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Jlenok [28]
3 years ago
11

When Earth pulls on an object, that object also pulls on Earth. The values of these two forces are . This phenomenon can be expl

ained using Newton’s law. We see the object move but not Earth because there’s a difference in their .
Physics
2 answers:
Verizon [17]3 years ago
6 0

Answer:

When Earth pulls on an object, that object also pulls on Earth. The values of these two forces are [the same]. This phenomenon can be explained using Newton’s [third] law. We see the object move but not Earth because there’s a difference in their [mass].

Explanation:

According to Newton’s third law, Earth pulls the object with the same amount of force that the object pulls Earth. The forces act in opposite directions. If there’s a large difference in mass between Earth and the object, the object moves, but Earth doesn’t.

algol133 years ago
5 0
The values of these two forces are equal. Your weight on Earth is equal to the Earth's weight on you. When you and the Earth fall toward each other, your acceleration is greater than the Earth's acceleration, because your mass is less than the Earth's mass.
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Packages having a mass of 6 kgkg slide down a smooth chute and land horizontally with a speed of 3 m/sm/s on the surface of a co
nalin [4]

Answer:

t = 1.02 s

Explanation:

The computation of the time required is shown below:

The package speed for belt is

= 3 -  1

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Moreover, the decelerative force would be acted on the block i.e u.m.g

So, the decelerative produced

= 0.2 × 9.81

= 1.962 m/s^2

And, final velocity = 0

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V = 0 = final velocity

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5 0
3 years ago
A ferry is travelling at 10 m/s relative to the river towards the south. The river is flowing at 2 m/s relative to
mixer [17]

Answer:

11 m/s south

Explanation:

The velocity of the passenger relative to the river bank is equal to the velocity of the passenger relative to the ferry, plus the velocity of the ferry relative to the river, plus the velocity of the river relative to the river bank.

v_passenger,bank = v_passenger,ferry + v_ferry,river + v_river,bank

If we take north to be positive and south to be negative:

v = 1.0 m/s + (-10 m/s) + (-2 m/s)

v = -11 m/s

v = 11 m/s south

8 0
3 years ago
How do you solve <img src="https://tex.z-dn.net/?f=4x%5E%7B3%7D" id="TexFormula1" title="4x^{3}" alt="4x^{3}" align="absmiddle"
Vlad1618 [11]

Hello There!

Here's a explanation!

Let's solve your equation step-by-step.

4x^3=2x^-^1

4x^3=\frac{2}{x}

Step 1: Multiply both sides by x.

4x^4=2

\frac{4x^4}{4} =\frac{2}{4}

(Divide both sides by 4).

x^4=\frac{1}{2}

x=+(\frac{1}{2} )^(^\frac{1}{4} ^)

Take the root.

ANSWER!

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AnimeVines

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3 years ago
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A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth.
Maksim231197 [3]

Answer:

Explanation:

We shall apply law of  conservation of mechanical energy for projectile being thrown .

Total energy on the surface = total energy at height h required

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\frac{-GM}{R} + \frac{m\times(.351\times\sqrt{2GM})^2 }{2R } = \frac{-GMm}{R+h} + 0

\frac{-GMm}{R} +\frac{1}{2}\times  \frac{-2GMm}{R} \times0.123=\frac{-GMm}{R+h}

\frac{0.877GMm}{R} =\frac{-GMm}{R+h}

h = .14 R

b )

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\frac{-0.649GMm}{R} = \frac{-GMm}{R+h}

h = .54 R

c ) least initial mechanical energy required at launch if the projectile is to escape Earth

= GMm / R + 1/2 m (2GM/R)

= 0

5 0
3 years ago
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