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3 years ago

If the mass of the spacecraft were doubled, how much would the force of gravity would change?

Physics

1 answer:

vodka [1.7K]3 years ago

5 0

F=mg
if m doubled, F would double as well

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A sports car moving at constant speed travels 164 m in 13.77 s. If it then brakes and comes to a stop in 3.6 s. What is its acce

Sergio039 [100]

Answer:

a = - 0.3376 g's

Explanation:

The sports car has a constant speed when travelling. Covered 164 m in 13.77 s. Thus, speed = 164/13.77 m/s

It brakes and now comes to a stop in 3.6 s.

Thus final velocity = 0 m/s

Formula for acceleration is;

a = (v - u)/t

a = (0 - (164/13.77))/3.6

a = -3.308 m/s²

In terms of g's", where 1.00 g = 9.80 m/s², we have;

a = -3.308/9.8 g's

a = - 0.3376 g's

4 0

3 years ago

A parallel circuit is constructed with three different branches. An ammeter is placed in each of the branches to measure the bra

neonofarm [45]

Answer:

Current in the battery will be 11.7 A

Explanation:

We have given that the there is parallel circuit with three different branches.

Current in each branch is $I_1=2.8A$ , $I_2=4.1A$ , $I_2=4.8A$

We have to find the current I in the battery

It is known that in parallel circuit current is sum of current in each branches

So current in the battery will be equal to

$I=I_1+I_2+I_3$

$I=2.8+4.1+4.8=11.7A$

Therefore current in the battery will be 11.7 A

8 0

3 years ago

What is a charge-coupled device (CCD), and how is it used in astronomy?

Sedaia [141]

Answer:

Charge-coupled device (CCD) is a device that receives and transfers an electrical charge to the next region

Explanation:

Charge-coupled device (CCD) is a device that receives and transfers an electrical charge to the next region where it can be modified like changing it to a electronic value.

In astronomy, high-powered telescopes can be used with CCD device image sensor cameras. The imaging system can concentrate for a number of hours on one place in space once the Earth's rotation synchronizes with the telescope.

6 0

4 years ago

A 1.0-kilogram rubber ball traveling east at 4.0 meters per second hits a wall and bounces back toward the west at 2.0 meters pe

Mrrafil [7]

Answer:

8 J and 2 J

Explanation:

Given that,

Mass of the rubber ball, m = 1 kg

Initial speed of the rubber ball, u = 4 m/s (in east)

Final speed of the rubber ball, v = -2 m/s (in west)

We need to find the kinetic energy of the ball before it hits the wall, the kinetic energy of the ball after it bounces off the wall.

Initial kinetic energy,

$K_i=\dfrac{1}{2}mv^2\\\\K_i=\dfrac{1}{2}\times 1\times (4)^2\\\\K_i=8\ J$

Final kinetic energy,

$K_f=\dfrac{1}{2}mv^2\\\\K_f=\dfrac{1}{2}\times 1\times (2)^2\\\\K_f=2\ J$

So, the initial kinetic energy is 8 J and the final kinetic energy is 2 J.

8 0

3 years ago

Need help on this question ASAP please and thank youu!

hichkok12 [17]

Answer:

- 500kg

Explanation:

momentum calculator online

8 0

3 years ago