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Vedmedyk [2.9K]

3 years ago

10

What is the net force?

1 answer:

vladimir2022 [97]3 years ago

4 0

The person is standing on the floor of the elevator, and he isn't
floating up or smashing through the floor.

So the person's acceleration is exactly the same as the elevator's
acceleration.

Now, Newton's second law of motion says:

F = M · a

   (the net force on the person) = (the person's mass) x (his acceleration)

   Net force = (80 kg) x (0.5 m/s²)

   = 40 newtons .

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A satellite at a particular point along an elliptical orbit has a gravitational potential energy of 5100 MJ with respect to Eart

serious [3.7K]

To solve this problem we will apply the theorem given in the conservation of energy, by which we have that it is conserved and that in terms of potential and kinetic energy, in their initial moment they must be equal to the final potential and kinetic energy. This is,

$E_{initial} = E_{final}$

$PE_{initial}+KE_{initial} = PE_{final}+KE_{final}$

Replacing the 5100MJ for satellite as initial potential energy, 4200MJ for initial kinetic energy and 5700MJ for final potential energy we have that

$KE_{final} = (PE_{initial}+KE_{initial} )-PE_{final}$

$KE_{final} = (5100+4200)-5700$

$KE_{final} = 3600MJ$

Therefore the final kinetic energy is 3600MJ

5 0

4 years ago

What is carbon dating?

ratelena [41]

Answer:

Im pretty sure its number 2

Explanation:

7 0

3 years ago

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Is there more than one kind of balance used in a laboratory

Inessa [10]

Maybe there is people who used

4 0

4 years ago

He starter motor of a car engine draws a current of 170 AA from the battery. The copper wire to the motor is 6.00 mmmm in diamet

sweet-ann [11.9K]

Answer:

129.2 C

0.33758239177 mm

Explanation:

n = Number density = $8.46\times 10^{28}\ electrons/m^3$

i = Current = 170 A

t = Time taken = 0.76 s

d = Diameter = 6 mm

Charge is given by

$q=it\\\Rightarrow q=170\times 0.76\\\Rightarrow q=129.2\ C$

The charge passing throught the motor is 129.2 C

Current density

$J=\dfrac{i}{A}\\\Rightarrow J=\dfrac{170}{\dfrac{\pi}{4}\times (6\times 10^{-3})^2}\\\Rightarrow J=6012520.07236\ A/m^2$

Drift velocity is given by

$v_d=\dfrac{J}{ne}\\\Rightarrow v_d=\dfrac{6012520.07236}{8.46\times 10^{28}\times 1.6\times 10^{-19}}\\\Rightarrow v_d=0.000444187357592\ m/s$

Distance traveled

$s=v_dt\\\Rightarrow s=0.000444187357592\times 0.76\\\Rightarrow s=0.00033758239177\ m=0.33758239177\ mm$

The electron traveled 0.33758239177 mm

8 0

3 years ago

50 points for this please help

Softa [21]

but the points are only 25

7 0

3 years ago

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