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3 years ago

Describe an experiment to show that liquids expand and contract at different rates

Physics

1 answer:

Arte-miy333 [17]3 years ago

8 0

Answer:

Depend on the rate of temperature applied.

Explanation:

Liquid expands on heating.

When liquid is heated then there is an increase in the average speed of it's molecules as a result molecules starts moving faster and hence kinetic energy of the liquid also get increased. This kinetic energy is directly proportional to the temperature applied. So, increase in temperature result in increase in kinetic energy which reduces the inter-molecular force of fluid and as a result liquid expands.

Liquid contracts on cooling.

Fall in temperature results in reduced kinetic energy which in turn also decreases the kinetic energy of the molecules of liquid thereby increasing the inter-molecular force of attraction of fluid that's why liquid contracts on cooling.

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The origin of an x axis is placed at the center of a nonconducting solid sphere of radius R that carries a charge +qsphere distr

MA_775_DIABLO [31]

Answer:

$q=49Q/64$

and

$x =16R/15$

Explanation:

See attached figure.

$E_{Q}=$ E due to sphere

$E_{q}=$ E due to particule

$E_{total}=E_{Q}-E_{q}=0$ (1)

according to the law of gauss and superposition Law:

$E_{Q}=E_{1}+E_{2}=E_{2}$ ; electric field due to the small sphere with r1=R/4

$E_{Q}=kq_{2}/(r_{1}^{2})=$

$q_{2}=density*4/3*pi*r_{1}^{3}=Q/(4/3*pi*R^{3})*4/3*pi*r_{1}^{3}=Q*r_{1}^{3}/R^{3}$

then: $E_{Q}=kq_{2}/(r_{1}^{2})=k*Q*r_{1}^{3}/(R^{3}*r_{1}^{2}) = kQ/(4*R^{2})$ (2)

on the other hand, for the particule:

$E_{q}=kq/(r_{p}^{2})$

$r_{p}=2R-R/4=7R/4$ ⇒ $E_{q}=16kq/(49R^{2})$ (3)

We replace (2) y (3) in (1):

$E_{total}=E_{Q}-E_{q}=0=kQ/(4*R^{2}) - 49kq/(16R^{2})$

$q=49Q/64$

--------------------

if R<x<2R AND $E_{total}=E_{Q}-E_{q}=0$

$E_{total}=E_{Q}-E_{q}=0=kQ/(x^{2}) - kq/(2R-x^{2})$

remember that $q=49Q/64$

then:

$Q(2R-x^{2})=49/64*x^{2}$

solving:

$x_{1} =16R/15$

$x_{2} =16R$

but: R<x<2R

so : $x =16R/15$

7 0

3 years ago

What is your angular position 75 seconds after the wheel starts turning, measured counterclockwise from the top? Express your an

AveGali [126]

Complete Question

A Ferris wheel on a California pier is 27 m high and rotates once every 32 seconds in the counterclockwise direction. When the wheel starts turning, you are at the very top.

What is your angular position 75 seconds after the wheel starts turning, measured counterclockwise from the top? Express your answer as an angle between 0∘ and 360∘. Express your answer in degrees.

Answer:

$\phi=123.75$

Explanation:

From the question we are told that:

Height $h=27m$

Period $T=32sec$

Time $t=75sec$

Generally the equation for angular velocity is mathematically given by

$\omega=\frac{2 \pi}{T}$

$\omega=\frac{2 \pi}{32}$

$\omega=0.196rad/s$

Therefore

$\theta=\omega t$

$\theta=0.196rad/s*75sec$

$\theta=843.75 \textdegree$

Therefore

$\phi=\theta-2(360)$

$\phi=123.75$

6 0

3 years ago

If the observed test value of a hypothesis test is outside of the established critical value(s), a researcher would __________.

sashaice [31]

I just had this question, the awnser is A.

6 0

3 years ago

Read 2 more answers

A cord of mass 0.65 kg is stretched between two supports 28 m apart. if the tension in the cord is 150 n, 2 how long will it tak

MaRussiya [10]

Ans: Time <span>taken by a pulse to travel from one support to the other = 0.348s
</span>
Explanation:
First you need to find out the speed of the wave.

Since
Speed = v = $\sqrt{ \frac{T}{\mu} }$

Where
T = Tension in the cord = 150N
μ = Mass per unit length = mass/Length = 0.65/28 = 0.0232 kg/m

So
v = $\sqrt{ \frac{150}{0.0232} }$ = 80.41 m/s

Now the time-taken by the wave = t = Length/speed = 28/80.41=0.348s

5 0

3 years ago

Water from a fire hose is directed horizontally against at a rate of 50.0 kg/s and a speed of 42.0 m/s.

ss7ja [257]

Answer:

Force, |F| = 2100 N

Explanation:

It is given that,

Water from a fire hose is directed horizontally against at a rate of 50.0 kg/s, $\dfrac{m}{t}=50\ kg/s$