<u>Answer
</u>
A. 1 and 2
<u>Explanation
</u>
At point 1 we have the highest potential energy and the kinetic energy is zero.
At 2 the potential energy is minimum and the kinetic energy is maximum.
The law of conservation of energy says that energy cannot be created nor destroyed. So, the change in P.E = Change in K.E.
P.E = height × gravity × mass. The height referred here is the perpendicular height. Gravity and mass are constant in this case.
From the diagram it can be seen clearly that the vertical height from 2 to 1 is much greater than from 4 to 3.
This shows that the change in P.E is greater between 1 and 2 and so is kinetic energy.
Answer:
1.635×10^-3m
Explanation:
Young modulus is the ratio of the tensile stress of a material to its tensile strain.
Young modulus = Tensile stress/tensile strain
Tensile stress = Force/Area
Given force = 130N
Area = Πr² = Π×(1.55×10^-3)²
Area = 4.87×10^-6m²
Tensile stress = 130/4.87×10^-6 = 8.39×10^7N/m²
Tensile strain = extension/original length
Tensile strain = e/3.9
Substituting in the young modulus formula given young modulus to be 2×10¹¹N/m²
2×10¹¹N/m² = 8.39×10^7/{e/3.9)}
2×10¹¹ = (8.39×10^7×3.9)/e
2×10¹¹e = 3.27×10^8
e = 3.27×10^8/2×10¹¹
e = 1.635×10^-3m
The stretch of the steel wire will be
1.635×10^-3m
Answer:
Explanation:
capacitance of sphere 2 will be 4.5 times sphere 1
a ) when spheres are in contact they will have same potential finally . So
V_1 / V_2 = 1
b )
Charge will be distributed in the ratio of their capacity
charge on sphere1 = q x 1 / ( 1 + 4.5 )
= q / 5.5
fraction = 1 / 5.5
c ) charge on sphere 2
= q x 4.5 / 5.5
fraction = 4.5 / 5.5
d ) surface charge density of sphere 1
= q /( 5.5 x A ) where A is surface area
surface charge density of sphere 2
= q x 4.5 /( 5.5 x 4.5² A ) where A is surface area
= q /( 5.5 x 4.5 A )
q_1/q_2 = 4.5
Answer:
hope it helps......
Explanation:
Pressure, in the physical sciences, the perpendicular force per unit area, or the stress at a point within a confined fluid.
Answer
given,
I = 0.140 kg ·m²
decrease from 3.00 to 0.800 kg ·m²/s in 1.50 s.
a) 
τ = -1.467 N m
b) angle at which fly wheel will turn
θ = 20.35 rad
c) work done on the wheel
W = τ x θ
W = -1.467 x 20.35 rad
W = -29.86 J
d) average power of wheel
