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2 years ago

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An Olympic high diver (mass = 70 kg) stands on a platform 20 m above the surface of the water. What is the diver's gravitational

potential energy?

1 answer:

raketka [301]2 years ago

4 0

Answer: 13720

Explanation: <em>the equation is....</em>

<h3>(70 kg) (9.8 m/ $s^{2}$ ) (20 m) = <em><u>13720</u></em></h3>

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A key falls from a bridge that is 32 m above the water. It falls directly into a model boat, moving with constant velocity, that

FinnZ [79.3K]

Answer:

Speed of the boat, v = 4.31 m/s

Explanation:

Given that,

Height of the bridge, h = 32 m

The model boat is 11 m from the point of impact when the key was released, d = 11 m

Firstly, we will find the time needed for the boat to get in this position using second equation of motion as :

$s=ut+\dfrac{1}{2}at^2$

Here, u = 0 and a = g

$t=\sqrt{\dfrac{2s}{g}}$

$t=\sqrt{\dfrac{2\times 32}{9.8}}$

t = 2.55 seconds

Let v is the speed of the boat. It can be calculated as :

$v=\dfrac{d}{t}$

$v=\dfrac{11\ m}{2.55\ s}$

v = 4.31 m/s

So, the speed of the boat is 4.31 m/s. Hence, this is the required solution.

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3 years ago

HOW MANY PIES DOSE IT TAKE TO GET TO THE MOON

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write an essay describing and predicting the effects of fitness-related stress management techniques on the body.

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This is something u are going to have to do

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3 years ago

Interest groups representing businesses and investors are often among the most successful lobbying groups in foreign policy. Why

alisha [4.7K]

The members of these groups make up the majority of voters in many districts thus this be considered a problem.

<u>Option: D</u>

<u>Explanation:</u>

Interest groups play a key role in US politics. Such organizations are made up of wealthy and powerful members who often seek to impose some form of leverage in politicians to promote their goals and agendas. Across the years via many campaigns, they have understood how to speak and manipulate elected leaders and apply leverage to get the kind of legislation that is in their favor. Here the majority of voters in several districts are standing due to group members, as we recognize the interest group belongs to a body in which it uses different methods of lobbying to influence others.

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3 years ago

A mass is oscillating with amplitude A at the end of a spring.

Dmitry_Shevchenko [17]

A) $x=\pm \frac{A}{2\sqrt{2}}$

The total energy of the system is equal to the maximum elastic potential energy, that is achieved when the displacement is equal to the amplitude (x=A):

$E=\frac{1}{2}kA^2$ (1)

where k is the spring constant.

The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:

$E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2$ (2)

where x is the displacement, m the mass, and v the speed.

We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:

$U=\frac{1}{3}K$

Using (2) we can rewrite this as

$U=\frac{1}{3}(E-U)=\frac{1}{3}E-\frac{1}{3}U\\U=\frac{E}{4}$

And using (1), we find

$U=\frac{E}{4}=\frac{\frac{1}{2}kA^2}{4}=\frac{1}{8}kA^2$

Substituting $U=\frac{1}{2}kx^2$ into the last equation, we find the value of x:

$\frac{1}{2}kx^2=\frac{1}{8}kA^2\\x=\pm \frac{A}{2\sqrt{2}}$

B) $x=\pm \frac{3}{\sqrt{10}}A$

In this case, the kinetic energy is 1/10 of the total energy:

$K=\frac{1}{10}E$

Since we have

$K=E-U$

we can write

$E-U=\frac{1}{10}E\\U=\frac{9}{10}E$

And so we find:

$\frac{1}{2}kx^2 = \frac{9}{10}(\frac{1}{2}kA^2)=\frac{9}{20}kA^2\\x^2 = \frac{9}{10}A^2\\x=\pm \frac{3}{\sqrt{10}}A$

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3 years ago