Answer:
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Explanation:
Answer:
Explanation:
At constant pressure , work done by gas = P x ΔV where P is pressure and ΔV is change in volume
ΔV = 9.2 - 5.6 = 3.6 L
3.6 L = 3.6 x 10⁻³ m³
ΔV = 3.6 x 10⁻³ m³
P = 3.7 x 10³ Pa
So work done
= 3.7 x 10³ x 3.6 x 10⁻³ J
= 13.32 J .
( c ) is the answer , because work is done by the gas so it will be positive.
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.
You should calculate 40 kg and the radius 3mm.
Answer:
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