Answer:
meters
Explanation:The question ask for the maximum value of the function f(t) which can be find by find the maxima of the function
The maxima of the function occurs when the slope is zero. i.e.
Hence the maxima occurs at t=1.63 seconds
The maximum value of f is
hence maximum height is found to be
meters
Answer:
a) the magnitude of the force is
F= Q() and where k = 1/4πε₀
F = Qqs/4πε₀r³
b) the magnitude of the torque on the dipole
τ = Qqs/4πε₀r²
Explanation:
from coulomb's law
E =
where k = 1/4πε₀
the expression of the electric field due to dipole at a distance r is
E(r) = , where p = q × s
E(r) = where r>>s
a) find the magnitude of force due to the dipole
F=QE
F= Q()
where k = 1/4πε₀
F = Qqs/4πε₀r³
b) b) magnitude of the torque(τ) on the dipole is dependent on the perpendicular forces
τ = F sinθ × s
θ = 90°
note: sin90° = 1
τ = F × r
recall F = Qqs/4πε₀r³
∴ τ = (Qqs/4πε₀r³) × r
τ = Qqs/4πε₀r²
Answer:
The speed at which the ball rolled off the end of the table is 3.3 m/s
Explanation:
Hi there!
Please, see the attached figure for a graphical description of the problem. Notice that the origin of the frame of reference is located at the edge of the table.
The position vector of the ball can be calculated as follows:
r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)
Where:
r = position vector.
x0 = initial horizontal position.
v0x = initial horizontal velocity.
t = time.
y0 = initial vertical position.
v0y = initial vertical velocity.
g = acceleration due to gravity.
When the ball reaches the ground, its position will be:
r final = (3, -4)
Then:
3 = x0 + v0x · t
-4 = y0 + v0y · t + 1/2 · g · t²
Since the origin of the frame of reference is located at the edge of the table, x0 and y0 = 0. v0y is also 0 ( see the initial velocity vector in the figure to elucidate why). Then:
3 m = v0x · t
-4 m = 1/2 · g · t²
We can solve for "t" in the equation of the y-component and use it in the equation of the x-component to obtain v0x:
-4 m = 1/2 · g · t²
-4 m = -1/2 · 9.8 m/s² · t²
8 m / 9.8 m/s² = t²
t = 0.9 s
Then:
3 m = v0x · 0.9s
3 m/ 0.9 s = v0x
v0x = 3.3 m/s
The speed at which the ball roll off the end of the table is 3.3 m/s
