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2 years ago

Plz help urgent i will give brainliest

Physics

1 answer:

In-s [12.5K]2 years ago

5 0

Answer:

The answer is D. Balanced forces

Hope it helps.......... pls mark as brainliest

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how do you Compute the approximate acceleration of gravity for an object above the earth's surface, assigning accel gravity with

solong [7]

The approximate acceleration of gravity for an object above the earth's surface is 9.8 m/s².

<h3>Acceleration due to gravity </h3>

The acceleration due to gravity of an object above the Earth's surface is calculated from Newton's second law of motion and Newton's law of universal gravitation.

<h3>Force between the object and Earth</h3>

$F_g = \frac{GmM}{R^2}$

<h3>Weight of the object</h3>

$F_g = mg\\$

Solve (1) and (2)

mg = GmM/R²

g = GM/R²

where;

M is mass of Earth
G is gravitational constant
R is radius of the Earth = 6,371 km

g = (6.673 x 10⁻¹¹ x 5.98 x 10²⁴)/(6,371 x 10³)²

g = 9.8 m/s²

Learn more about acceleration due to gravity here: brainly.com/question/88039

#SPJ1

7 0

1 year ago

PLZ HELP!

MrMuchimi

Answer:

The answer is ball D

Explanation:

because I said it was

8 0

3 years ago

Read 2 more answers

A flare is launched from a life raft with an initial velocity of 192 ft/sec. How many seconds will it take for the flare to retu

DIA [1.3K]

We use the formula,

$h= ut- 16 t^2$

Here, h is the variable represents the height of the flare in feet when it returns to the sea so, h = 0 and u is the initial velocity of the flare, in feet per second and its value of 192 ft/sec.

Substituting these values in above equation, we get

$0 = 192 t - 16 t^2 \\\\ 16 t( 12 - t ) =0 \\\\ t = 12 s$ .

Here, t= 0 neglect because it is the time when the flare is launched.

Thus, flare return to the sea in 12 s.

8 0

3 years ago

Freight car A with a gross weight of 200,000 lbs is moving along the horizontal track in a switching yard at 4 mi/hr. Freight ca

zhenek [66]

Answer: a) 4.7 mi/hr. b) 86,500 lbs. mi²/Hr²

Explanation:

As in any collision, under the assumption that no external forces exist during the very small collision time, momentum must be conserved.

If the collision is fully inelastic, both masses continue coupled each other as a single mass, with a single speed.

So, we can write the following:

p₁ = p₂ ⇒m₁.v₁ + m₂.v₂ = (m₁ + m₂). vf

Replacing by the values, and solving for vf, we get:

vf = (200,000 lbs. 4 mi/hr + 100,000 lbs. 6 mi/hr) / 300,000 lbs = 4.7 mi/hr

If the track is horizontal, this means that thre is no change in gravitational potential energy, so any loss of energy must be kinetic energy.

Before the collision, the total kinetic energy of the system was the following:

K₁ = 1/2 (m₁.v₁² + m₂.v₂²) = 3,400,000 lbs. mi² / hr²

After the collision, total kinetic energy is as follows:

K₂ = 1/2 ((m₁ + m₂) vf²) = 3,313,500 lbs. mi²/hr²

So we have an Energy loss, equal to the difference between initial kinetic energy and final kinetic energy, as follows:

DE = K₁ - K₂ = 86,500 lbs. mi² / hr²

This loss is due to the impact, and is represented by the work done by friction forces (internal) during the impact.

8 0

2 years ago

What happens when two forces act in the same direction?

Wewaii [24]

Using the picture provided the forces are added together, because they are putting force on an object the same direction, thus the forces are added.

5 0

2 years ago

Read 2 more answers