Answer:
K = m g (A - A2)
Explanation:
In a block spring system the total energy is the sum of the potential energy plus the kinetic energy, for maximum elongation all the energy is potential
Em = U₀ = m g A
For when the system is at an ele
Elongation A2 less than A, energy has two parts
Em = K + U₂
K = Em –U₂
We substitute
K = m g A - m gA2
K = m g (A - A2)
Kinetic Energy = 1/2 * mv²
Kinetic Energy = 14 J, v = 2.3 m/s , m = ?
14 = 1/2 * m* 2.3²
14 = 0.5*m*2.3*2.3
m = 14 / (0.5*2.3*2.3)
m = 5.29 kg.
Mass = 5.29 kg.
It would be the first option.
Explanation-
The number of protons is equal to the atomic number the number of neutrons is the mass minus the atomic number.
You have effectively got two capacitors in parallel. The effective capacitance is just the sum of the two.
Cequiv = ε₀A/d₁ + ε₀A/d₂ Take these over a common denominator (d₁d₂)
Cequiv = ε₀d₂A + ε₀d₁A / (d₁d₂) Cequiv = ε₀A( (d₁ + d₂) / (d₁d₂) )
B) It's tempting to just wave your arms and say that when d₁ or d₂ tends to zero C -> ∞, so the minimum will occur in the middle, where d₁ = d₂
But I suppose we ought to kick that idea around a bit.
(d₁ + d₂) is effectively a constant. It's the distance between the two outer plates. Call it D.
C = ε₀AD / d₁d₂ We can also say: d₂ = D - d₁ C = ε₀AD / d₁(D - d₁) C = ε₀AD / d₁D - d₁²
Differentiate with respect to d₁
dC/dd₁ = -ε₀AD(D - 2d₁) / (d₁D - d₁²)² {d2C/dd₁² is positive so it will give us a minimum} For max or min equate to zero.
-ε₀AD(D - 2d₁) / (d₁D - d₁²)² = 0 -ε₀AD(D - 2d₁) = 0 ε₀, A, and D are all non-zero, so (D - 2d₁) = 0 d₁ = ½D
In other words when the middle plate is halfway between the two outer plates, (quelle surprise) so that
d₁ = d₂ = ½D so
Cmin = ε₀AD / (½D)² Cmin = 4ε₀A / D Cmin = 4ε₀A / (d₁ + d₂)
Answer:
B 14.5 m/s to the east
Explanation:
We can solve this problem by using the law of conservation of momentum.
In fact, if the system is isolated, the total momentum of the system must be conserved.
Here the total momentum before the stuntman reaches the skateboard is:
where
M = 72.0 kg is the mass of the stuntman
v = 15.0 m/s is his initial velocity (to the east)
The total momentum after the stuntmen reaches the skateboard is:
where
m = 2.50 kg is the mass of the skateboard
v' is the final velocity of the stuntman and the skateboard
Since momentum must be conserved, we have
And solvign for v',
And since the sign is the same as v, the direction is the same (to the east).
