Answer:
a. Springs oscillate with the same frequency
Explanation:
As they both are in the same height at equilibrium, so
weight of ball must be balanced with spring force, that is
k×x=mg
k= stiffness constant of spring
x=distance stretched
g= acceleration due to gravity
so, we can write
k/m=g/x
as the g is a constant and they stretched to same distance x so the g/x term becomes constant and
and k/m is same for both the springs so they will oscillate at the same frequency.
hence option a is correct.
Answer:
The correct answer to the following question will be Option A (moment arm; pivot point).
Explanation:
- The moment arm seems to be the duration seen between joint as well as the force section trying to act mostly on the joint. Each joint that is already implicated in the workout seems to have a momentary arm.
- The moment arm extends this same distance from either the pivot point to just the position of that same pressure exerted.
- The pivotal point seems to be the technical indicators required to fully measure the appropriate demand trends alongside different time-frames.
The other three choices are not related to the given situation. So that option A is the appropriate choice.
Answer:
B
Explanation:
The capacitor is a component which has the ability to store energy in the form of an electrical charge making a potential difference on those two metal plates
A capacitor consists of two or more parallel conductive (metal) plates. They are electrically seperated by an insulating material (ex: air, mica,ceramic etc.) which is called as Dielectric Layer
Due to this insulating layer, DC current can not flow through the capacitor.But it allows a voltage to be present across the plates in the form of an electrical charge.
Prior to touching the bar magnet, the magnetic domains in the nail were pointing in random directions. When Taylor touched the nail to the bar magnet the magnetic fields of the magnetic domains aligned and it became a temporary magnet.
We are given: Final velocity (
)=20 m/s .
Time t= 2.51 s and
distance s = 82.9 m.
We know, equation of motion
Let us plug values of final velocity, and time in above equation.
Subtracting 2.51a from both sides, we get
-----------equation(1)
Using another equation of motion
Plugging values of vi =20-2.51a, t=2.51 and distnace s=82.9 in this equation.
We get,
Now, we need to solve it for a.
20-20+2.51a=165.8a.
-163.29a=0
a=0.
So, the acceleration would be 0 m/s^2.
