Answer:
v = √k/m x
Explanation:
We can solve this exercise using the energy conservation relationships
starting point. Fully compressed spring
Em₀ = = ½ k x²
final point. Cart after leaving the spring
= K = ½ m v²
Em₀ = Em_{f}
½ k x² = ½ m v²
v = √k/m x
I believe the answer you are looking for is 1.4 N.
0.145*9.8=1.42
So you round 1.42 to 1.4
Hope this helps!
The correct answer
to this question would be:
<span><span>
A. </span>
No part of your vehicle will extend out into
the traffic lane.</span>
This kind of maneuver only shows your skill
to handle the vehicle in tight spaces, ability to judge distance, and showing control
of the vehicle as you turn into a straight-in parking space.
<span> </span>
Nothing will happen. The mercury and steel will not move. It’s just like filling a glass halfway with cement and then pouring water on top of it.
Explanation:
Mass of the diskshaped grindstone, m = 1.1 kg
Radius of disk, r = 0.09 m
Angular velocity,
Time, t = 42.4 s
We need to find the frictional torque exerted on the grindstone. Torque in the rotational kinematics is given by :
I is moment of inertia of disk,
So, the frictional torque exerted on the grindstone is .