Answer:
a) 6636 km
b) 0.0154
Explanation:
The height above the earth at its furthest point is 368 km
The height above the earth at its closest point is 164 km
Radius of the Earth is 6370 km
The distance of the satellite from the center of the earth to the furthest point is 6370 + 368 km = 6738 km
The distance of the satellite from the center of the earth to the closest point is 6370 + 164 = 6534 km
If we add together the sum of the distance of the satellite from the furthest and its closest distance, it is equal to the 2 major semi axis.
Basically,
2a = R + r
a = (R + r) / 2
a = (6738 + 6534) / 2
a = 13272 / 2
a = 6636 km
Eccentricity, e = (a - r) / a
Eccentricity, e = (6636 - 6534) / 6636
Eccentricity, e = 102 / 6636
Eccentricity, e = 0.0154
Answer:
The quantity of motion is the measure of the same, arise from the velocity and quantity of matter conjointly. In other words, rather than defining the quantity of motion of a given object as simply the kinematic velocity v of the object, he defined it as the product mv, where m is the mass of the object.
Explanation:
Answer:
Explanation:
Given that
x= 150 ft
y= 14 ft
From the diagram
When ,x= 150 ft and y= 14 ft
z=150.74 ft
By differentiating with respect to time t
Here x is constant that is why
Now by putting the values in the above equation we get
Therefore the distance between balloon and observer increasing with 0.65 ft/s.
Answer:
The average induced emf in the coil is 0.0286 V
Explanation:
Given;
diameter of the wire, d = 11.2 cm = 0.112 m
initial magnetic field, B₁ = 0.53 T
final magnetic field, B₂ = 0.24 T
time of change in magnetic field, t = 0.1 s
The induced emf in the coil is calculated as;
E = A(dB)/dt
where;
A is area of the coil = πr²
r is the radius of the wire coil = 0.112m / 2 = 0.056 m
A = π(0.056)²
A = 0.00985 m²
E = -0.00985(B₂-B₁)/t
E = 0.00985(B₁-B₂)/t
E = 0.00985(0.53 - 0.24)/0.1
E = 0.00985 (0.29)/ 0.1
E = 0.0286 V
Therefore, the average induced emf in the coil is 0.0286 V
