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Kisachek [45]
3 years ago
8

Which phase of matter is made up of particles that are packed relatively close together, with an indefinite shape but a definite

volume? gas liquid plasma solid?
Physics
1 answer:
Bezzdna [24]3 years ago
5 0

property of gas

gas molecules are very loosely bonded with each other so the force between the molecules is very weak

Due to this weak force the molecules of gas can attain any shape and any volume.

So gas moles have no finite shape and no finite volume

Property of Solids

Solids are tightly packed and having very strong bond between the molecules

If all molecules are tightly bonded then they have always a finite shape and finite volume so in this case we can not change the volume or shape of the solids as it is fixed.

Property of Liquids

Liquid molecules are moderately bonded together and since the force is more than the force between the gas molecules so liquids have finite volume but indefinite shape

liquid attains the shape of the container but its volume is fixed

So in above all examples Liquid is correct answer

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tia_tia [17]
I think you need to add more.. but I may know where you are leading

Was he 200 m away and made the trip in 200 seconds?

If yes...
2 m/s was his speed and 0 velocity
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3 years ago
Technician A says that low pressure smoke installed in the fuel system can be used to check for leaks. Technicians B says that n
pickupchik [31]

Answer:

A. Technician A only.B.

Explanation: The fuel system of a vehicle is made up of the fuel pump,the fuel filter,the injector or carburettor and the fuel tank. The main function of the fuel system is supply fuel to the engine of a vehicle. In order to check for leakage in the fuel system a small pressure smoke is usually installed in the fuel system.

Nitrogen at low pressure has also been used to check for leakage in the fuel system of a vehicle.

4 0
3 years ago
A tall cylinder contains 30 cm of water. Oil is carefully poured into the cylinder, where it floats on top of the water, until t
Anastaziya [24]

Answer:

The gauge pressure is  P_g  =  2058 \ P_a

Explanation:

From the question we are told that

       The height of the water contained is  h_w  =  30 \ cm  =  0.3 \ m

        The height of liquid in the cylinder is  h_t  =  40 \ cm  = 0.4 \ m

       

At the bottom of the cylinder the gauge pressure is  mathematically represented as

        P_g  =  P_w + P_o

Where  P_w is the pressure of water which is mathematically represented as

      P_w  =  \rho_w  *  g * h_w

Now  \rho_w is the density of water with a constant values of  \rho_w  = 1000 \ kg /m^3

   substituting values

      P_w  = 1000 *  9.8 *  0.3

     P_w  =  2940 \  Pa

While P_o is the pressure of oil which is mathematically represented as

          P_o  =  \rho_o *  g *  (h_t -h_w )

Where \rho _o is the density of oil with a constant value

         \rho _o  = 900 \ kg / m^3

substituting values

       P_o  =  900 *  9.8 * (0.4 - 0.3)

       P_o  =  882 \ Pa

Therefore

      P_g  =  2940 - 882

      P_g  =  2058 \ P_a

6 0
3 years ago
When light is directed on a metal surface, the kinetic energies of the photoelectrons a) are random b) vary with the frequency o
jekas [21]

Answer:

b) vary with the frequency of the light

Explanation:

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K.E=(hv -W•)

Where K.E is the Kinectic energy

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ν =frequency of the radiation

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Then, we can see that K.E is proportional linearly to "v" in the equation above.

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5 0
2 years ago
If a 20 g cannonball is shot from a 5 kg cannon with a velocity of 100
balu736 [363]

Strange as it may seem, the statement in the question appears to be <em>TRUE</em>.  

-- Before the shot, neither the cannon nor the ball is moving, so their combined momentum is zero.  

-- Since momentum is conserved, we know immediately that their combined momentum AFTER the shot also has to be zero.

-- (20g is rather puny for a "cannonball" ... about the same weight as four nickels. But we'll take your word for it and just do the Math and the Physics.)

-- Momentum = (mass) x (velocity)

After the shot, the momentum of the cannonball is

(0.02 kg) x (100 m/s ==> that way)

Momentum of the ball = 2 kg-m/s ==> that way.

-- In order for both of them to add up to zero, the momentum of the cannon must be (2 kg-m/s this way <==) .

Momentum of cannon = (5 kg) x (V m/s this way <==)

2 kg-m/s this way <== = (5 kg) x (V m/s this way <==)

Divide each side by (5 kg):

V m/s  =  (2/5) m/s this way <==

Speed of recoil of the cannon = <em>-- 0.4 m/s</em>

3 0
3 years ago
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