Answer:
Explanation:
Given:
Charge = <em>q</em>
Electric field strength =
weight of the droplet = <em>mg</em>
The charge is suspended motionless. This is because the electric force on the charge is balanced by the weight of the droplet.
electric force on charged droplet,
This is balanced by the weight,
Equating the two:
Hi there! :)
Use the following kinematic equation to solve for the final velocity:
In this instance, the runner started from rest, so the initial velocity is 0 m/s. We can rewrite the equation as:
Plug in the given acceleration and time:
Answer:
12164.4 Nm
Explanation:
CHECK THE ATTACHMENT
Given values are;
m1= 470 kg
x= 4m
m2= 75kg
Cm = center of mass
g= acceleration due to gravity= 9.82 m/s^2
The distance of centre of mass is x/2
Center of mass(1) = x/2
But x= 4 m
Then substitute, we have,
Center of mass(1) = 4/2 = 2m
We can find the total torque, through the summation of moments that comes from both the man and the beam.
τ = τ(1) + τ(2)
But
τ(1)= ( Center of m1 × m1 × g)= (2× 470× 9.81)
= 9221.4Nm
τ(2)= X * m2 * g = ( 4× 75 × 9.81)= 2943Nm
τ = τ(1) + τ(2)
= 9221.4Nm + 2943Nm
= 12164.4 Nm
Hence, the magnitude of the torque about the point where the beam is bolted into place is 12164.4 Nm
