Ask question

Ask question

All categories

3 years ago

6

Two satellites of equal mass are orbiting the planet Mars. Satellite B is closer to Mars than Satellite A. How does the gravitat

ional pull of Mars affect each of these satellites?

1 answer:

Gnoma [55]3 years ago

7 0

The object in the lower orbit will move faster in its orbit, and will have a shorter orbital period. None of that depends on their masses either.

You might be interested in

A manometer is used to measure the air pressure in a tank. the fluid used has a specific gravity of 1.25, and the differential h

BartSMP [9]

Specific Gravity of the fluid = 1.25
Height h = 28 in
Atmospheric Pressure = 12.7 psia
Density of water = 62.4 lbm/ft^3 at 32F
Density of the Fluid = Specific Gravity of the fluid x Density of water = 1.25 x 62.4
Density of the Fluid p = 78 lbm/ft^3
Difference in pressure as we got the differential height, dP = p x g x h dP = (78 lbm/ft^3) x (32.174 ft/s^2) x (28/12 ft) [ 1 lbf / 32.174 ft/s^2] [1 ft^2 /
144in^2]
Difference in pressure = 1.26 psia
(a) Pressure in the arm that is at Higher
P = Atmospheric Pressure - Pressure difference = 12.7 - 1.26 = 11.44 psia
(b) Pressure in the tank that is at Lower
P = Atmospheric Pressure + Pressure difference = 12.7 + 1.26 = 13.96psia

4 0

3 years ago

A group of students decides to set up an experiment in which they will measure the specific heat of a small amount of metal. The

lesya692 [45]

I think the answer is C

7 0

3 years ago

Which letter on the diagram below represents the trough of a wave?

Nikitich [7]

A would be the wavelength, C would be a crest, D would be the amplitude, leaving B which is the trough.

8 0

3 years ago

Read 2 more answers

A parallel-plate capacitor has square plates that are 8.00cm on each side and 4.20mm apart. The space between the plates is comp

Sergeu [11.5K]

Answer:

$U=1.29\times 10^{-7}\ J$

Explanation:

Given that

a= 8 cm (square)

A= a ² = 64 cm²

d= 4.2 mm

d₁= 2.1 mm ,K₁= 4.7

d₂=2.1 mm , K₂=2.6

We know that capacitance given as

$C_1=\dfrac{K_1\varepsilon _oA}{d_1}$

$C_1=\dfrac{4.7\times 8.85\times 10^{-12}\times 64\times 10^{-4}}{2.1\times 10^{-3}}$

$C_1=1.26\times 10^{-10}\ F$

$C_2=\dfrac{K_2\varepsilon _oA}{d_2}$

$C_2=\dfrac{2.6\times 8.85\times 10^{-12}\times 64\times 10^{-4}}{2.1\times 10^{-3}}$

$C_2=0.701\times 10^{-10}\ F$

Net capacitance

$C=\dfrac{C_1C_2}{C_1+C_2}$

$C=\dfrac{1.26\times 10^{-10}\times 0.701\times 10^{-10}}{1.26\times 10^{-10}+0.701\times 10^{-10}}\ F$

$C=4.5\times 10^{-11}\ F$

We know that stored energy given as

$U=\dfrac{CV^2}{2}$

V= 76 V

$U=\dfrac{4.5\times 10^{-11}\times 76^2}{2}\ J$

$U=1.29\times 10^{-7}\ J$

3 0

3 years ago

Ann puts a beaker on the scale which shows the mass to be 1.65kg. The actual mass was 1.72 kg. what was the percent error in Ann

mojhsa [17]

Measured reading is 1.65 kg

actual reading was 1.72 kg

error in the reading is 1.72 - 1.65 = 0.07 kg

Now percentage error is given as

$error = \frac{error}{true reading}*100$

$error = \frac{0.07}{1.72}*100$

$error = 4.1$

<em>so this is -4.1% error as the measured reading is less than the actual value</em>

8 0

3 years ago