = 60°

Distance covered, d = 5 m

Tension in the rope, T = 40 N

Coefficient of kinetic friction, \mu = 0.2

Let the Normal reaction = N

The weight of the block acting downwards = mg

The vertical resolution of the 40 N force, f_{y} = 40sin \theta

\sum f(y) = 0

N + 40 sin \theta - mg = 0\\N = -40sin60 + 10*9.81 = 0\\N = 63.46 N

\sum f(x) = 0

40 cos 60 - f_{r} - ma = 0\\ f_{r} = \mu N\\ f_{r} = 0.2 * 63.46\\ f_{r} = 12.69 N\\40cos 60 - 12.69-10a = 0\\7.31 = 10a\\a = 0.731 m/s^{2}

v^{2}  = u^{2} + 2as\\u = 0 m/s\\v^{2}  =  2 * 0.731 * 5\\v^{2}  = 7.31\\v = \sqrt{7.31} \\v = 2.704 m/s

Power, P = Fvcos \theta

P = 40 *2.704 cos60\\P = 54.074 W

7 0
3 years ago
Help for brainlist easy science plz
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Answer:

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