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1 year ago

A pendulum has a mass of 3kg and is lifted to a height of .3m. What is the maximum speed of the pendulum

Physics

1 answer:

Kryger [21]1 year ago

7 0

Given data

*The given mass of the pendulum is m = 3 kg

*The given height is h = 0.3 m

The formula for the maximum speed of the pendulum is given as

$v_{\max }=\sqrt[]{2gh}$

*Here g is the acceleration due to the gravity

Substitute the values in the above expression as

$\begin{gathered} v_{\max }=\sqrt[]{2\times9.8\times0.3} \\ =2.42\text{ m/s} \end{gathered}$

Hence, the maximum speed of the pendulum is 2.42 m/s

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What is required to cause acceleration

Vesna [10]

Constant force. Basicly in no gravity you have to keep puching smth for it to accelerate and once you stop it just would remain at the same speed. On earth the fact that something is falling faster and faster means it keeps beeing constantly pulled by gravity so the same principle applies

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4 years ago

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What is the weight in Newtons of a ball with a mass of 7.77 kg?

rusak2 [61]

Answer:

76.1N

Explanation:

Given parameters:

Mass of the ball = 7.77kg

Unknow:

Weight of balloon = ?

Solution:

Weight is the vertical force applied on a body.

Weight = mass x acceleration due gravity

So;

Weight = mass x acceleration due to gravity

So;

Weight = 7.77 x 9.8 = 76.1N

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3 years ago

If you were on a ship at sea, and a tsunami passed under your ship, what would probably be your reaction? explain.

swat32

<span>Extremely powerful single waves have no effect on ships at sea since the depth of water allows the energy to be distributed over hundreds and thousands of feet. In deep water, the bigger the wave, the faster it moves and the slower the surface changes height. As the wave gets into shallow waters, it slows down and can start to pile up to large heights.</span>

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3 years ago

Consider a model of a hydrogen atom in which an electron is in a circular orbit of radius r = 5.92×10−11 m around a stationary p

DaniilM [7]

Answer:

2.068 x 10^6 m / s

Explanation:

radius, r = 5.92 x 10^-11 m

mass of electron, m = 9.1 x 10^-31 kg

charge of electron, q = 1.6 x 10^-19 C

As the electron is revolving in a circular path, it experiences a centripetal force which is balanced by the electrostatic force between the electron and the nucleus.

centripetal force = $\frac{mv^{2}}{r}$

Electrostatic force = $\frac{kq^{2}}{r^{2}}$

where, k be the Coulombic constant, k = 9 x 10^9 Nm^2 / C^2

So, balancing both the forces we get

$\frac{kq^{2}}{r^{2}}=\frac{mv^{2}}{r}$

$v=\sqrt{\frac{kq^{2}}{mr}}$

$v=\sqrt{\frac{9\times 10^{9}\times1.6\times 10^{-19}\times 1.6\times 10^{-19}}{9.1\times 10^{-31}\times 5.92\times10^{-11}}}$

v = 2.068 x 10^6 m / s

Thus, the speed of the electron is give by 2.068 x 10^6 m / s.

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3 years ago

In an extrasolar planetary system containing a single planet, the parent star is measured to move about its center of mass every

Mademuasel [1]

Answer:

Orbital Time Period is 24 years

Explanation:

This can be explained by the definition of time period.

Time period can be defined as the time taken by an object to complete one cycle, here, time taken to complete one revolution.

Also, we know that an extra solar planet which is also called as an exo planet is that planet which is outside our solar system and orbits any star other than our sun. The system in consideration is extra solar system with a single planet.

Therefore, the time taken by the parent star to move about its mass center is the orbital time period that is 24 years.

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3 years ago