As an object accelerates i.e., change it's velocity(either direction or speed), the position of the object depends on two factor; If the acceleration was direction based then it might have a zero displacement for eg: if it travels in circle. or it might have a net displacement if it travels in a straight line, quantitatively
where,
s = displacement
u = initial velocity
v = final velocity
a = acceleration
t = time
Now, for the hypothesis;
There is no direct relationship between fan speed and acceleration but anyways generally speaking if we do have a relationship that with more fan speed we have a larger displacement of air i.e., a more force i.e., greater acceleration
Thus, it can be said, well not exactly scientific, that with a greater fan speed there will be greater acceleration. if fan speed is increased then acceleration will be more.
:)
Answer:
3.974 Joule
Explanation:
Diameter of ring = 7.7 cm
a = Distance from the center = d/2 = 3.85 cm = 0.0385 m
Q = Charge = 5 mC
q = Charge to move = 3.4 mC
k = Coulomb constant = 9×10⁹ Nm²/C²
Work done will be equal to Potential energy when mass is at center
∴ Work to move a tiny 3.4 mC charge from very far away to the center of the ring is 3.974 Joule
Answer:
Explanation:
a )
in the regions r < R₁
charge q inside sphere of radius R₁ = 0
Applying gauss's law for electric field E at distance r <R₁
electric flux through Gaussian surface of radius r = 4π r² E
4π r² E = q / ε₀ = 0 / ε₀
E = 0
Applying gauss's law for electric field E at distance R₁ < r < R₂ .
charge q inside sphere of radius R₁ = q₁
Applying gauss's law for electric field E at distance R₁ < r < R₂
electric flux through Gaussian surface of radius r = 4π r² E
4π r² E = q₁ / ε₀
E = q₁ / 4πε₀
in the regions r> R₂
charge q inside sphere of radius R₂ = (q₁ + q₂)
Applying gauss's law for electric field E at distance r > R₂
electric flux through Gaussian surface of radius r = 4π r² E
4π r² E = (q₁ + q₂) / ε₀
E = (q₁ + q₂) /4π ε₀
b )
For electric flux to be zero at r > R₂
(q₁ + q₂) /4π ε₀ = 0
q₁ + q₂ = 0
q₁ / q₂ = - 1 .