Answer:
Linear and rotational Kinetic Energy + Gravitational potential energy
Explanation:
The ball rolls off a tall roof and starts falling.
Let us first consider the potential energy or more specifically gravitational potential energy (
;
= mass of the ball,
= acceleration due to gravity,
= height of the roof). This energy comes because someone or something had to do work to take the ball to the top of the roof against the force of gravity. The potential energy is naturally maximum at the top and minimum when the ball finally reaches the ground.
Now, the ball starts to roll and falls off the roof. It shall continue rotating because of inertia (Newton's first law). This contributes to the rotational kinetic energy (
;
=moment of inertia of the ball &
= angular velocity).
Finally comes the linear kinetic energy or simply, kinetic energy (
) which is caused due to the velocity
of the ball.
Answer:
.5 grams
Explanation:
1 gram is equal to 1000 milligrams (mg)
Answer:
a) 
b) 
c) 
d) 
e)
&
f) 
Explanation:
From the question we are told that:
Stretch Length 
Mass 
Total stretch length
a)
Generally the equation for Force F on the spring is mathematically given by


b)Generally the equation for Max Velocity of Mass on the spring is mathematically given by

Where
A=Amplitude

And

Therefore


c)
Generally the equation for Max Acceleration of Mass on the spring is mathematically given by



d)
Generally the equation for Total mechanical energy of Mass on the spring is mathematically given by



e)
Generally the equation for the period T is mathematically given by



Generally the equation for the Frequency is mathematically given by


f)
Generally the Equation of time-dependent vertical position of the mass is mathematically given by

Where
'= signify order of differentiation
Answer:
T= 4.24sec
Explanation:
We are going to use the formula below to calculate.

Where T is period
L is length of rod
g is acceleration due to gravity =
From the problem, the rod is pivoted at 1/4L which means that three quarter of the rod was used for the oscillation. lets call this

= 4.4625m
thus
T= 4.24sec
Recall this gas law:
= 
P₁ and P₂ are the initial and final pressures.
V₁ and V₂ are the initial and final volumes.
T₁ and T₂ are the initial and final temperatures.
Given values:
P₁ = 475kPa
V₁ = 4m³, V₂ = 6.5m³
T₁ = 290K, T₂ = 277K
Substitute the terms in the equation with the given values and solve for Pf:

<h3>P₂ = 279.2kPa</h3>