To solve this problem we will apply the concepts related to load balancing. We will begin by defining what charges are acting inside and which charges are placed outside.
PART A)
The charge of the conducting shell is distributed only on its external surface. The point charge induces a negative charge on the inner surface of the conducting shell:
. This is the total charge on the inner surface of the conducting shell.
PART B)
The positive charge (of the same value) on the external surface of the conducting shell is:
The driver's net load is distributed through its outer surface. When inducing the new load, the total external load will be given by,
Answer:-2.86*10⁻⁴
Explanation: Use the equation change in volume = (change in pressure * original volume) / Bulks Modulus. ΔV = (-Δp*V₀) / B
Plugging in your numbers, you should get ΔV = (-2.29*10⁷*1) / (8*10¹⁰) = -2.86*10⁻⁴
ΔP = P₂-P₁ ----> ΔP = 2.30*10⁷ - 1.00*10⁵ = 2.29*10⁷
When a swimmer pushes through water to swim they are propelled forward because of the water resistance against the hand and feet. It's A. The water doesn't automatically push the swimmer forward. It releases a reaction after the swimmer pushes through the water.
Answer:
A = 2.36m/s
B = 3.71m/s²
C = 29.61m/s2
Explanation:
First, we convert the diameter of the ride from ft to m
10ft = 3m
Speed of the rider is the
v = circumference of the circle divided by time of rotation
v = [2π(D/2)]/T
v = [2π(3/2)]/4
v = 3π/4
v = 2.36m/s
Radial acceleration can also be found as a = v²/r
Where v = speed of the rider
r = radius of the ride
a = 2.36²/1.5
a = 3.71m/s²
If the time of revolution is halved, then radial acceleration is
A = 4π²R/T²
A = (4 * π² * 3)/2²
A = 118.44/4
A = 29.61m/s²
Answer:
space , small amount of gravity is found in the space ,infact we can say that there is no gravity in the space
