a. The force applied would be equal to the frictional
force.
F = us Fn
where, F = applied force = 35 N, us = coeff of static
friction, Fn = normal force = weight
35 N = us * (6 kg * 9.81 m/s^2)
us = 0.595
b. The force applied would now be the sum of the
frictional force and force due to acceleration
F = uk Fn + m a
where, uk = coeff of kinetic friction
35 N = uk * (6 kg * 9.81 m/s^2) + (6kg * 0.60 m/s^2)
uk = 0.533
Answer:
4.39 x 10^-19 J
Explanation:
q1 = 1.6 x 10^-19 C
q2 = - 1.6 x 10^-19 C
r1 = 3 x 10^-10 m
r2 = 7 x 10^-10 m
The formula for the potential energy is given by
U1 = k q1 q2 / r1 = - (9 x 10^9 x 1.6 x 10^-19 x 1.6 x 10^-19) / (3 x 10^-10)
U1 = - 7.68 x 10^-19 J
U2 = k q1 q2 / r2 = - (9 x 10^9 x 1.6 x 10^-19 x 1.6 x 10^-19) / (7 x 10^-10)
U2 = - 3.29 x 10^-19 J
Change in potential energy is
U2 - U1 = - 3.29 x 10^-19 + 7.68 x 10^-19 = 4.39 x 10^-19 J
Answer:
Option B. 6.25 J/S
Explanation:
Data obtained from the question include:
t (time) = 2secs
F (force) = 50N
d (distance) = 0.25m
P (power) =?
The power can be obtained by using the formula P = workdone/time.
P = workdone / time
P = (50 x 0.25)/ 2
P = 6.25J/s
Answer:
24°
Explanation:
sin(34°)/sin(x)=v2/v1
x=arcsin(2,1*sin(34°)/2,8)=24°
