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devlian [24]
4 years ago
9

Match the physical properties on the left with their descriptions on the right. 1. luster ability to dissolve in another substan

ce 2. freezing point the shininess of a material 3. viscosity temperature at which a liquid becomes a solid 4. solubility able to be pulled into a wire 5. ductility the measure of a substance's resistance to flow
Physics
1 answer:
Serjik [45]4 years ago
7 0

Answer :

1) Luster →  the shininess of a material

2) Freezing point → temperature at which a liquid becomes a solid

3) Viscosity → the measure of a substance's resistance to flow

4) Solubility → ability to dissolve in another substance

5) Ductility → able to be pulled into a wire

Explanation :

Luster : It is defined as the material which has the shining property.

Freezing point : It is the temperature at which the phase changes from liquid state to solid state at low temperature.

Viscosity : It measures the resistance of the liquids to flow .

Solubility : It is defined as the ability of a substance (solute) to dissolve into a liquid (solvent).

Ductility : It is defined as the ability of a metal into a thin wire without breaking.

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In my trigonometry class, we were assigned a problem on Angular and Linear Velocity.
Rzqust [24]

1) 0.0011 rad/s

2) 7667 m/s

Explanation:

1)

The angular velocity of an object in circular motion is equal to the rate of change of its angular position. Mathematically:

\omega=\frac{\theta}{t}

where

\theta is the angular displacement of the object

t is the time elapsed

\omega is the angular velocity

In this problem, the Hubble telescope completes an entire orbit in 95 minutes. The angle covered in one entire orbit is

\theta=2\pi rad

And the time taken is

t=95 min \cdot 60 =5700 s

Therefore, the angular velocity of the telescope is

\omega=\frac{2\pi}{5700}=0.0011 rad/s

2)

For an object in circular motion, the relationship between angular velocity and linear velocity is given by the equation

v=\omega r

where

v is the linear velocity

\omega is the angular velocity

r is the radius of the circular orbit

In this problem:

\omega=0.0011 rad/s is the angular velocity of the Hubble telescope

The telescope is at an altitude of

h = 600 km

over the Earth's surface, which has a radius of

R = 6370 km

So the actual radius of the Hubble's orbit is

r=R+h=6370+600=6970 km = 6.97\cdot 10^6 m

Therefore, the linear velocity of the telescope is:

v=\omega r=(0.0011)(6.97\cdot 10^6)=7667 m/s

4 0
3 years ago
A ball moving 2 ft/s rolls off a table (on earth) that is 32 inches high. How long will it take the ball to hit the floor answer
mart [117]

Answer:When the ball rolls off the edge of the table, it will continue moving forward at 2.0 m/s until it hits the floor.

Explanation:This is what I would say is the answer bc I had to do reasearch on a lot of this for my work this year so if its not im veery sorry

3 0
3 years ago
A conducting spherical shell with inner radius ‘a’ and outer radius ‘b’ has a positive charge Q located at its center. The total
kifflom [539]

Answer:deeeznuts

Explanation:

4 0
3 years ago
Identify the type of chemical reaction that is described.
Anastasy [175]

Water breaks down into  hydrogen and oxygen: It is a decomposition reaction as a single substance decomposes to give two products.

H_2O \rightarrow H_2 +\frac {1}{2}O_2

Leaves make starch using  chlorophyll and carbon dioxide: Synthesis reaction: as the synthesis reaction involves two or more than two reactants which join together to result into a single main product along with the formation of simple by products.

6CO_2 + 6H_2O \rightarrow C_6H_{12}O_6 + 6O_2

Food burns in oxygen gas  and releases a lot of energy: Combustion: Combustion process involves the use of oxygen to give products along with release of energy.

C_6H_{12}O_6 + 6O_2\rightarrow 6CO_2 + 6H_2O

Adding vinegar (acid) to baking soda (alkali) gives  a product that is neither  acidic nor alkaline: Neutralization: acetic acid in vinegar reacts with soda (base) to give salt (neutral) and .

CH_3COOH + NaHCO_3\rightarrow CH_3COONa +H_2O + CO_2





6 0
4 years ago
Objects with masses of 135 kg and a 435 kg are separated by 0.500 m. Find the net gravitational force exerted by these objects o
IrinaVladis [17]

Answer:

The net gravitational force on the mass is 1.27\times 10^{-5}

Explanation:

We have by Newton's law of gravity the force of attraction between masses m_{1},m_{2}

F_{att}=G\frac{m_{1}m_{2}}{r^{2}}

Applying vales we get

Force of attraction between 135 kg mass and 38 kg mass is

F_{1}=6.67\times 10^{-11}\frac{135\times 38}{(0.25)^{2}}\\\\F_{1}=5.47\times 10^{-6}N

Force of attraction between 435 kg mass and 38 kg mass is

F_{2}=6.67\times 10^{-11}\frac{435\times 38}{(0.25)^{2}}\\\\F_{2}=1.76\times 10^{-5}N

Thus the net force on mass 38.0 kg is F_{2}-F_{1}=1.76\times 10^{-5}-5.47\times 10^{-6}\\\\F_{2}-F_{1}=1.27\times 10^{-5}

8 0
4 years ago
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