Answer:
3.125 m
Explanation:
We are given that
Mass of box=m=11.2 kg
Speed of box=u=3.5m/s
Coefficient of kinetic friction=
Final velocity,v=0
a.We have to find the horizontal force applied by worker to maintain the motion.
According to question
Horizontal force=F=

Substitute the values
Horizontal force=
b.According to work-energy theorem






Hence, the box slide before coming to rest=3.125 m
The answer is D. If you aren't consistent with your drop positions, then your data may be invalid. To be frank: it basically screws over the experiment.
A)We know the formula of the angular speed is ω = 2π / TWhere T is the time period.When second hand completes one revolution then the time taken is 60s.So T = 60sThen the angular speed of the second hand is ω= 2π / (60s) = 0.1047 rad/sb)When the minute hand completes one revolution the time taken is T = 1 hr = 3600sThen the angular speed of the minute hand is ω =(2π) / (3600s) = 0.001745 rad/sc)When the hour hand completes one revolution then the timeperiod is T = 12hrs = (12)(3600)sThen the angular speed of the hour hand is ω =(2π) / [(12)(3600)s] = 1.45444 x 10^-4 rad/s
Answer:
The net magnetic field ta the center of square is
.
Explanation:
Current, I = 12 A , side ,a = 10 cm = 0.1 m
Let the magnetic field due to the one side is B.
The magnetic field is given by

Net magnetic field at the center of the square is
B' = 4 B
