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3 years ago

What does 9.8 m/s/s have to do with acceleration?

Physics

2 answers:

ruslelena [56]3 years ago

6 0

9.8 is the gravitational field strength on earth, perhaps re-read the question

Stells [14]3 years ago

3 0

The more force acts on an object, the greater acceleration it has.

The force that Earth's gravity exerts on every falling object makes the object fall with acceleration of 9.8 m/s/s. It's the same for every object on Earth. The size or mass or weight of the object doesn't matter.

The number is different on other bodies, like the Moon, or Mars.

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A car traveling with 500,000 J of kinetic energy is brought to a kinetic energy of

allsm [11]

Answer:

33,333.33 N

Explanation:

Given that :

Initial kinetic energy = 500,000 J

Final kinetic energy = 100,000 J

Using the relation :

Force * time = change in momentum (Newton's law)

Force (F) * 0.12 = (500,000 - 100,000)

0.12F = 400,000 J

Force = (400,000 J) / 0.12s

Force = 33333.333

Force = 33,333.33 N

3 0

3 years ago

An object weighs 100 newtons on earth.What is its weight on the moon?

Pachacha [2.7K]

Answer:

a) The gravitational acceleration at the surface of the Moon is g moon=1.67 m/s

The ratio of weights (for a given mass ) is the ratio of g-values, so

moon

=(100N)(1.67/9.8)=17N.

(b) For the force on that object caused by Earth's gravity to equal 17 N, then the free fall acceleration at its location must be

=1.67m/s

. Thus , .

= r 2

⇒

=1.5×10

So the object would need to be a distance of r/R

=2.4 "radii" from Earth's center.

4 0

3 years ago

Read 2 more answers

An electric charge, A, is placed carefully between two other charges, B and C, and experiences no net electric force. Do B

Brrunno [24]

Answer:

I do not have enough information to tell

Explanation:

This is deduced due to the fact that if the net force due to B and C on A is zero, the charges on B and C could either be positive or negative depending on the charge on A.

5 0

3 years ago

Could anyone help me with this?

tatiyna

Answer:

Answer is option C..

Mark as brainlist

3 0

3 years ago

A seagull flying horizontally over the ocean at a constant speed of 2.60 m/s carries a small fish in its mouth. It accidentally

Ivenika [448]

(a) +2.60 m/s

The motion of the fish dropped by the seagul is a projectile motion, which consists of two independent motions:

- a horizontal uniform motion, at constant speed

- a vertical motion, at constant acceleration (acceleration of gravity, $g=-9.8 m/s^2$ , downward)

In this part we are only interested in the horizontal motion. As we said the horizontal component of the fish's velocity does not change, therefore its value when the fish reaches the ocean is equal to its initial value, which is the speed at which the seagull was flying (because it was flying horizontally):

$v_x = +2.60 m/s$

(b) -17.2 m/s

The vertical component of the fish's velocity instead follows the equation:

$v_y = u_y +gt$

where

$u_y = 0$ is the initial vertical velocity, which is zero

$g=-9.8 m/s^2$ is the acceleration of gravity

t is the time

Since the fish reaches the ocean at t = 1.75 s, we can substitute this time into the formula to find the final vertical velocity:

$v_y = 0+(-9.8)(1.75)=-17.2 m/s$

where the negative sign indicates the direction (downward).

(c)

The horizontal component of the fish's velocity would increase

The vertical component of the fish's velocity would stay the same.

As we said from part (a) and (b):

- The horizontal component of the fish's velocity is constant during the motion and it is equal to the initial velocity of the seagull -> so if the seagull's initial speed increases, the horizontal velocity of the fish will increase too

- The vertical component of the fish's velocity does not depend on the original speed of the seagull, therefore it is not affected.

4 0

3 years ago