Answer:
his speed is 5cm a second
Answer:
Explanation:
Given that,
Number of turn N = 40
Diameter of the coil d= 11cm = 0.11m
Then, radius = d/2 = 0.11/2 =0.055m
r = 0.055m
Then, the area is given as
A =πr²
A = π × 0.055²
A = 9.503 × 10^-3 m²
Magnetic Field B = 0.35T
Magnetic field reduce to zero in 0.1s, t = 0.1s
so we want to find induce electric field. To find the electric field,(E) we need to find the electric potential (V).
E.M.F is given as
ε = —N • dΦ/dt
Where magnetic flux is given as
Φ = BA
Then, ε = —N • dΦ/dt
ε = —N • dBA/dt
ε = —NBA/t
Then, its magnitude is
ε = NBA/t
Inserting the values of N, B, A and t
ε = 40×0.35×9.503×10^-3/0.1
ε = 1.33 V
Then, using the relationship between Electric field and electric potential
V = Ed
ε = E•d
E = ε/d
E = 1.33/0.11
E = 12.09 V/m
<span> the answer is 1,500 & 1,700
</span>
Answer:
d = (75 i ^ + 93 j ^ + 27 k ^) m
, d2 = (900 i ^ + 1116 j ^ + 324 k ^) m
Explanation:
The two objects are in circular orbit together, therefore with the same angular velocity, after the launch they move with the relative velocity, so we can use the kinematic relation
v = d / t
d = v t
Reduce time to units SI
t = 5 min (60 s / 1 min) = 300 s
X axis
x = vₓ t
x = 0.25 300
x = 75 m
Y axis
y =
t
y = 0.31 300
y = 93 m
Z axis
z=
t
z = 0.09 300
z = 27 m
d = (75 i ^ + 93 j ^ + 27 k ^) m
For the time of 1 h
t2 = 1 h (3600s / 1 h) = 3600
x2 = 900 m
y2 = 1116 m
z2 = 324 m
d2 = (900 i ^ + 1116 j ^ + 324 k ^) m