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TiliK225 [7]
3 years ago
11

Do action and reaction cancel each other out?

Physics
1 answer:
galben [10]3 years ago
8 0

Answer:

Action and reaction forces don't cancel each other out because they act on separate objects. ... Action and reaction forces are always equal in magnitude, so it's not possible to exert more force on an object than it can exert back.

Explanation:

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If a projectile is launched vertically from the Earth with a speed equal to the escape speed, how high above the earth's surface
Alecsey [184]

Answer: h = 3R

Explanation:

Using the law of conservation of energy,

Total energy at the beginning of the launch would be equal to total energy at any point.

kinetic energy + gravitational potential energy = constant

Initial energy of the projectile =\frac{1}{2}mv_e^2-\frac{GMm}{R}... (1)

where R is the radius of the Earth, M is the mass ofthe Earth, m is the mass of the projectile.

escape velocity, v_e=\sqrt{\frac{2GM}{R}}

Total energy at height h above the Earth where speed of the projectile is half the escape velocity:

\frac{1}{2}m(v_e/2)^2-G\frac{Mm}{R+h} ...(2)

(1)=(2)

⇒\frac{1}{2}m(v_e/2)^2-G\frac{Mm}{R+h} = \frac{1}{2}mv_e^2-\frac{GMm}{R}

⇒G\frac{Mm}{R}-G\frac{Mm}{R+h}= \frac{1}{2}m \frac{3}{4}v_e^2 =\frac{1}{2}m \frac{3}{4} (\sqrt{\frac{2GM}{R}})^2

⇒\frac{GM(R+h-R)}{R(R+h)} = \frac{3}{4}(\frac{GM}{R})

⇒\frac{h}{R+h} = \frac{3}{4}

⇒h = 3R

Thus, at height equal to thrice radius of Earth, the speed of the projectile would reduce to half of escape velocity.  

4 0
3 years ago
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What is a natural resource?
sladkih [1.3K]

Answer:

A natural resource is something found in nature that people use.

Explanation:

4 0
4 years ago
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1. Summarize What are the four conditions of a<br>purely competitive market?​
bearhunter [10]

Answer:

Explanation:

1. Many buyers and sellers participate in the market.

2. Sellers offer identical products.

3. Buyers and sellers are well informed about products.

4. Sellers are able to enter and exit the market freely.

5 0
3 years ago
The drawings show three charges that have the same magnitude but may have different signs. In all cases the distance d between t
Radda [10]

Answer:

a)F = 6816.5680 N

b) F' = 0 N

c)F''=28195.5N

Explanation:

Magnitude of charges M=1.6\muC

Distance d=2.6

Generally the equation for  Net Force is mathematically given by

For First Drawing

F =\frac{ k q^2}{d^2}+\frac{k q^2}{d^2}  

F =\frac{2* 9*10^9* (1.6*10^-6)^2}{ (2.6*10^-3^2}

F = 6816.5680 N

For second Drawing

F' =\frac{ k q^2}{d^2 }-\frac{ k q^2}{ d^2}  

F' = 0 N

For Third Drawing

F'' =\frac{ k q^2}{d^2} * \sqrt (2)

F'' = 9*10^9* (6.4*10^-6)^2 * \frac{\sqrt(2)}{(4.3*10^-3)^2}

F''=28195.5N

4 0
3 years ago
How much time does it take for a car to accelerate from 3.44 m/s to 20.9 m/s if the average acceleration is 6.00 m/s^3
miv72 [106K]

Do you go to GA Connexus Academy??

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