Answer:
Qsinθ/4πε₀R²θ
Explanation:
Let us have a small charge element dq which produces an electric field E. There is also a symmetric field at P due to a symmetric charge dq at P. Their vertical electric field components cancel out leaving the horizontal component dE' = dEcosθ = dqcosθ/4πε₀R² where r is the radius of the arc.
Now, let λ be the charge per unit length on the arc. then, the small charge element dq = λds where ds is the small arc length. Also ds = Rθ.
So dq = λRdθ.
Substituting dq into dE', we have
dE' = dqcosθ/4πε₀R²
= λRdθcosθ/4πε₀R²
= λdθcosθ/4πε₀R
E' = ∫dE' = ∫λRdθcosθ/4πε₀R² = (λ/4πε₀R)∫cosθdθ from -θ to θ
E' = (λ/4πε₀R)[sinθ] from -θ to θ
E' = (λ/4πε₀R)[sinθ]
= (λ/4πε₀R)[sinθ - sin(-θ)]
= (λ/4πε₀R)[sinθ + sinθ]
= 2(λ/4πε₀R)sinθ
= (λ/2πε₀R)sinθ
Now, the total charge Q = ∫dq = ∫λRdθ from -θ to +θ
Q = λR∫dθ = λR[θ - (-θ)] = λR[θ + θ] = 2λRθ
Q = 2λRθ
λ = Q/2Rθ
Substituting λ into E', we have
E' = (Q/2Rθ/2πε₀R)sinθ
E' = (Q/θ4πε₀R²)sinθ
E' = Qsinθ/4πε₀R²θ where θ is in radians
A diver having mass m climbs up the diving board.
We know that Gravitational potential energy is given as <span>P<span>EG</span>=mgΔh</span>
What changes is his Gravitational potential energy due to change of height <span>Δh</span> with reference to the ground/water level.
While standing on the diving board his velocity is zero. As such kinetic energy is also zero.
Once he jumps off the springboard we see he gets additional energy from the springboard and falls down under action of gravity g. Due to decrease of height above the ground level Gravitational potential energy decreases and gets converted in to his kinetic energy. <span>1/2m<span>v2</span></span>.
While in air he encounters air resistance. Some of his energy is spent in overcoming this resistance. Gets converted in to kinetic and thermal energy of surrounding air and his body.
Once diver reaches the water, we see water splashing and hear noise of splash. Thereafter the diver comes to rest. Now his potential energy becomes zero. And converted kinetic energy has been converted in to kinetic energy, heat energy and sound energy of water.
As such energy transformation equation looks like
<span><span>Gravitational PE+Elastic PE of springboard</span><span>→Kinetic energy of air and water+Sound energy of splash+thermal energy</span></span>
Answer:
can you show a graph but if not i believe the answer is x=6m
Explanation:
Answer:
W = 1884J
Explanation:
This question is incomplete. The original question was:
<em>Consider a motor that exerts a constant torque of 25.0 N.m to a horizontal platform whose moment of inertia is 50.0kg.m^2 . Assume that the platform is initially at rest and the torque is applied for 12.0rotations . Neglect friction.
</em>
<em>
How much work W does the motor do on the platform during this process? Enter your answer in joules to four significant figures.</em>
The amount of work done by the motor is given by:
Where I = 50kg.m^2 and ωo = rad/s. We need to calculate ωf.
By using kinematics:
But we don't have the acceleration yet. So, we have to calculate it by making a sum of torque:
=> 
Now we can calculate the final velocity:
Finally, we calculate the total work:
Since the question asked to "<em>Enter your answer in joules to four significant figures.</em>":
W = 1884J
