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2 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]2 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 / $s^{2}$

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A bug walks exactly halfway around the edge of a circular cupcake with a diameter of 5 cm what is the distance he traveled and w

adelina 88 [10]

The circumference of a circle is (pi) x (diameter)

The circumference of the cupcake is (pi) x (5 cm)

Halfway around is (1/2) x (pi) x (5 cm) = (2.5 pi cm) = about 7.85 cm

The 'why' appears up above, in the first 2 lines of this solution.

7 0

3 years ago

A soccer ball with mass 0.450 kg is initially moving with speed 2.20 m/s. A soccer player kicks the ball, exerting a constant fo

Alinara [238K]

Answer:

0.187 m

Explanation:

We'll begin by calculating the acceleration of the ball. This can be obtained as follow:

Mass (m) = 0.450 Kg

Force (F) = 38 N

Acceleration (a) =?

F = m × a

38 = 0.450 × a

Divide both side by 0.450

a = 38 / 0.450

a = 84.44 m/s²

Finally, we shall determine the distance. This can be obtained as follow:

Initial velocity (u) = 2.20 m/s.

Final velocity (v) = 6 m/s

Acceleration (a) = 84.44 m/s²

Distance (s) =?

v² = u² + 2as

6² = 2.2² + (2 × 84.44 × s)

36 = 4.4 + 168.88s

Collect like terms

36 – 4.84 = 168.88s

31.52 = 168.88s

Divide both side by 168.88

s = 31.52 / 168.88

s = 0.187 m

Thus, the distance is 0.187 m

6 0

2 years ago

Calculate the number of electrons in a small, electrically neutral silver pin that has a mass of 7.1 g. Silver has 47 electrons

Assoli18 [71]

Answer: The number of electrons in given amount of silver are $1.87\times 10^{24}$

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

We are given:

Given mass of silver = 7.1 g

Molar mass of silver = 107.87 g/mol

Putting values in above equation, we get:

$\text{Moles of silver}=\frac{7.1g}{107.87g/mol}=0.066mol$

Number of electrons in 1 atom of silver = 47

According to mole concept:

1 mole of an element contains $6.022\times 10^{23}$ number of particles

So, 0.066 moles of silver will contain = - $(0.066\times 47\times 6.022\times 10^{23})=1.87\times 10^{24}$ number of electrons

Hence, the number of electrons in given amount of silver are $1.87\times 10^{24}$

7 0

2 years ago

two engines are turned on for 763 s at a moment when the velocity of the craft has x and y components of v0x = 6380 m/s and v0y

svet-max [94.6K]

Answer:

Explanation:

Given

initial velocity component of engines is

$v_0_x=6380 m/s$

$v_0_y=6770 m/s$

time period of engine running=763 s

Displacement in $x=4.50\times 10^6$

$y=7.27\times 10^6$

Using $s=ut+\frac{at^2}{2}$ in x and y direction

$x=v_0_x\times t+\frac{at^2}{2}$

$4.50\times 10^6=6380\times 763+\frac{a\times 763^2}{2}$

$4.50\times 10^6-4.86\times 10^6=\frac{a\times 763^2}{2}$

$a=-1.23 m/s^2$

In y direction

$y=v_0_y\times t+\frac{a't^2}{2}$

$7.27\times 10^6=6770\times 763+\frac{a\times 763^2}{2}$

$7.27\times 10^6-5.16\times 10^6=\frac{a\times 763^2}{2}$

$a=7.24 m/s^2$

x component $=-1.23 m/s^2$

y component $=7.24 m/s^2$

3 0

2 years ago

Read 2 more answers

n a downhill ski race, surprisingly, little advantage is gained by getting a running start. (This is because the initial kinetic

Annette [7]

Answer:

Explanation:

a ) starting from rest , so u = o and initial kinetic energy = 0 .

Let mass of the skier = m

Kinetic energy gained = potential energy lost

= mgh = mg l sinθ

= m x 9.8 x 70 x sin 30

= 343 m

Total kinetic energy at the base = 343 m + 0 = 343 m .

b )

In this case initial kinetic energy = 1/2 m v²

= .5 x m x 2.5²

= 3.125 m

Total kinetic energy at the base

= 3.125 m + 343 m

= 346.125 m

c ) It is not surprising as energy gained due to gravitational force by the earth is enormous . So component of energy gained due to gravitational force far exceeds the initial kinetic energy . Still in a competitive event , the fractional initial kinetic energy may be the deciding factor .

7 0

3 years ago