Explanation:
Given that,
A ball is tossed straight up with an initial speed of 30 m/s
We need to find the height it will go and the time it takes in the air.
At its maximum height, its final speed, v = 0 and it will move under the action of gravity. Using equation of motion :
v = u +at
Here, a = -g
v = u -gt
i.e. u = gt
So, the time for upward motion is 3.06 seconds. It means that it will in air for 3.06×2 = 6.12 seconds
Let d is the maximum distance covered by it.
Putting all values
Hence, it will go to a height of 45.91 m and it will in the air for 6.12 seconds.
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