To solve this problem we will use the general system of simple harmonic movement and compare this structure with the given value. From this equation we can find the phase. Our given value is
Data:
t = 2 s
The expression given from the theory for the harmonic movement is:
Where,
= Phase Angle
A = Amplitude
Here the Phase angle is given as
Comparing we have,
Replacing the time of t, we have that
Therefore the phase of motion at t=2s is
Answer:
0.98 g/m
Explanation:
Note: Since Tension and frequency are constant,
Applying,
F₁²M₁ = F₂²M₂............... Equation 1
Where F₁ = Frequency of the G string, F₂ = Frequency of the A string, M₁ = mass density of the G string, M₂ = mass density of the A string.
make M₂ the subject of the equation
M₂ = F₁²M₁/F₂²............... Equation 2
From the question,
Given: F₁ = 196 Hz, M₁ = 0.31 g/m, F₂ = 110 Hz
Substitute these values into equation 2
M₂ = 196²(0.31)/110²
M₂ = 0.98 g/m
Answer:
<h2>300 J</h2>
Explanation:
The potential energy of a body can be found by using the formula
PE = mgh
where
m is the mass
h is the height
g is the acceleration due to gravity which is 10 m/s²
From the question we have
PE = 2 × 10 × 15
We have the final answer as
<h3>300 J</h3>
Hope this helps you
From our physics class, we know that Work is a product of
Force and Displacement. The force must be the force in the same direction with
the displacement. So force and displacement must be parallel.
A. How much work in J does the string do on the boy if the
boy stands still?
Since we know that: Work = Force * Displacement
and in this case displacement is equal to zero, therefore
Work must also be equal to zero.
Work = 0
B. How much work does the string do on the boy if the boy
walks a horizontal distance of 11m away from the kite?
Now are given a displacement of 11 m which is a horizontal
distance. Therefore we must first calculate the horizontal component of force:
Fx = 4.5 N * cos 30
Fx = 3.897 N
So work is:
Work = 3.897 N * 11 m
<span>Work = 42.87 J ~ 43 J</span>