All categories

3 years ago

What is the force of gravity (from the Earth) on the 700kg satellite if it’s 10km above the Earths surface?

Physics

1 answer:

joja [24]3 years ago

3 0

Answer:

The force of gravity on a 700 kg satellite if its 10 km above Earth's surface is given by

= ${\frac{(6.674\times 10^{-11}N. m^2/kg^2)(5.97\times 10^{24}kg) }{(10\times10^3)^2}$ = 3984378 m / $s^{2}$

Explanation:

The force of gravity on a 700 kg satellite if its 10 km above Earth's surface is given by

= ${\frac{(6.674\times 10^{-11}N. m^2/kg^2)(5.97\times 10^{24}kg) }{(10\times10^3)^2}$ = 3984378 m / $s^{2}$

You might be interested in

zepelin [54]

A I believe it would be true

8 0

3 years ago

If you have a 1.0 m aqueous solution of NaCl, by how much will it increase the water’s boiling point, if KB = 0.512 °C/m? In oth

fgiga [73]

Answer: The elevation in boiling point is 1.024°C.

Explanation:

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=ik_b\times m$

where,

i = Van't Hoff factor = 2 (for NaCl)

$\Delta T_b$ = change in boiling point = ?

$k_b$ = boiling point constant = $0.512^oC/m$

m = molality = 1.0 m

Putting values in above equation, we get:

$\Delta Tb=2\times 0.512^oC/m\times 1.0m\\\\\Delta Tb=1.024^oC$

Hence, the elevation in boiling point is 1.024°C.

5 0

3 years ago

Calculate the acceleration of a car if it's velocity increases from 15 m/s to 75m/s in 5 seconds.

Andreas93 [3]

initial velocity=u=15m/s
Final velocity=v=75m/s
Time=t=5s

$\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}$

$\\ \sf\longmapsto Acceleration=\dfrac{75-15}{5}$

$\\ \sf\longmapsto Acceleration=\dfrac{60}{5}$

$\\ \sf\longmapsto Acceleration=12m/s^2$

7 0

3 years ago

Consider the displacement vectors Ā=(i +6j)m, B = (3i– 7j)m,

andrezito [222]

Answer:

Dx = -0.5

Dy = -0.25

Explanation:

Two vectors are given in rectangular components form as follows:

A = i + 6j

B = 3i - 7j

It is also given that:

A - B - 4D = 0

so, we solve this to find D vector:

(i + 6j) - (3i - 7j) - 4D = 0

- 2i - j = 4D

D = - (2/4)i - (1/4)j

D = - (1/2)i - (1/4)j

D = - 0.5i - 0.25j

Therefore,

Dx = -0.5

Dy = -0.25

8 0

3 years ago

The air in a room has a pressure of 1 atm, a dry-bulb temperature of 24°C, and a wet-bulb temperature of 17°C. Using the psychro

Licemer1 [7]

Answer:

Given that

Dry-bulb temperature(T) =24°C

Wet-bulb temperature(Tw) = 17°C

Pressure ,P = 1 atm

As we know that psychrometric chart are drawn at constant pressure.

From the diagram

ω= specific humidity

Lets take these two lines Dry-bulb temperature(T) line and Wet-bulb temperature(Tw) cut at point P

From chart at point P

Specific humidity,ω = 0.00922 kg/kg

The enthalpy ( h)

h=47.59 KJ/kg

The relative humidity, RH

RH= 49.58 %

Specific volume ,

v= 0.853 m³/kg

5 0

3 years ago