Answer:
Explanation:
Given that,
The mutual inductance of the two coils is
M = 300mH = 300 × 10^-3 H
M = 0.3 H
Current increase in the coil from 2.8A to 10A
∆I = I_2 - I_1 = 10 - 2.8
∆I = 7.2 A
Within the time 300ms
t = 300ms = 300 × 10^-3
t = 0.3s
Second Coil resistance
R_2 = 0.4 ohms
We want to find the current in the second coil,
The same induced EMF is in both coils, so let find the EMF,
From faradays law
ε = Mdi/dt
ε = M•∆I / ∆t
ε = 0.3 × 7.2 / 0.3
ε = 7.2 Volts
Now, this is the voltage across both coils,
Applying ohms law to the second coil, V=IR
ε = I_2•R_2
0.72 = I_2 • 0.4
I_2 = 0.72 / 0.4
I_2 = 1.8 Amps
The current in the second coil is 1.8A
Answer:
y = 10.2 m
Explanation:
It is given that,
Charge, 
It is placed at a distance of 9 cm at x axis
Charge, 
It is placed at a distance of 16 cm at x axis
We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,
Here,
So,
Squaring both sides,
So, at a distance of 10.2 m on the y axis the electric potential equals 0.
Answer: 56.72 ft/s
Explanation:
Ok, initially we only have potential energy, that is equal to:
U =m*g*h
where g is the gravitational acceleration, m the mass and h the height.
h = 50ft and g = 32.17 ft/s^2
when the watermelon is near the ground, all the potential energy is transformed into kinetic energy, and the kinetic energy can be written as:
K = (1/2)*m*v^2
where v is the velocity.
Then we have:
K = U
m*g*h = (m/2)*v^2
we solve it for v.
v = √(2g*h) = √(2*32.17*50) ft/s = 56.72 ft/s
solution:
y = v0t + ½at²
1150 = 79t + ½3.9t²
0 = 3.9t² + 158t - 2300
from quadratic equations and eliminating the negative answer
t = (-158 + v158² -4(3.9)(-2300)) / 2(3.9)
t = 11.37 s to engine cut-off
the velocity at that time is
v = v0 + at
v = 79 + 3.9(11.37)
v = 123.3 m/s
it rises for an additional time
v = gt
t = v/g
t = 123.3 / 9.8
t = 12.59 s
gaining more altitude
y = ½vt
y = 123.3(12.59) /2
y = 776 m
for a peak height of
y = 776 + 1150
