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3 years ago

1. Water boils at 100°C at sea level. If the water in this experiment did not boil at 100°C, what could be the reason?

Physics

2 answers:

umka2103 [35]3 years ago

8 0

Answer:

It may not be at the sea level

Explanation:

The reason here is water only boils at sea level. This means that if you move water to a different height, say top of a mountain, the boiling temperature of water would change. This is due to the pressure drop at high place. The drop of pressure would make it harder to transform water liquid to gas, thus requiring more temperature.

yulyashka [42]3 years ago

8 0

Answer:

The experiment is not performed at sea level.

And also there may be presence of impurities in the water used for the experiment

Explanation:

The boiling point of pure water at sea level is 100°C, but the boiling point can be affected mainly by two factors;

i) Pressure

ii) impurities

At higher altitudes ( above sea level) the atmospheric pressure is reduced (the higher you go the lower the atmospheric pressure), thereby leading to reduced boiling point of water (at lower pressure water boils faster). While if the altitude is lower than sea level atmospheric pressure is higher so water boils at higher temperatures (above 100°C).

Secondly, presence of impurities(e.g salt) in water increases the boiling point of water in most cases.

Therefore, if water does not boil at 100°C it may be as a result of the elevation (altitude) or the water may contain impurities.

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Imagine a system where a block rests on an inclined plane. The block is then given an initial push so that it starts sliding dow

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Answer:

statement - 'The work done by friction is equal to the sum of the work done by the gravity and the initial push' is correct.

Explanation:

The statement ''The work done by friction is equal to the sum of the work done by the gravity and the initial push" is correct.

The above statement is correct because, the initial push will tend to slide down the block thus the work done by the initial push will be in the downward direction. Also, the gravity always acts in the downward direction. thus, the work done done by the gravity will also be in the downward direction

here, the downward direction signifies the downward motion parallel to the inclined plane.

Now we know that the work done by the friction is against the direction of motion. Thus, the friction force will tend to move the block up parallel to the inclined plane.

Hence, for the block to stop sliding the the above statement should be true.

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3 years ago

A long non-conducting hollow cylinder with inner radius, ri and outer radius, r2 is charged with a uniform positive charge with

IgorLugansk [536]

Answer:yup

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Two cyclists start on a race between points A and D on two different routes. Cyclist X takes the route passing through the equid

ollegr [7]

The displacement is the shortest distance between two points, which is 546.41. The displacement for both is 546.41 meters

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3 years ago

Read 2 more answers

21) A youngster having a mass of 50.0 kg steps off a 1.00 m high platform. If she keeps her legs fairly rigid and comes to rest

zlopas [31]

Answer:

-22,150 N

Explanation:

When the youngster jumps off the platform, during the fall her initial potential energy is converted into kinetic energy, according to the law of conservation of energy. Therefore, we can write:

$mgh=\frac{1}{2}mu^2$

where the term on the left is the potential energy while the term on the right is the kinetic energy, and where

m = 50.0 kg is the mass of the youngster

$g=9.8 m/s^2$ is the acceleration due to gravity

h = 1.00 m is the heigth of the platform

u is the speed of the youngster as she reaches the floor

Solving for u,

$u=\sqrt{2gh}=\sqrt{2(9.8)(1.00)}=4.43 m/s$

Then, when the youngster hits the floor, the force exerted on her during the deceleration is given by:

$F=\frac{\Delta p}{\Delta t}=\frac{m(v-u)}{\Delta t}$

where $\Delta p$ is her change in momentum, and where

m is the mass

v = 0 is the final velocity (she comes to a stop)

u = 4.43 m/s is the initial velocity

$\Delta t=10.0 ms =0.010 s$ is the duration of the collision

Substituting,

$F=\frac{(50.0)(0-4.43)}{0.010}=-22150 N$

And the negative sign means the direction of the force is opposite to the motion (so, upward).

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3 years ago

A 150 kg line backer sacks the 120 kg quarterback. With what force is the quarterback sacked if the line backer has an accelerat

Gekata [30.6K]

Answer:

The force required to move the quarterback with linebacker is <u>1215 N</u>

Explanation:

$\text { Mass of linebacker } \mathrm{m}_{2}=150 \mathrm{kg}$

$\text { Mass of quarterback } \mathrm{m}_{2}=120 \mathrm{kg}$

$\text { Moved at an acceleration }(a)=4.5 \mathrm{m} / \mathrm{s}^{2}$

Using Newton's second law, it is established that F = Ma

Where F is net force acting on the system, a is the acceleration and M is mass of the two object $\left(m_{1}+m_{2}\right)$

Now consider both $\mathrm{m}_{1} \text { and } \mathrm{m}_{2}$ as a system, so net force acting on the system is $\text { Force }=\left(m_{1}+m_{2}\right) a$

Substitute the given values in the above formula,

$\text { Force }=(150+120) \mathrm{kg} \times 4.5 \mathrm{m} / \mathrm{s}^{2}$

$\text { Force }=270 \mathrm{kg} \times 4.5 \mathrm{m} / \mathrm{s}^{2}$

Force = 1215 N

<u>1215 N </u>is the force required to move the quarterback with linebacker.

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3 years ago