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3 years ago

Which state of matter can be compressed?

Physics

2 answers:

aivan3 [116]3 years ago

6 0

D is the answer I believe

MA_775_DIABLO [31]3 years ago

3 0

Gases............In gases, the atoms are much more spread out than in solids or liquids, and the atoms collide randomly with one another. A gas will fill any container, but if the container is not sealed, the gas will escape. Gas can be compressed much more easily than a liquid or solid

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The sediment deposited by glaciers is called __________ .

nalin [4]

Answer:

The sediment deposited by glaciers is called Glacial deposition.

3 0

3 years ago

a stone is thrown vertically upwards with an initial velocity of 20 metre per second find the maximum height it reaches and the

ziro4ka [17]

Answer:

y = 20 [m]

Explanation:

To solve this type of problem we must use the following equation of kinematics.

$v_{f}^{2} =v_{o}^{2}-2*g*y$

where:

Vf = final velocity = 0

Vo = initial velocity = 20 [m/s]

g = gravity acceleration = 10 [m/s²]

y = elevation [m]

Now the key to solving this problem is to know that when the stone reaches the maximum height its speed is zero. Therefore the final velocity term is zero.

$0 = (20)^{2}-2*10*y\\20*y=400\\y=20 [m]$

Another important fact is that the sign of gravitational acceleration was taken as negative, since it acts in the opposite direction to the movement of the stone.

4 0

3 years ago

To demonstrate an elastic collision, a teacher places a tennis ball on the floor. She wants to hit the ball so that the followin

ExtremeBDS [4]

Those conditions are met when you have an elastic collision of equal masses.
So the answer would be a tennis ball.
Let us write momentum conservation law for this problem. The left-hand side is before the collision and right-hand side is after the collision:
$m_1v_1=m_2v_1$
What this equation means is that all momentum of the first ball is transferred to the second ball and velocities before and after the collision are the same.
There is no need to write energy conservation law. We can see that above equation holds true only if $m_1=m_2$ .

8 0

3 years ago

At what distance above the surface of the earth is the acceleration due to the earth's gravity 0.995 m/s2 if the acceleration du

gregori [183]

If we assume the Earth to be of uniform density and spherical, then Newton's inverse square law of g-field can be applied.

i.e. g = GM/r², which means g is inversely proportional to r²
Or, gr² = constant.

The Earth's radius is approximately 6.4x10^6 m

So, 0.975 x (R')² = 9.80 x (6.4x10^6)², where R' is the distance from Earth's centre.
R' = 2.03x10^7 m

Distance above Earth's surface = 2.03x10^7 - 6.4x10^6 = 1.39x10^7

4 0

4 years ago

Read 2 more answers

Suppose you take a 50gram ice cube from the freezer at an initial temperature of -20°C. How much energy would it take to complet

notsponge [240]

Answer:

The amount of energy required is $152.68\times 10^{3}Joules$

Explanation:

The energy required to convert the ice to steam is the sum of:

1) Energy required to raise the temperature of the ice from -20 to 0 degree Celsius.

2) Latent heat required to convert the ice into water.

3) Energy required to raise the temperature of water from 0 degrees to 100 degrees

4) Latent heat required to convert the water at 100 degrees to steam.

The amount of energy required in each process is as under

1) $Q_1=mass\times S.heat_{ice}\times \Delta T\\\\Q_1=50\times 2.05\times 20=2050Joules$

where

$'S.heat_{ice}$ ' is specific heat of ice = $2.05J/^{o}C\cdot gm$

2) Amount of heat required in phase 2 equals

$Q_2=L.heat\times mass\\\\\therefore Q_{2}=334\times 50=16700Joules$

3) The amount of heat required to raise the temperature of water from 0 to 100 degrees centigrade equals

$Q_3=mass\times S.heat\times \Delta T\\\\Q_1=50\times 4.186\times 100=20930Joules$

where

$'S.heat_{water}$ ' is specific heat of water= $4.186J/^{o}C\cdot gm$

4) Amount of heat required in phase 4 equals

$Q_4=L.heat\times mass\\\\\therefore Q_{4}=2260\times 50=113000Joules\\\\$ $\\\\\\\\$ Thus the total heat required equals $Q=Q_{1}+Q_{2}+Q_{3}+Q_{4}\\\\Q=152.68\times 10^{3}Joules$

5 0

3 years ago