How would the force of a test charge change if the electric field is doubled?
2 answers:
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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
Answer:
Explanation:
First of all, let's remind that:
- The kinetic energy of an object is given by , where m is the mass and v is the speed
- The momentum of an object is given by
- The inertia of an object is proportional to its mass, so we can write , where k just indicates a constant of proportionality
In this problem, we have:
- (the two objects have same kinetic energy)
- (A has three times the momentum of B)
Re-writing both equation we have:
If we divide first equation by second one we get
And if we substitute it into the first equation we get
So, B has 9 times more mass than A, and so B has 9 times more inertia than A, and their ratio is: