Base in your question about the magnetic field of the Earth near the equator where as its almost horizontally to the north and has magnitude of B=0.5x10^-4t, the answer is <span>Velocity of electron will be westwards.</span>
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.
Below is the solution:
W done by Normal = 0. (make the incline flat, Normal force goes directly up: no work done)
<span>W done by gravity = w*displacement = (11kg*9.8) * 7.5sin(35) = -463J </span>
<span>W done by friction is the opposite of the work done by weight because the object is not moving. Therefore W done by friction = 463J</span>
v^2-u^2=2 x a x d
25^2-0^2=2 x a x 70
625-0=140 x a
625=140a
a=625/140
a=4.46 m/s^2
im not very sure but i think this is how you do this
Answer:
a₂ = m₁ / m₂ a₁
Explanation:
For this exercise we note that the attraction between the two stars is an action and reaction force, therefore it has the same magnitude, but it is applied to each of the bodies
Let's apply Newton's second law on the star 1
F₁ = m₁ a₁
Newton's second law in star 2
F₂ = m₂ a₂
| F₁ | = | F₂ |
m₁ a₁ = m₂ a₂
a₂ = m₁ / m₂ a₁
I believe the answer is A.
Hope This Helps! Have A Nice Day!!
