Answer:
The energy, that is dissipated in the resistor during this time interval is 153.6 mJ
Explanation:
Given;
number of turns, N = 179
radius of the circular coil, r = 3.95 cm = 0.0395 m
resistance, R = 10.1 Ω
time, t = 0.163 s
magnetic field strength, B = 0.573 T
Induced emf is given as;
where;
ΔФ is change in magnetic flux
ΔФ = BA = B x πr²
ΔФ = 0.573 x π(0.0395)² = 0.002809 T.m²
According to ohm's law;
V = IR
I = V / R
I = 3.0848 / 10.1
I = 0.3054 A
Energy = I²Rt
Energy = (0.3054)² x 10.1 x 0.163
Energy = 0.1536 J
Energy = 153.6 mJ
Therefore, the energy, that is dissipated in the resistor during this time interval is 153.6 mJ
Answer:
Explanation:
given data:
pressure 1 MPa
diameter of pipe = 30 cm
average velocity = 10 m/s
area of pipe
A = 0.070 m2
WE KNOW THAT mass flow rate is given as
for pressure 1 MPa, the density of steam is = 4.068 kg/m3
therefore we have
Answer:
Explanation:
Using the kinematics equation to determine the velocity of car B.
where;
initial velocity
= constant deceleration
Assuming the constant deceleration is = -12 ft/s^2
Also, the kinematic equation that relates to the distance with the time is:
Then:
The distance traveled by car B in the given time (t) is expressed as:
For car A, the needed time (t) to come to rest is:
Also, the distance traveled by car A in the given time (t) is expressed as:
Relating both velocities:
t = 2.25 s
At t = 2.25s, the required minimum distance can be estimated by equating both distances traveled by both cars
i.e.
d + 104.625 = 114.75
d = 114.75 - 104.625
d = 10.125 ft
Answer:
we know that
A worker currently makes \$425.00 per week
remember that
1\ year=52\ weeks
so
by proportion find the amount that the worker will earn in one year
\frac{425}{1} \frac{\$}{week} =\frac{x}{52} \frac{\$}{weeks} \\ \\x=52*425 \\ \\x=\$22,100
therefore
the answer is
\$22,100
Explanation:
Answer:
Explanation:
Given that:
x(t) = 10 sin(10t) . sin (15t)
the objective is to find the power and the rms value of the following signal square.
Recall that:
sin (A + B) + sin(A - B) = 2 sin A.cos B
x(t) = 10 sin(15t) . cos (10t)
x(t) = 5(2 sin (15t). cos (10t))
x(t) = 5 × ( sin (15t + 10t) + sin (15t-10t)
x(t) = 5sin(25 t) + 5 sin (5t)
From the knowledge of sinusoidial signal Asin (ωt), Power can be expressed as:
For the number of sinosoidial signals;
Power can be expressed as:
As such,
For x(t), Power
For the number of sinosoidial signals;
For x(t), the RMS value is as follows: