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SCORPION-xisa [38]
2 years ago
13

Five difference between elastic collision and inelastic collision?​

Physics
1 answer:
olga_2 [115]2 years ago
8 0

Answer:

Elastic Collision

Inelastic Collision

The total kinetic energy is conserved. The total kinetic energy of the bodies at the beginning and the end of the collision is different.

Momentum does not change. Momentum changes.

No conversion of energy takes place. Kinetic energy is changed into other energy such as sound or heat energy.

Highly unlikely in the real world as there is almost always a change in energy. This is the normal form of collision in the real world.

An example of this can be swinging balls or a spacecraft flying near a planet but not getting affected by its gravity in the end.

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<span>Describe the relationship of attractive forces between molecules and the ability of a solvent to dissolve a substance. Solvents can dissolve a substance only if the attraction of the solvent molecules is greater than the attraction between the molecules of the substance.</span>
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3 years ago
A weight lifter applies an upward force of 1100 N while lowering a dumbbell
Paul [167]

Answer:

A

Explanation:

work = force \times distance

work = 1100 \times 0.5

= 550 \: j

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3 years ago
A 15-ft3 tank contains oxygen initially at 14.7 psia and 80°F. A paddle wheel within the tank is rotated until the pressure insi
Nikolay [14]

Explanation:

Equation for energy balance will be as follows.

         \Delta E_{system} = E_{in} - E_{out}

        \Delta U = W_{in} - Q_{out}

Hence,    W_{in} = Q_{out} + mC_{v} (T_{2} - T_{1})

Therefore, we will calculate the final temperature as follows.

            \frac{P_{1}V}{T_{1}} = \frac{P_{2}V}{T_{2}}

       T_{2} = \frac{20 psia}{14.7}(638 R)

                   = 868.03 R

Now, we will calculate the mass as follows.

             m = \frac{P_{1}V}{RT_{1}}

                 = \frac{14.7 psia \times 15 ft^{3}}{0.3353 psi ft^{3}/lbm R \times 638 R}

                 = 1.031 lbm

Hence,

        W_{in} = Q_{out} + mC_{v} (T_{2} - T_{1})

Putting the values into the above equation as follows.

            W_{in} = Q_{out} + mC_{v} (T_{2} - T_{1})

    W_{in} = 20 Btu + 1.031 lbm (\frac{0.160 Btu}{lbm R})(735 - 540)R

            W_{in} = 655.2 Btu

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6 0
3 years ago
Use this free body diagram to help you find the magnitude of the force F1 needed to keep this block in static equilibrium 15.3 N
bija089 [108]

Do you have a picture of the diagram?

5 0
3 years ago
Sam, whose mass is 75 kg , takes off across level snow on his jet-powered skis. The skis have a thrust of 160 N and a coefficien
Volgvan

Answer:

top speed = 17.25

Total height = 281.19 m

Explanation:

given data

mass = 75 kg

thrust = 160 N

coefficient of kinetic friction = 0.1

solution

we get here frictional force acting that is

frictional force = \mu *m*g   .............1

frictional force = 0.1 × 75 × 9.8

frictional force = 73.5 N

and

Net force acting will be F = 160 - 73.5  N

F = 86.5 N

so

Acceleration in the First 15 second  will be

F = ma .........2

86.5 = 75 × a

a = 1.15 m/s²

and

now After 15 second the velocity will be  as

v = u + at   ..........3

here u is 0

so v will be

V = 1.15 × 15

v = 17.25

and

now we get travels distance S  in 15 s

s = u × t + 0.5 × a × t²  

here u is 0

so distance s will be

s = 0.5 × a × t²  

s = 0.5 ×  1.15 × 15²  

s = 129.37 m

and

now  acceleration acting is

F =  \mu *m*g  

m a =  \mu *m*g

a = \mu* g

a = - 0.98

here it is negative it mean downward nature of acceleration

and

now we get distance s by this formula

V² - u² = 2 a s    

here v velocity is 0  and

u initial velocity is 17.25 m/s

put here value

0 - 17.25² = 2 × (-0.98) × s    

solve it we get

s = 151.82 m

so

Total height is

Total height = 129.37 m + 151.82 m

Total height = 281.19 m

7 0
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