Call the capacitance C.
<span>Note the energy in a capacitor with voltage V is E =½CV². </span>
<span>Initial energy = ½C(12)² = 72C </span>
<span>40% of energy is delivered, so 60% remains.in the capacitor. </span>
<span>Remaining energy = (60/100) x 72C =43.2C </span>
<span>If the final potential difference is X, the energy stored is ½CX² </span>
<span>½CX² = 43.2C </span>
<span>X² = 2 x 43.2 = 86.4 </span>
<span>X = 9.3V</span>
The magnitude of the change in momentum of the block between zero and 4. 0 seconds is
dp=1.6m/s
<h3>What is the magnitude of the change in momentum of the block between zero and 4. 0 seconds?</h3>
Generally, the equation for the kinematics equation is mathematically given as
x=x+ut+0.5at^2
Therefore
1.6=0+0+0.5*a*1^2
a=0.2m/s^2
In conclusion, change in momentum
dP=mv-mu
2.0*0.8-0
dp=1.6m/s
Read more about Force
brainly.com/question/13370981
Answer: arrange systematically in groups; separate according to type, class, etc.
Explanation:
i hooked this up btw
The potential energy of the block is given by:
V = m*g*h
m mass
g = 9.81m/s²
h height
The potential energy of a spring is given by:
V = 0.5 * k * x²
k spring constant
x compression of the spring
If the block starts from rest it has potential energy, but no kinetic energy. As it slides down the incline potential energy is converted into kinetic energy. When the block hits the spring the kinetic energy is converted into spring's potential energy. If the spring is fully compressed and the block is at rest again, the block has transferred all its energy into the spring. No energy is lost. So we can write:
m * g * h = 0.5 * k * x²
m = 0.5 kg
g = 9.81 m/s²
h = 2.5m * sin 37° = 1,5 m
x = 0,6 m
Solve for k.
k = 2 * m * g * h / x² = 40.8 N/m
