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faltersainse [42]
3 years ago
5

What is the melting point of substance A?

Physics
1 answer:
Misha Larkins [42]3 years ago
3 0

Answer:

Solids are easily recognized by their ability to retain a fixed shape and definite volume. Particles making

up a solid are held together in a rigid form. They are not free to move about or slide past one another and

the solid does not have the ability to flow. (Although the particles of a solid do not move position to position, they do have motion in that they are constantly vibrating.

To change the temperature of a solid, heat energy must be added. The amount of heat energy that changes

the temperature of 1.0 g of a solid by 1.0°C is called its specific heat (c). Each substance has its own

specific heat. The specific heat of ice is 2.1 Joules/g°C. In other words we must supply 1.0 gram of ice

with 2.1 Joules of heat energy to raise its temperature by 1.0 °C.

The general equation for calculating heat energy to change the temperature of a solid is:

Heat = Mass x Specific Heat (solid) x Temperature Change

Q = m c DT

10 g 10 g 10 g 10 g 10 g 10 g

Calculate the heat necessary to change 10 g of ice(s) at -20 °C to 10 g of ice(s) at 0°C. (A-B)

Q = mc∆T = (10 g) (2.1 J/g°C) (20°C) = 420 J

If you continue to add heat energy once the temperature of the ice reaches 0°C , the heat absorbed is called

the heat of fusion (Lf). This heat is used to cause a change of phase (from a solid to a liquid). This heat is

increasing the potential energy of the molecules of the solid. No temperature change takes place. Each

substance has its own heat of fusion. The heat of fusion for ice is 340 Joules/g. Exactly the same amount

of heat is given up when 1.0 g of water is changed to ice. This heat is called the heat of crystallization.

The general equation for calculating heat energy to change a solid to a liquid is:

Heat = Mass x Heat of Fusion

Q = m Lf

Calculate the heat necessary to change 10 g of ice(s) at 0°C to 10 g of water(l) at 0°C.(B-C)

Explanation:

Q = mLf = (10 g)( 340 J/g) = 3400 J

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Ammonia enters the expansion valve of a refrigeration system at a pressure of 10 bar and a temperature of 22oC and exits at 2.0
KonstantinChe [14]

Answer:

The  the quality of the refrigerant at the exit of the expansion valve is 0.179.

Explanation:

Given that,

Initial pressure = 10 bar

Temperature = 22°C

Final pressure = 2.0 bar

We using the value of h

h = 293.4\ kJ/kg

The refrigerant during expansion undergoes a throttling process

Therefore, h_{1}=h_{2}

We need to calculate the quality of the refrigerant at the exit of the expansion valve

At 2.0 bar,

The property of ammonia

h_{f}=47.8 kJ/kg

h_{g}=1417.7 kJ/kg

Using formula

h_{2}=h_{f}+x(h_{g}-h_{f})

Put the value into the formula

293.4 =47.8+x(1417.7-47.8)

x=\dfrac{293.4-47.8}{1417.7-47.8}

x=0.179

Hence, The  the quality of the refrigerant at the exit of the expansion valve is 0.179.

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3 years ago
Explain two ways you could reduce the friction between two surfaces
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Physics help please
zhuklara [117]

Answer: 37.981 m/s

Explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the ball has two components: <u>x-component</u> and <u>y-component.</u> Being their main equations as follows:

<u>x-component: </u>

x=V_{o}cos\theta t   (1)

Where:

x=52 m is the point where the ball strikes ground horizontally

V_{o} is the ball's initial speed

\theta=0 because we are told the ball is thrown horizontally

t is the time since the ball is thrown until it hits the ground

<u>y-component: </u>

y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}   (2)

Where:

y_{o}=120m  is the initial height of the ball

y=0  is the final height of the ball (when it finally hits the ground)

g=-9.8m/s^{2}  is the acceleration due gravity

Knowing this, let's start by finding t from (2):

<u></u>

0=y_{o}+V_{o}sin(0\°) t+\frac{gt^{2}}{2}   (3)

0=y_{o}+\frac{gt^{2}}{2}  

t=\sqrt{\frac{-2 y_{o}}{g}}   (4)

t=\sqrt{\frac{-2 (120 m)}{-9.8m/s^{2}}}   (5)

t=4.948 s   (6)

Then, we have to substitute (6) in (1):

x=V_{o}cos(0\°) t   (7)

And find V_{o}:

V_{o}=\frac{x}{t}   (8)

V_{o}=\frac{52 m}{4.948 s}   (9)

V_{o}=10.509 m/s   (10)

On the other hand, since we are dealing with constant acceleration (due gravity) we can use the following equation to find the value of the ball's final velocity V:

V=V_{o} + gt (11)

V=10.509 m/s + (-9.8 m/s^{2})(4.948 s) (12)

V=-37.981 m/s (13) This is the ball's final velocity, and the negative sign indicates its direction is downwards.

However, we were asked to find the <u>ball's final speed</u>, which is the module of the ball's final vleocity vector. This module is always positive, hence the speed of the ball just before it strikes the ground is 37.981 m/s (positive).

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