Answer:
60N
Explanation:
in this case the minimum amount of force required must be equal to the friction Force. i.e <u>Newton</u><u>'s</u><u> </u><u>first</u><u> </u><u>law</u><u> of</u><u> </u><u>mot</u><u>ion</u><u>.</u>
therefore the maximum amount of frictional force is equal to the applied force which is 60N.
because of the net force acting on the object is zero the object is in constant motion . i.e equal and opposite force must be applied so that the object is in constant velocity therefore the total frictional force must be 60N
Answer:
595391.482946 m/s

Explanation:
E = Energy = 1.85 keV
I = Current = 5.15 mA
e = Charge of electron = 
t = Time taken = 1 second
m = Mass of proton = 
Velocity of proton is given by

The speed of the proton is 595391.482946 m/s
Current is given by

Number of protons is

The number of protons is 
Answer:
a.
b.
Explanation:
We are given that




a.We have to find the angle


b. We have to find the speed 
According to law of conservation of momentum



Answer:
9517.2 lbm
Explanation:
Electricity consumption = 14000 kWh/year
Fuel consumption = 900 gal/year
Amount of CO₂ produced per gallon = 26.4 lbm/gal
Amount of CO₂ produced per kWh = 1.54 lbm/kWh
Amount of CO₂ produced in one year

Reduction would be

The reduction in the amount of CO₂ produced is 9517.2 lbm