All categories

4 years ago

A satellite orbits the Earth. Which changes would decrease the force of Earth's gravity on the satellite?

Physics

1 answer:

iren2701 [21]4 years ago

7 0

The correct answer.
1.Remove mass from the satellite
2.Increase the distance above Earth that the satellite orbits.

You might be interested in

Question 4 (2 points)

Igoryamba

The answer is a) Teres Major Muscle

7 0

2 years ago

Read 2 more answers

A bobsled has a momentum of 4000 kg•m/s to the south. Friction on the

mina [271]

Answer:

1500 north

Explanation:

8 0

3 years ago

Your friends sit in a sled in the snow. If you apply a force of 95 N to them, they have an acceleration of 0.8 m/s2. What is the

ss7ja [257]

F=ma
m=F/a=95/0.8= 118.75kg
your friend is pretty heavy XD

5 0

3 years ago

Read 2 more answers

6- A metal block measures 10 cm x 2 cm x 2 cm. what is its volume? how many blocks each

zalisa [80]

You need 5 blocks of the smaller object to contain the same amount of volume of the bigger object

3 0

4 years ago

A hollow cylinder with an inner radius of 5 mm and an outer radius of 26 mm conducts a 4-A current flowing parallel to the axis

bearhunter [10]

Answer:

B = 38.2μT

Explanation:

By the Ampere's law you have that the magnetic field generated by a current, in a wire, is given by:

$B=\frac{\mu_o I_r}{2\pi r}$ (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

r: distance from the center of the cylinder, in which B is calculated

Ir: current for the distance r

In this case, you first calculate the current Ir, by using the following relation:

$I_r=JA_r$

J: current density

Ar: cross sectional area for r in the hollow cylinder

Ar is given by $A_r=\pi(r^2-R_1^2)$

The current density is given by the total area and the total current:

$J=\frac{I_T}{A_T}=\frac{I_T}{\pi(R_2^2-R_1^2)}$

R2: outer radius = 26mm = 26*10^-3 m

R1: inner radius = 5 mm = 5*10^-3 m

IT: total current = 4 A

Then, the current in the wire for a distance r is:

$I_r=JA_r=\frac{I_T}{\pi(R_2^2-R_1^2)}\pi(r^2-R_1^2)\\\\I_r=I_T\frac{r^2-R_1^2}{R_2^2-R_1^2}$ (2)

You replace the last result of equation (2) into the equation (1):

$B=\frac{\mu_oI_T}{2\pi r}(\frac{r^2-R_1^2}{R_2^2-R_1^2})$

Finally. you replace the values of all parameters:

$B=\frac{(4\pi*10^{-7}T/A)(4A)}{2\PI (12*10^{-3}m)}(\frac{(12*10^{-3})^2-(5*10^{-3}m)^2}{(26*10^{-3}m)^2-(5*10^{-3}m)^2})\\\\B=3.82*10^{-5}T=38.2\mu T$

hence, the magnitude of the magnetic field at a point 12 mm from the center of the hollow cylinder, is 38.2μT

8 0

3 years ago