Answer:
The puck moves a vertical height of 2.6 cm before stopping
Explanation:
As the puck is accelerated by the spring, the kinetic energy of the puck equals the elastic potential energy of the spring.
So, 1/2mv² = 1/2kx² where m = mass of puck = 39.2 g = 0.0392 g, v = velocity of puck, k = spring constant = 59 N/m and x = compression of spring = 1.3 cm = 0.013 cm.
Now, since the puck has an initial velocity, v before it slides up the inclined surface, its loss in kinetic energy equals its gain in potential energy before it stops. So
1/2mv² = mgh where h = vertical height puck moves and g = acceleration due to gravity = 9.8 m/s².
Substituting the kinetic energy of the puck for the potential energy of the spring, we have
1/2kx² = mgh
h = kx²/2mg
= 59 N/m × (0.013 m)²/(0.0392 kg × 9.8 m/s²)
= 0.009971 Nm/0.38416 N
= 0.0259 m
= 2.59 cm
≅ 2.6 cm
So the puck moves a vertical height of 2.6 cm before stopping
T<span>he equation to be used here to determine the distance between two equipotential points is:
V = k * Q / r
where v is the voltage of the point, k is a constant, Q is charge of the point measured in coloumbs and r is the distance.
In this case, we can use ratio of proportions to determine the distance between the two points. in this respect,
Point 1:
V = k * Q / r = 290
r = k*Q/290 ; kQ = 290r
Point 2:
V = k * Q / R = 41
R = k*Q/41
from equation 10 kQ = 290r
R = 290/(41)= 7.07 m
The distance between the two points then is equal to 7.07 m.
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Answer:
Power, P = 924.15 watts
Explanation:
Given that,
Length of the ramp, l = 12 m
Mass of the person, m = 55.8 kg
Angle between the inclined plane and the horizontal, 
Time, t = 3 s
Let h is the height of the hill from the horizontal,
h = 5.07 m
Let P is the power output necessary for a person to run up long hill side as :
P = 924.15 watts
So, the minimum average power output necessary for a person to run up is 924.15 watts. Hence, this is the required solution.
