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Sergio [31]
4 years ago
7

Dry

Physics
1 answer:
AURORKA [14]4 years ago
7 0
A
All components of health
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What type of wave is formed by a sound wave?
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The answer is c
sound wave produced longitudinal wave
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What is the difference between a physical change and a chemical change
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A physical change in something doesn't change what the it is. For example, if you break glass, it will still be glass. In a chemical change where there is a chemical reaction, a new thing is formed and energy is either given off or absorbed. For example, when you burn a log. The carbon in the log is reacting to the oxygen to create ashe and smoke
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A shell of mass m and speed v explodes into two identical fragments. If the shell was moving horizontally (the positive x direct
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Answer:

The velocity of the other fragment immediately following the explosion is v .

Explanation:

Given :

Mass of original shell , m .

Velocity of shell , + v .

Now , the particle explodes into two half parts , i.e  \dfrac{m}{2} .

Since , no eternal force is applied in the particle .

Therefore , its momentum will be conserved .

So , Final momentum = Initial momentum

mv=\dfrac{mv}{2}+\dfrac{mu}{2}\\\\u=v

The velocity of the other fragment immediately following the explosion is v .

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3 years ago
A Hooke's law spring is mounted horizontally over a frictionless surface. The spring is then compressed a distance d and is used
zloy xaker [14]

Answer:

The compression is \sqrt{2} \  d.

Explanation:

A Hooke's law spring compressed has a potential energy

E_{potential} = \frac{1}{2} k (\Delta x)^2

where k is the spring constant and \Delta x the distance to the equilibrium position.

A mass m moving at speed v has a kinetic energy

E_{kinetic} = \frac{1}{2} m v^2.

So, in the first part of the problem, the spring is compressed a distance d, and then launch the mass at velocity v_1. Knowing that the energy is constant.

\frac{1}{2} m v_1^2 = \frac{1}{2} k d^2

If we want to double the kinetic energy, then, the knew kinetic energy for a obtained by compressing the spring a distance D, implies:

2 * (\frac{1}{2} m v_1^2) = \frac{1}{2} k D^2

But, in the left side we can use the previous equation to obtain:

2 * (\frac{1}{2} k d^2) = \frac{1}{2} k D^2

D^2 =  \frac{2 \ (\frac{1}{2} k d^2)}{\frac{1}{2} k}

D^2 =  2 \  d^2

D =  \sqrt{2 \  d^2}

D =  \sqrt{2} \  d

And this is the compression we are looking for

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4 years ago
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