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2 years ago

Determine the amount of potential energy of a 5-newton book that is moved to three different shelves on a

Physics

2 answers:

JulijaS [17]2 years ago

4 0

Multiply 5 newton by each height. 1st is 5 joules, 2nd is 7.5, while 3rd is 10 joules.

Elza [17]2 years ago

3 0

Yo sup??

potential energy is given by

U=mgh

mg=5 N

U1=5*1=5 J

U2=5*1.5=7.5 J

U3=5*2=10 J

Hope this helps.

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A pipe is open at both ends. The pipe has resonant frequencies of 528 Hz and 660HZ (among others).

yawa3891 [41]

To develop this problem it is necessary to apply the oscillation frequency-related concepts specifically in string or pipe close at both ends or open at both ends.

By definition the oscillation frequency is defined as

$f = n\frac{v}{2L}$

Where

v = speed of sound

L = Length of the pipe

n = any integer which represent the number of repetition of the spectrum (n)1,2,3...)(Number of harmonic)

Re-arrange to find L,

$f = n\frac{v}{2L}\\L = \frac{nv}{2f}$

The radius between the two frequencies would be 4 to 5,

$\frac{528Hz}{660Hz}= \frac{4}{5}$

$4:5$

Therefore the frequencies are in the ratio of natural numbers. That is

$4f = 528\\f = \frac{528}{4}\\f = 132Hz$

Here f represents the fundamental frequency.

Now using the expression to calculate the Length we have

$L = \frac{nv}{2f}\\L = \frac{(1)343m/s}{2(132)}\\L = 1.29m$

Therefore the length of the pipe is 1.3m

For the second harmonic n=2, then

$L = \frac{nv}{2f}\\L = \frac{(2)343m/s}{2(132)}\\L = 2.59m$

Therefore the length of the pipe in the second harmonic is 2.6m

7 0

2 years ago

A charge Q is to be divided into two parts, labeled 'q' and 'Q-q? What is the relationship of q to Q if, at any given distance,

Lisa [10]

Answer:

Explanation:

The two charges are q and Q - q. Let the distance between them is r

Use the formula for coulomb's law for the force between the two charges

$F = \frac{Kq_{1}q_{2}}{r^{2}}$

So, the force between the charges q and Q - q is given by

$F = \frac{K\left ( Q-q \right )q}}{r^{2}}$

For maxima and minima, differentiate the force with respect to q.

$\frac{dF}{dq} = \frac{k}{r^{2}}\times \left ( Q - 2q \right )$

For maxima and minima, the value of dF/dq = 0

So, we get

q = Q /2

Now $\frac{d^{2}F}{dq^{2}} = \frac{-2k}{r^{2}}$

the double derivate is negative, so the force is maxima when q = Q / 2 .

6 0

3 years ago

Why do outer planets take so long to orbit the sun

Hoochie [10]

Answer:

Because the gravitational pull is weaker, And the outer planets are further away so they have more distance to cover in orbit.

5 0

2 years ago

Read 2 more answers

Yo-Yo man releases a yo-yo from rest and allows it to drop, as he keeps the top end of the string stationary. The mass of the yo

patriot [66]

To solve this problem it is necessary to use the conservation equations of both kinetic, rotational and potential energy.

By definition we know that

$KE + KR = PE$

Where,

KE =Kinetic Energy

KR = Rotational Kinetic Energy

PE = Potential Energy

In this way

$\frac{1}{2} mv^2 +\frac{1}{2} I\omega^2 = mgh$

Where,

m = mass

v= Velocity

I = Moment of Inertia

$\omega =$ Angular velocity

g = Gravity

h = Height

We know as well that $\omega = v/r$ for velocity (v) and Radius (r)

Therefore replacing we have

$\frac{1}{2} mv^2 +\frac{1}{2} I\omega^2 = mgh$

$[tex]h= \frac{1}{2} \frac{v^2}{g} +\frac{1}{2} \frac{I}{mg}(\frac{v}{r})^2$ [/tex]

$h= \frac{1}{2}v^2 ( \frac{1}{g} +\frac{I}{mg}\frac{1}{r^2} )$

$h= \frac{1}{2}0.75^2 ( \frac{1}{9.8} +\frac{2.9*10^{-5}}{(0.056)(9.8)}\frac{1}{(0.0064)^2} )$

$h = 0.3915m$

Therefore the height must be 0.3915 for the yo-yo fall has a linear speed of 0.75m/s

6 0

3 years ago

A cube has a density of 1900 kg/m 3 kg/m3 while at rest in the laboratory. What is the cube's density as measured by an experime

Deffense [45]

Answer:

4847.94844926 kg/m³

Explanation:

$\rho'$ = Actual density of cube = 1900 kg/m³

$\rho$ = Density change due to motion

v = Velocity of cube = 0.92c

c = Speed of light = $3\times 10^8\ m/s$

Relativistic density is given by

$\rho=\dfrac{\rho'}{\sqrt{1-\dfrac{v^2}{c^2}}}\\\Rightarrow \rho=\dfrac{1900}{\sqrt{1-\dfrac{0.92^2c^2}{c^2}}}\\\Rightarrow \rho=\dfrac{1900}{\sqrt{1-0.92^2}}\\\Rightarrow \rho=4847.94844926\ kg/m^3$

The cube's density as measured by an experimenter in the laboratory is 4847.94844926 kg/m³

3 0

2 years ago