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2 years ago

If a certain mass of mercury has a volume of 0.002 m3 at a temperature of 20°C, what will be the volume at 50°C?

Physics

2 answers:

Lina20 [59]2 years ago

6 0

Answer : The correct option is, (A) $0.002010812m^3$

Solution : Given,

Volume of mercury at $20^oC$ is $0.002m^3$

As mercury is a liquid. So, we have to apply the volume of expansion of liquid.

Formula used for the volume expansion of liquid,

$V_{T}=v_{1}[1+\gamma (T_{2}-T_{1})]$

or,

$V_{2}=V_{1}[1+\gamma (T_{2}-T_{1})]$

where,

$V_{T}$ = volume of liquid at temperature $T^oC$

$V_{1}$ = volume of liquid at temperature $20^oC$

$V_{2}$ = volume of liquid at temperature $50^oC$

$\gamma$ = volume expansion coefficient of mercury at $20^oC$ is 0.00018 per centigrade

Now put all the given values in the above formula, we get

$V_{2}=0.002[1+0.00018(50-20)]=0.0020108m^3$

Therefore, the volume of mercury at $50^oC$ is, $0.0020108m^3$

Natalka [10]2 years ago

5 0

By using the combined gas law which says that P1V1/T1 = P2V2/T2, assuming constant pressure, the volume at the required temperature can be obtained.((0.002 m3)(50+273.15))/(20+273.15) = volume 2 at 50 degrees C.
This gives the answer 0.0022048 m3

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1. 1500j of work was done to move a box 20m. What force was applied to the box ?

Fantom [35]

Answer:

1. 75N

2. 67,983 J (=67.98 kJ)

Explanation:

1. Work = Force x Distance

we are given that Work = 1,500J and Distance = 20m

hence,

Work = Force x Distance

1,500 = Force x 20

Force = 1,500 ÷ 20 = 75N

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we are given that mass = 165 kg and change in height = 42m

assuming that gravity, g = 9.81 m/s²

Potential Energy, PE = mass x gravity x change in height

Potential Energy, PE = 165 x 9.81 x 42 = 67,983 J (=67.98 kJ)

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2 years ago

Boyles law<br> squeezing a balloon is one way to burst it. Why? ...?

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3 years ago

A Ferris wheel on a California pier is 27 m high and rotates once every 32 seconds in the counterclockwise direction. When the w

Leto [7]

A) 140 degrees

First of all, we need to find the angular velocity of the Ferris wheel. We know that its period is

T = 32 s

So the angular velocity is

$\omega=\frac{2\pi}{T}=\frac{2\pi}{32 s}=0.20 rad/s$

Assuming the wheel is moving at constant angular velocity, we can now calculate the angular displacement with respect to the initial position:

$\theta= \omega t$

and substituting t = 75 seconds, we find

$\theta= (0.20 rad/s)(75 s)=15 rad$

In degrees, it is

$15 rad: x = 2\pi rad : 360^{\circ}\\
x=\frac{(15 rad)(360^{\circ})}{2\pi}=860^{\circ} = 140^{\circ}$

So, the new position is 140 degrees from the initial position at the top.

B) 2.7 m/s

The tangential speed, v, of a point at the egde of the wheel is given by

$v=\omega r$

where we have

$\omega=0.20 rad/s$

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Substituting into the equation, we find

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2 years ago

ALWAYS use significant figure rules. Remember that these rules apply to all numbers that are measurements. A person walks along

Elza [17]

To find the total number of miles traveled by a person, we add the distance that he has traveled: 3.0 + 5.00 + 4.000.

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3 years ago

The spin cycle of a clothes washer extracts the water in clothing by greatly increasing the water's apparent weight so that it i

andreyandreev [35.5K]

The apparent weight of a 1.1 g drop of water is 4.24084 N.

<h3>What is Apparent Weight?</h3>

According to physics, an object's perceived weight is a characteristic that describes how heavy it is. When the force of gravity acting on an object is not counterbalanced by a force of equal but opposite normality, the apparent weight of the object will differ from the actual weight of the thing.
By definition, an object's weight is equal to the strength of the gravitational force pulling on it. It follows that even a "weightless" astronaut in low Earth orbit, with an apparent weight of zero, has almost the same weight that he would have if he were standing on the ground; this is because the gravitational pull of low Earth orbit and the ground are nearly equal.

Solution:

N = Speed of rotation = 1250 rpm

D = Diameter = 45 cm

r = Radius = 22.5 cm

M = Mass of drop = 1.1 g

Angular speed of the water = $\omega = \frac{2\pi N}{60}$

$\omega = \frac{2\pi \times 1250}{60}$

$\omega = 130.89 rad/s$

Apparent weight is given by

$W _a = M\omega^{2}R$

$W_a = 1.1 \times 10^-^3\times (130.89)^2\times 0.225$

$W_a$ = 4.24084 N

Know more about Apparent weight brainly.com/question/14323035

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Question:

The spin cycle of a clothes washer extracts the water in clothing by greatly increasing the water's apparent weight so that it is efficiently squeezed through the clothes and out the holes in the drum. In a top loader's spin cycle, the 45-cm-diameter drum spins at 1250 rpm around a vertical axis. What is the apparent weight of a 1.1 g drop of water?

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1 year ago