solution:
the spring force applied by a spring with spring constant k can be expressed as
where k acts as the spring constant
and x indicates the spring's deformation
to determine the work completed by the spring
the amount of work done by the spring when moving from x=0 to x=L
substituting the limits x=0 and x=L
we derive the work done in terms of k and L
Answer:
Explanation:
We define the linear charge density as:
Where L is the length of the rod, in this scenario the semicircle's length is L = πr
The potential at the center created by a differential element of charge is:
where k denotes Coulomb's constant
r signifies the distance from dq to the center of the circle
Thus.
The potential at the semicircle's center
The force exerted on the car during the stop measures 6975 N.
Explanation: Given that the mass (m) is 930 kg, speed (s) at 56 km/h converts to 15 m/s, and the stopping time (t) is 2 s, we compute the force using F = m * a. Here, acceleration (a) can be obtained through a = s/t. The total force calculation confirms that F = 930 kg * (15 m/s) / 2 s results in 6975 N.
Answer:
x₂=2×1
Explanation:
According to the work-energy theorem, we can assume that the gravitational potential energy at the lowest point of compression is zero since the kinetic energy change is 0;
mgx-(kx)²/2 =0 where m refers to the object's mass, g indicates the acceleration due to gravity, k denotes spring constant, and x represents the spring's compression.
mgx=(kx)²/2
x=2mg/k----------------compression when the object is at rest
However, ΔK.E =-1/2mv²⇒kx²=mv² -----------where v symbolizes the object's velocity and K.E signifies kinetic energy
Thus, if kx²=mv² then
v=x *√(k/m) ----------------where v=0
<pDoubling v results in multiplying x *√(k/m) by 2, leading to x₂ being double x₁
The essential principle for this question is Ohm’s Law: V=IR, I=V/R, R=V/I. Therefore, the answer is (3) Resistance, as it is inversely related to Current (I=V/R).
