Answer:
3875J
Explanation:
Energy is defined as the power × time
And it's defined as
Power = IV - I- current and V- voltage
Now quantity of electricity; Q = I × t
Where I is current and t is time
Now Energy = I ×V×t = V× I×t = V× Q;
where Q is quantity of electricity 775C and V is 5.0volt
Hence 775 × (5) =3875J
a. 4.52 m/s south
Velocity is a vector, whose magnitude is defined as the ratio between the displacement of the object and the time taken for the displacement to occur:

where
d is the displacement
t is the time
Velocity is a vector, so it also has a direction, which corresponds to that of the displacement.
For the ball in this problem,
d = 9.5 m south
t = 2.1 s
Substituting, we find:

and the directiion is the same as the displacement (south).
b. 4.52 m/s north
For this part, we must keep in mind that the speed is the magnitude of the velocity; however, speed is a scalar, so it does not have a direction.
Here we are told that the tennis ball travels at constant speed, so on its way back from Liam to Katie the ball's velocity is still the same as before, therefore

However, this time the direction is opposite to before, since the ball is travelling in the opposite direction.
Therefore, the ball's velocity when Liam returns Katie's service is
4.52 m/s north
Answer and explanation:
A.
Muon travelled straight down towards the earth. Therefore the tree moves up in the rest frame of muon (option a)
B.
In muon rest frame it travels Zero meters
C.
Distance, d = Velocity, v * Time, s


D.
Distance from the top of the mountain to the tree is the same as the distance travelled by the tree in the muons rest frame
that is same as in part C which is 594m
E.
Using lorentz contraction
In the rest frame of someone standing on the mountain
the distance is given by




d' = 258.92m
F.
in the rest frame of someone standing on the mountain,
muon moves straight down
Answer:
P = 0.6814 MPa
Explanation:
For this exercise we must use the Betuliano relation
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ rho v₂² + rho g y₂
where we use the subscript 1 for the first pipe and the subscript 2 for the second pipe
indicate that the two pipes are at the same level therefore
y₁ = y₂
P₂ = P₁ + ½ ro (v₁² - v₂²)
let's calculate
P₂ = 13,785 10⁵ + ½ 997 (41,221² - 55,654²)
P₂ = (13.785 - 6.97) 105
P₂ = 6.814 10⁵ Pa
let's reduce to Mpa
P₂ = 6.814 10⁵ Pa (1 MPa / 10⁶ Pa)
P₂ = 6.814 10-1 MPa
P = 0.6814 MPa