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
Q = mcΔT
<span>q = 55.8g x 0.450J/gC x 23.5C </span>
<span>q = 590. J ................ to three significant digits
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Explanation:
Load=800N
Effort=200N
1. Mechanical Advantage = LOAD/EFFORT
= 800N/200N
= 4
2 Velocity Ratio = no. Of pulleys =5
3. Efficiency = Mechanical advantage / velocity ratio × 100%
= (4/5)×100%
=80%
4. output work= load×load distance
= 800N × 5m
= 4 × 1000J
5. Efficiency = (output work/input work) ×100%
Or, 80% = (4000J/input work) ×100%
Or, 80%/100% = 4000J/inputwork
Or, 4/5 = 4000J/inputwork
Or, input work =4000J × 5/4
Input work = 5×1000J
I hope it helped! ;-)
Oh my gosh ! Resisting the force of gravity always DOES involve doing work.
If no work is being done, then you're NOT resisting the force of gravity.
Example:
-- ball rolling on the floor . . . no work
-- ball rolling up a ramp . . . work being done
-- ball rolling down a ramp . . . work being done, BY gravity
