3 kilometers, it is just 5/60 or 1/12 multiplied by 36.
Answer:
(a) A = 0.650 m
(b) f = 1.3368 Hz
(c) E = 17.1416 J
(d) K = 11.8835 J
U = 5.2581 J
Explanation:
Given
m = 1.15 kg
x = 0.650 cos (8.40t)
(a) the amplitude,
A = 0.650 m
(b) the frequency,
if we know that
ω = 2πf = 8.40 ⇒ f = 8.40 / (2π)
⇒ f = 1.3368 Hz
(c) the total energy,
we use the formula
E = m*ω²*A² / 2
⇒ E = (1.15)(8.40)²(0.650)² / 2
⇒ E = 17.1416 J
(d) the kinetic energy and potential energy when x = 0.360 m.
We use the formulas
K = (1/2)*m*ω²*(A² - x²) (the kinetic energy)
and
U = (1/2)*m*ω²*x² (the potential energy)
then
K = (1/2)*(1.15)*(8.40)²*((0.650)² - (0.360)²)
⇒ K = 11.8835 J
U = (1/2)*(1.15)*(8.40)²*(0.360)²
⇒ U = 5.2581 J
Answer:
A title
Explanation:
Because this is middle school.
Answer:
The inducerd emf is 1.08 V
Solution:
As per the question:
Altitude of the satellite, H = 400 km
Length of the antenna, l = 1.76 m
Magnetic field, B = 
Now,
When a conducting rod moves in a uniform magnetic field linearly with velocity, v, then the potential difference due to its motion is given by:
Here, velocity v is perpendicular to the rod
Thus
e = lvB (1)
For the orbital velocity of the satellite at an altitude, H:
where
G = Gravitational constant
= mass of earth
= radius of earth
Using this value value in eqn (1):
