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3 years ago

Newton’s law of universal gravitation stars that every object in the universe attracts every other object

Physics

1 answer:

LenaWriter [7]3 years ago

4 0

Yes it does ! Uh huh. Right you are. Truer words are seldom written.

You have quoted the law quite accurately but also incompletely.

Do you have a question to ask ?

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If earth's mass were half its actual value but its radius stayed the same, the escape velocity of earth would be:________

siniylev [52]

If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be $V_e = \sqrt{\dfrac{GM}{r}}$ .

<h3>What is an escape velocity?</h3>

The ratio of the object's travel distance over a specific period of time is known as its velocity. As a vector quantity, the velocity requires both the magnitude and the direction. the slowest possible speed at which a body can break out of the gravitational pull of a certain planet or another object.

The formula to calculate the escape velocity of earth is given below:-

$V_e=\sqrt{\dfrac{2GM}{r}}$

Given that earth's mass was half its actual value but its radius stayed the same. The escape velocity will be calculated as below:-

$V_e=\sqrt{\dfrac{2GM}{r\times 2}}$

$V_e = \sqrt{\dfrac{GM}{r}}$ .

Therefore, If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be $V_e = \sqrt{\dfrac{GM}{r}}$ .

To know more about escape velocity follow

brainly.com/question/14042253

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8 0

1 year ago

How do the tension of the cord and the force of gravity affect a pendulum?

LuckyWell [14K]

Answer:

<em>Force of gravity may not affect a pendulum during its equilibrium state</em>. But the gravity can affect the pendulum when a force occurs in any direction of the bob connected to the cord that makes a swing sideways. The gravity of pendulum never stops, it always accelerates. So the gravity affects the pendulum acceleration and speed.

<em>Similarly the tension in the cord will not affect the pendulum</em><em> </em>but if change in the length of the pendulum while keeping other factors constant changes the length of the period of pendulum. longer pendulum swings with lower frequency than shorter pendulums.

6 0

3 years ago

Read 2 more answers

Strength of electromagnets

Triss [41]

Lines of Force around an Electromagnet. ... The magnetic field strength of an electromagnet is therefore determined by the ampere turns of the coil with the more turns of wire in the coil the greater will be the strength of the magnetic field.

6 0

3 years ago

A T-shirt cannon can shoot a 0.085 kg T-shirt at nearly 30 m/s. The T-shirt cannon has a mass of 33 kg. If the initial net momen

IgorLugansk [536]

Answer:

Approximately $0.077\; {\rm m\cdot s^{-1}}$ (assuming that external forces on the cannon are negligible.)

Explanation:

If an object of mass $m$ is moving at a velocity of $v$ , the momentum $p$ of that object would be $p = m\, v$ .

Momentum of the t-shirt:

$\begin{aligned} p(\text{t-shirt}) &= m(\text{t-shirt}) \, v(\text{t-shirt}) \\ &= 0.085\; {\rm kg} \times 30\; {\rm m \cdot s^{-1}} \\ &= 2.55 \; {\rm kg \cdot m \cdot s^{-1}} \end{aligned}$ .

If there is no external force (gravity, friction, etc.) on this cannon, the total momentum of this system should be conserved. In other words, if $p(\text{cannon})$ denote the momentum of this cannon:

$p(\text{t-shirt}) + p(\text{cannon}) = 0$ .

$p(\text{cannon}) = -p(\text{t-shirt}) = -2.55\; {\rm kg \cdot m \cdot s^{-1}}$ .

Rewrite $p = m\, v$ to obtain $v = (p / m)$ . Since the mass of this cannon is $m(\text{cannon}) = 33\; {\rm kg}$ , the velocity of this cannon would be:

$\begin{aligned} v(\text{cannon}) &= \frac{p(\text{cannon})}{m(\text{cannon})} \\ &= \frac{-2.55\; {\rm kg \cdot m \cdot s^{-1}}}{33\; {\rm kg}} \\ &\approx 0.077\; {\rm m \cdot s^{-1}}\end{aligned}$ .

8 0

1 year ago

A certain capacitor, in series with a resistor, is being charged. At the end of 10 ms its charge is half the final value. The ti

vichka [17]

To solve this problem we will apply the expression of charge per unit of time in a capacitor with a given resistance. Mathematically said expression is given as

$q(t) = e^{-\frac{t}{(R*C)}}$