Momentum of car
Given: Mass m= 1,400 Kg; V = 6.0 m/s
Formula: P = mv
P = (1,400 Kg)(6.0 m/s)
P = 8,400 Kg.m/s
Velocity of the rider to have the same momentum as a car.
Mass of rider and bicycle m = 100 Kg
P = mv
V = P/m
V = 8,400 Kg.m/s/100 Kg
V = 84 m/s
Answer:
recall that heat absorbed released is given by
Q = mc*(T2 - T1)
where
m = mass (in g)
c = specific heat capacity (in J/g-k)
T = temperature (in C or K)
*note: Q is (+) when heat is absorbed and (-) when heat is released.
substituting,
Q = (480)*(0.97)*(234 - 22)
Q = 98707 J = 98.7 kJ
Explanation:
Answer:
D. two positively charged objects
Answer:
5.1*10^3 J/m^3
Explanation:
Using E = q/A*eo
And
q =75*10^-6 C
A = 0.25
eo = 8.85*10^-12
Energy density = 1/2*eo*(E^2) = 1/2*eo*(q/A*eo)^2 = [q^2] / [2*(A^2)*eo]
= [(75*10^-6)^2] / [2*(0.25)^2*8.85*10^-12]
= 5.1*10^3 J/m^3
