Answer:
The maximum speed of sonic at the bottom of the hill is equal to 19.85m/s and the spring constant of the spring is equal to (497.4xmass of sonic) N/m
Energy approach has been used to sole the problem.
The points of interest for the analysis of the problem are point 1 the top of the hill and point 2 the bottom of the hill just before hitting the spring
The maximum velocity of sonic is independent of the his mass or the geometry. It is only depends on the vertical distance involved
Explanation:
The step by step solution to the problem can be found in the attachment below. The principle of energy conservation has been applied to solve the problem. This means that if energy disappears in one form it will appear in another.
As in this problem, the potential and kinetic energy at the top of the hill were converted to only kinetic energy at the bottom of the hill. This kinetic energy too got converted into elastic potential energy .
x = compression of the spring = 0.89
Explanation:
Below is an attachment containing the solution.
Answer:
0
Explanation:
m = Mass of person
g = Acceleration due to gravity = 9.81 m/s²
d = Vertical height from the ground
F = Force = Weight = mg
Net work done would be
Hence, the work done on the person by the gravitational force is 0
Answer:
False.
Explanation:
The statement shown in the question above is false and this can be confirmed by Newton's law on universal gravitation. According to Newton, the gravitational force exerted on any body is proportional to its weight, but the distance that the object travels when falling is disproportionate. In addition, if the force resulting from the weight of the object and its displacement has an angle of 0º, the weight force of that object will provide an increase in kinetic energy.
Answer:
The image height is 3.0 cm
Explanation:
Given;
object distance, 
= 15.0 cm
image distance, 
= 5.0 cm
height of the object, 
= 9.0 cm
height of the image, 
= ?
Apply lens equation;
Therefore, the image height is 3.0 cm. The negative values for image height indicate that the image is an inverted image.
