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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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Scilla [17]

a) At a position of 2.0m, the Initial energy is all made up of the potential energy=m*g*hi<span>
and meanwhile at 1.5 all its energy is also potential energy=m*g*hf 

The percentage of energy remaining is E=m*g*hi/m*g*hf x 100 

and since mass and gravity are constant so it leaves us with just E=hi/hf 
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b) Here use the equation vf^2=vi^2+2gd 

<span>where g is gravity, vf is the final velocity and vi is the initial velocity while d is the distance travelled

so in here we are looking for the vi so let us isolate that variable 
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therefore,</span></span>

vi =sqrt(-2gd) <span>
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6 0
3 years ago
You are an astronaut in space far away from any gravitational field, and you throw a rock as hard as you can. The rock will:
Nesterboy [21]

Answer:

the rock will continue at the same speed unless it is affected by another force such as gravity and so if you threw it it will continue to move unless affected by a force

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What is the name of the ratio of the voltage applied to a circuit and the current in a circuit
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4 0
3 years ago
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liubo4ka [24]

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8 0
2 years ago
Find the momentum of a particl with a mass of one gram moving with half the speed of light.
joja [24]

Answer:

129900

Explanation:

Given that

Mass of the particle, m = 1 g = 1*10^-3 kg

Speed of the particle, u = ½c

Speed of light, c = 3*10^8

To solve this, we will use the formula

p = ymu, where

y = √[1 - (u²/c²)]

Let's solve for y, first. We have

y = √[1 - (1.5*10^8²/3*10^8²)]

y = √(1 - ½²)

y = √(1 - ¼)

y = √0.75

y = 0.8660, using our newly gotten y, we use it to solve the final equation

p = ymu

p = 0.866 * 1*10^-3 * 1.5*10^8

p = 129900 kgm/s

thus, we have found that the momentum of the particle is 129900 kgm/s

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