(a) 0.448
The gravitational potential energy of a satellite in orbit is given by:

where
G is the gravitational constant
M is the Earth's mass
m is the satellite's mass
r is the distance of the satellite from the Earth's centre, which is sum of the Earth's radius (R) and the altitude of the satellite (h):
r = R + h
We can therefore write the ratio between the potentially energy of satellite B to that of satellite A as

and so, substituting:

We find

(b) 0.448
The kinetic energy of a satellite in orbit around the Earth is given by

So, the ratio between the two kinetic energies is

Which is exactly identical to the ratio of the potential energies. Therefore, this ratio is also equal to 0.448.
(c) B
The total energy of a satellite is given by the sum of the potential energy and the kinetic energy:

For satellite A, we have

For satellite B, we have

So, satellite B has the greater total energy (since the energy is negative).
(d) 
The difference between the energy of the two satellites is:

The density is 81.4 g/m3. Before you start plugging numbers into the density formula (D=M/V), you should convert 104 kg to grams, which ends up being 104,000 grams. Then you can plug in the 104,000 grams and 1,278 m3 into the formula. When you divide the mass by the volume, you get a really long decimal, which you can round to 81.4 g/m3, or whatever place your teacher wants you to round to.
Answer:
Obesity changes in the structure and function of the heart. It increases your risk of heart disease. The more you weigh, the more blood you have flowing through your body. The heart has to work harder to pump the extra blood.
Explanation:
Answer: magnitude of the magnetic field at a distance of 19.4 cm from the wire=4.29mT
Explanation:
According to Biot-Savart law, A magnetic field generated by a current carrying wire at a distance is represented as
B=μ₀I/ 2πr
B = magnetic field intensity 1000 mT =1T, 6.50mT = 6.50 X 10^-3T
μ₀ =permeability of free space 4π × 10−7 H/m
I = current intensity
r = radius, 100cm = 1m, 12.8 cm= 12.8 x 10^-2m
6.50 X 10^-3 = μ₀ x I/ 2 π X 12.8 X 10^-2
I =6.50 X 10 ^-3 X 2π X X 12.8 X 10^-2/ 4π × 10−7 H/m
I= 4160 A
when the magnetic field is at 19.4 cm from the wire
B=μ₀I/ 2πr
= 4π × 10−7 H/m x4160/ 2π x 19.4 x 10^-2
=0.004288
= 4.29x 10 ^-3T
= 4.29mT