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RoseWind [281]
4 years ago
10

If you could help me out that would be fantastic.?!!!

Physics
1 answer:
Volgvan4 years ago
3 0
It's either C. Or B.
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How are stars formed and classified?
Romashka-Z-Leto [24]
Stars form from an accumulation of gas and dust, which collapses due to gravity and starts to form stars. Stars are typically classified by their spectrum in what is known as the Morgan-Keenan or MK system.
6 0
3 years ago
During a particular thunderstorm, the electric potential difference between a cloud and the ground is vcloud - vground = 3.50 10
wariber [46]
<span>The change in the electron's potential energy is equal to the work done on the electron by the electric field. The electron's potential energy is the stored energy relative to the electron's position in the electric field. Vcloud - Vground represents the change in Voltage. This voltage quantity is given to be 3.50 x 10^8 V, with the electron at the lower potential. The formula for calculating the change in the electron's potential energy (EPE) is found by charge x (Vcloud - Vground) = (EPEcloud - EPE ground) where charge is constant = 1.6 x 10^-19. Filling in the known quantities results in the expression 1.6 x 10^-19 (3.50 x 10^8) = (EPEcloud - EPEground) = 5.6 x 10^-11. Therefore, the change in the electron's potential energy from cloud to ground is 5.6 x 10^-11 joules.</span>
8 0
3 years ago
At each corner of a square of side l there are point charges of magnitude Q, 2Q, 3Q, and 4Q.What is the magnitude and direction
bogdanovich [222]

Answer:

magnitude of force on charge 2Q  = \frac{KQ^{2} }{I^{2} }

Direction of force on charge = 61 ⁰

Explanation:

The magnitude on the force on the charge can be evaluated by finding the net force acting on the charge 2Q  i.e x-component of the net force and the y-component of the net force

║F║ = \sqrt{f_{x}^{2} + f_{y}^{2}    }  =  after considering the forces coming from Q, 3Q and 4Q AND APPLYING COULOMBS LAW

magnitude of force acting on 2Q = \frac{KQ^{2} }{I^{2} }

The direction of the force on charge 2Q is calculated as

tan ∅ = \frac{f_{y} }{f_{x} } = 1.8284

therefore ∅ = tan^{-1}  1.8284

= 61⁰

3 0
3 years ago
A ball A of mass 0.5 kg moving with a Velacity of 10 m/s a head on Collision with a ball B of mass 2kg moving with a Velocity of
Nesterboy [21]

Answer:

The common velocity v after collision is 2.8m/s²

Explanation:

look at the attachment above ☝️

3 0
2 years ago
Near the surface of the earth, objects in free fall (but not terminal velocity) experience
Mice21 [21]
D. Constant Acceleration
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3 years ago
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