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Sonbull [250]
3 years ago
12

What do astonomers use to calculate the age of the universe

Physics
1 answer:
makvit [3.9K]3 years ago
3 0

Answer: Stars

Explanation: Some stars have remained intact since the Big Bang. These stars are shown in something called the HR diagram with portrays the brightness and largeness of the star. Also, white dwarfs and black dwarfs can be traced back which may correspond with the initiation of the universe.

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Which "spheres" are interacting when water evaporates from plants
Novay_Z [31]
The plant grows in the solid part of earth, the lithosphere. When water evaporates from the plant, it enters the hydrosphere, the portion if earth on kand and in the air that contains water. The atmosphere is part of the hydrosphere.
8 0
3 years ago
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An object with an acceleration of –10 m/s² is _____.
Sidana [21]

The answer might be B because the value is negative, and negative could mean slowing down.

7 0
3 years ago
Calculate the energy of the green light emitted, per photon, by a mercury lamp with a frequency of 5.49 × 1014 hz.
Tcecarenko [31]
The energy of a photon is given by
E=hf
where
h=6.6 \cdot 10^{-34} Js is the Planck constant
f is the frequency of the photon

In our problem, the frequency of the light is 
f=5.49 \cdot 10^{14}Hz
therefore we can use the previous equation to calculate the energy of each photon of the green light emitted by the lamp:
E=hf=(6.6 \cdot 10^{-34}Js)(5.49 \cdot 10^{14} Hz)=3.62 \cdot 10^{-19} J
8 0
3 years ago
A 2.00-kg rock has a horizontal velocity of magnitude 12.0 m>s when it is at point P in Fig. E10.35. (a) At this instant, wha
zaharov [31]

Answer:

(A) L = 115.3kgm²/s

(B) dL/dt = 94.1kgm²/s²

Explanation:

The magnitude of the angular momentum of the rock is given by the foemula

L = mvrSinθ

We have been given θ = 36.9°, m = 2.0kg, v = 12.0m/s and r = 8.0m.

Therefore L = 2.00 × 12 × 8.0 × Sin 36.9° =

115.3 kgm²/s

(B) The magnitude of the rate of angular change in momentum is given by

dL /dt = d(mvrSinθ)/dt = mgrSinθ = 2.00 × 9.8 × 8.0× Sin36.9 = 94.1kgm²/s²

7 0
3 years ago
The quantity of charge q (in coulombs) that has passed through a surface of area 2.00 cm2 varies with time
Makovka662 [10]

Explanation:

We have,

Surface area, A=2\ cm^2=0.0002\ m^2

The current varies wrt time t as :

q(t) = 4t^3 + 5t + 6

(a) At t = 2 seconds, electrical charge is given by :

q(t) = 4t^3 + 5t + 6\\\\q(2) = 4(2)^3 + 5(2) + 6\\\\q=48\ C

(b) Current is given by :

I=\dfrac{dq}{dt}\\\\I=\dfrac{d(4t^3 + 5t + 6)}{dt}\\\\I=12t^2+5

Instantaneous current at t = 1 s is,

I=12(1)^2+5=17\ A

(c) Current is, I=12t^2+5

Current density is given by electric current per unit area.

J=\dfrac{I}{A}\\\\J=\dfrac{(12t^2+5)}{0.0002}\\\\J=5000(12t^2+5)\ A/m^2

Therefore, it is the required explanation.

7 0
3 years ago
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