For a simple harmonic motion, the position of the mass at any time t is given by

where
A is the amplitude of the motion (in this problem, A=17.5 cm)

is the angular frequency of the oscillator
t is the time
The angular frequency of the motion in the problem is given by

And so, we can find the position x of the mass (with respect to the equilibrium position) at time t=2.50 s:
Answer:
σ = 0.255*10^-3 C/m²
Explanation:
The Electric field Intensity act due to plate = σ/ε₀, where σ is surface charge density of plate.
At equilibrium ,
Upward force = downward force
Tcosθ = mg ----(1)
Assuming that the Forward force = backward force, then
Tsinθ = σq/ε₀
[ ∵ F = qE , ∴ F = qσ/ε₀ ] -----(2)
Dividing equation (2) by (1)
Tsinθ/Tcosθ = qσ/ε₀mg
⇒Tanθ = qσ/ε₀mg
σ = ε₀mg tanθ/q
Now substituting the values of
σ = (8.85*10^-12 * 1 * tan 30) / 2*10^-8
σ = (8.85*10^-12 * 0.5774) / 2*10^-8
σ = 5.11*10^-12 / 2*10^-8
σ = 0.255*10^-3 C/m²
Answer: Hence, ( 30,20 ) will not maximize the profit as it lies inside the solution region.
Explanation:
Because of gravity and friction.
Answer:
1)

2)

Explanation:
<u>Projectile Motion</u>
When an object is launched near the Earth's surface forming an angle
with the horizontal plane, it describes a well-known path called a parabola. The only force acting (neglecting the effects of the wind) is the gravity, which acts on the vertical axis.
The heigh of an object can be computed as

Where
is the initial height above the ground level,
is the vertical component of the initial velocity and t is the time
The y-component of the speed is

1) We'll find the vertical component of the initial speed since we have not enough data to compute the magnitude of 
The object will reach the maximum height when
. It allows us to compute the time to reach that point

Solving for 

Thus, the maximum heigh is

We know this value is 8 meters

Solving for 

Replacing the known values


2) We know at t=1.505 sec the ball is above Julie's head, we can compute



