Answer:
The magnitud of the torque doing by the foot at the point where the shaft is attached is 0.063 Nm
Explanation:
The torque is obtained by multiplying the longitude from the shaft attached point to force acting point by the force perpendicular component:
T = d * Fp
Notice that the perpindicular component is the total force magnitud times the sino of the angle respect the horizontal:
Fp = F*sin(a)
Replacing the values for the force and the angle:
Fp = 33N*sin(π/5) = 33N * 0.011 = 0.363 N
Taking the distance in meters:
T = 0.2m * 0.363 N = 0.063 Nm
Answer:
a=4m/s2
Explanation:
Vi= 20m/s
S=150 m
t= 5 sec
a=?
using 2nd equation of motion;
S= Vit+0.5at^{2}
putting values
150= (20)(5)+0.5(a)(
)
150-100= 12.5a
50/12.5 =a
4m/s2=a
As we know from work energy theorem that work done by external force is change in kinetic energy
So from all above we can say
in order to find work done we can say
here we have
so we will have
so from above equation we can say that final kinetic energy will be same as work done
so we will have kinetic energy as
If you know the amount of force acted on an object to find the amount actually being put onto it, you just add the amount of work done onto it, but if you are trying to find the difference between the amount of forces you just subtract the amount of force acting on the object.
Answer:
(a) 2.3 x 10^-12 N
(b) 1.6 x 10^5 m/s
Explanation:
q1 = 1.6 x 10^-19 C
q2 = 1.6 x 10^-19 C
r = 1 x 10^-8 m
(a) the electrostatic force is given by
F = K q1 x q2 / r^2
F = ( 9 x 10^9 x 1.6 x 10^-19 x 1.6 x 10^-19) / (1 x 10-^-8)^2
F = 2.3 x 10^-12 N
The force is attractive in nature because the nature of charge on electron and proton s opposite to each other.
(b) The electrostatic force is balanced by the centripetal force actig on the electron.
F = mv^2 / r
where, r be the radius of orbit and v be the velocity of electron in the orbit.
2.3 x 10^-12 = (9.1 x 10^-31 x v^2) / (1 x 10^-8)
v = 1.6 x 10^5 m/s
