Answer:
Red shift supports the big bang theory. ... The light from distant galaxies is red shifted (this tells us the galaxies are moving away from us) and the further away the galaxy the greater the red shift (this tells us that the more distant the galaxy the faster it is moving). Constellations look like they are moving because earth is rotating on it's axis.
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Answer:
vpg = 0.064 N
Explanation:
Upthrust = Volume of fluid displaced
upthrust liquid on the cube g=10ms−2
vpg =0.2 x 0.2 x 0.2 x0.8 x 10= 0.064N
vpg = 0.064 N
hope it helps.
Answer: Electrons move around the nucleus in fixed orbits of equal levels of energy
Explanation:
The statement that accurately represents the arrangement of electrons in Bohr’s atomic model is that the electrons move around the nucleus in fixed orbits of equal levels of energy.
It should be noted that the electrons have a fixed energy level when they travel around the nucleus in with energies which varies for different levels.
Higher energy levels are depicted by the orbits that are far from the nucleus. There's emission of light when the electrons then return back to a lower energy level.
the friction force provided by the brakes is 30000 N.
<h3>What is friction force?</h3>
Friction force is the force that opposes the motion between two bodies in contact.
To calculate the average friction force provided by the brakes, we apply the formula below.
Formula:
- K.E = F'd............. Equation 1
Where:
- K.E = Kinetic energy of the train
- F' = Friction force provided by the brakes
- d = distance
Make F' the subject of the equation
- F' = K.E/d............ Equation 2
From the question,
Given:
Substitute these values into equation 2
- F' = (8.1 ×10⁶)/270
- F' = 30000 N
Hence, the friction force provided by the brakes is 30000 N
Learn more about friction force here: brainly.com/question/13680415
<span>Radius = 4.6 m
Time for one complete rotation t = 5.5 s.
Distance = 2 x 3.14 x R = 2 x 3.14 x 4.6 m = 28.888.
Velocity V = distance / time = 28.888 / 5.5 s = 5.25 m/s
Force exerted by cat Fc = mV^2 / R = (mx 5.25^2) / 4.6 m
Force of the cat Fc = 6m, m being the mass.
Normal force = Us x m x g = Us x m x 9.81 = Us9.81m
equating the both forces => Us9.81m = 6m => Us = 6 / 9.81 => Us = 0.6116
So coefficient of static friction = 0.6116</span>
