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siniylev [52]
3 years ago
6

Match the following items.

Physics
1 answer:
Ray Of Light [21]3 years ago
8 0

Answer:

Je ne Sachez que Qu’est-ce que le

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A proton in the nucleus of an atom has an electrical charge of:<br> neutral<br> -<br> +<br> zero
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Answer:

proton is positively charged changechar

Explanation:

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What is the Orbital Notation for Radon
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A movable piston having a mass of 8.00 kg and a cross-sectional area of 5.00 cm2 traps 0.430 moles of an ideal gas in a vertical
svp [43]
We can assume the process to be adiabatic such that we can make use of the formula:
W = R (T2 - T1) / (γ - 1)
W = 8.314 (297 - 17) / (1.4 - 1)
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5 0
4 years ago
Consider a system to be two train cars traveling toward each other. What is the total momentum of the system before the train ca
Brut [27]

Let say the two train cars are of masses m_1 and m_2

now if the speed of two cars are v_1 and v_2

then we can say that the momentum of two cars before they collide is given by

P = m_1v_1 - m_2v_2

here two cars are moving in opposite direction so we can say that the net momentum is subtraction of two cars momentum.

Now since in these two car motion there is no external force on them while they collide

So the momentum of two cars are always conserved.

hence we can say that the final momentum of two cars will be same after collision as it is before collision

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5 0
3 years ago
Read 2 more answers
Assume that the driver begins to brake the car when the distance to the wall is d=107m, and take the car's mass as m-1400kg, its
Evgen [1.6K]

Answer:

Explanation:

a ) Let let the frictional force needed be F

Work done by frictional force = kinetic energy of car

F x 107 = 1/2 x 1400 x 35²

F = 8014 N

b )

maximum possible static friction

= μ mg

where μ is coefficient of static friction

= .5 x 1400 x 9.8

= 6860 N

c )

work done by friction for μ = .4

= .4 x 1400 x 9.8 x 107

= 587216 J

Initial Kinetic energy

= .5 x 1400 x 35 x 35

= 857500 J

Kinetic energy at the at of collision

= 857500 - 587216

= 270284 J

So , if v be the velocity at the time of collision

1/2 mv² = 270284

v = 19.65 m /s

d ) centripetal force required

= mv₀² / d which will be provided by frictional force

= (1400 x 35 x 35) / 107

= 16028 N

Maximum frictional force possible

= μmg

= .5 x 1400 x 9.8

= 6860 N

So this is not possible.

4 0
3 years ago
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