Answer:
(a) 3.44 x 10^-3 m^3/s
(b) 8.4 m/s
Explanation:
area of water line, A = 5.29 x 10^-3 m
number of holes, N = 15
Speed of water in line, V = 0.651 m/s
(a) Volume flow rate is given by
V = area of water line x speed of water in water line
V = 5.29 x 10^-3 x 0.651 = 3.44 x 10^-3 m^3/s
(b) area of one hole, a = 4.13 x 10^-4 m
Let v be the velocity of water in each hole
According to the equation of continuity
A x V = a x v
5.29 x 10^-3 x 0.651 = 4.1 x 10^-4 x v
v = 8.4 m/s
Answer:
1.) 11 km/s
2.) 9.03 × 10^-5 metres
Explanation:
Given that an electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric field, which has magnitude E = 48 N/C.
Electron q = 1.6×10^-19 C
Electron mass = 9.11×10^-31 Kg
(a) What is the speed of the electron 1.3 ns after entering this region?
E = F/q
F = Eq
Ma = Eq
M × V/t = Eq
Substitute all the parameters into the formula
9.11×10^-31 × V/1.3×10^-9 = 48 × 1.6×10^-19
V = 7.68×10^-18 /7.0×10^-22
V = 10971.43 m/s
V = 11 Km/s approximately
(b) How far does the electron travel during the 1.3 ns interval?
The initial velocity U = 64 km/s
S = ut + 1/2at^2
S = 64000×1.3×10^-6 + 1/2 × 8.4×10^12 × ( 1.3×10^-9)^2
S =8.32×10^-5 + 7.13×10^-6
S = 9.03 × 10^-5 metres
Answer:
sthe salt dissolves and the sand stays intact
Answer:
d = 120 [m]
Explanation:
In order to solve this problem, we must use the theorem of work and energy conservation. Where the energy in the final state (when the skater stops) is equal to the sum of the mechanical energy in the initial state plus the work done on the skater in the initial state.
The mechanical energy is equal to the sum of the potential energy plus the kinetic energy. As the track is horizontal there is no unevenness, in this way, there is no potential energy.
E₁ + W₁₋₂ = E₂
where:
E₁ = mechanical energy in the initial state [J] (units of Joules)
W₁₋₂ = work done between the states 1 and 2 [J]
E₂ = mechanical energy in the final state = 0
E₁ = Ek = kinetic energy [J]
E₁ = 0.5*m*v²
where:
m = mass = 60 [kg]
v = initial velocity = 12 [m/s]
Now, the work done is given by the product of the friction force by the distance. In this case, the work is negative because the friction force is acting in opposite direction to the movement of the skater.
W₁₋₂ = -f*d
where:
f = friction force = 36 [N]
d = distance [m]
Now we have:
0.5*m*v² - (f*d) = 0
0.5*60*(12)² - (36*d) = 0
4320 = 36*d
d = 120 [m]
