Answer:
Explanation:
Given that,
B(t) = B0 cos(ωt) • k
Radius r = a
Inner radius r' = a/2 and resistance R.
Current in the loop as a function of time I(t) =?
Magnetic flux is given as
Φ = BA
And the Area is given as
A = πr², where r = a/2
A = πa²/4
Then,
Φ = ¼ Bπa²
Φ(t) = ¼πa²Bo•Cos(ωt)
Then, the EMF is given as
ε(t) = -dΦ/dt
ε(t) = -¼πa²Bo • -ωSin(ωt)
ε(t) = ¼ωπa²Bo•Sin(ωt)
From ohms law,
ε = iR
Then, i = ε/R
I(t) = ¼ωπa²Bo•Sin(ωt) /R
This is the current induced in the loop.
Check attachment for better understanding
The total electric potential at the center of the square due to the four charges is V = √2Q/πÈa.
<h3>What do you mean by electric potential? </h3>
The amount of work needed to move a unit charge from a reference point to a specific point against an electric field. It's SI unit is volt.
V = kq/r
Where V represents electric potential, K is coulomb constant, q is Charge and r is distance between any two around charge to the point charge.
Electric potential at O due to four charges is given by,
V = 4KQ/ r
where, r = √2a/2 = a/√2
V = 4k × Q√2/a
V = √2Q/πÈa
The total electric potential at the center of the square due to the four charges is V = √2Q/πÈa.
To learn more about electric potential refer to:
brainly.com/question/12645463
#SPJ4
Answer:29.627 m
Explanation:
Given
Initial velocity of life preserver(u) is 1.6 m/s
it takes 2.3 s to reach the water
using equation of motion
v=u+at
v=24.163 m/s
Let s be the height of life preserver
s=29.627 m
Geothermal energy provides force that causes magma to erupt to the surface
Answer:
a) x = v₀² sin 2θ / g
b) t_total = 2 v₀ sin θ / g
c) x = 16.7 m
Explanation:
This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity
sin θ = 
/ vo
cos θ = v₀ₓ / vo
v_{oy} = v_{o} sin θ
v₀ₓ = v₀ cos θ
v_{oy} = 13.5 sin 32 = 7.15 m / s
v₀ₓ = 13.5 cos 32 = 11.45 m / s
a) In the x axis there is no acceleration so the velocity is constant
v₀ₓ = x / t
x = v₀ₓ t
the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
t = v_{o} sin θ / g
we substitute
x = v₀ cos θ (2 v_{o} sin θ / g)
x = v₀² /g 2 cos θ sin θ
x = v₀² sin 2θ / g
at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,
b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time
at the highest point the vertical speed is zero
v_{y} = v_{oy} - gt
v_{y} = 0
t = v_{oy} / g
t = v₀ sin θ / g
as the time to get on and off is the same the total time or flight time is
t_total = 2 t
t_total = 2 v₀ sin θ / g
c) we calculate
x = 13.5 2 sin (2 32) / 9.8
x = 16.7 m
