Answer:
The magnitude of the net force is √2F.
Explanation:
Since the two particles have the same charge Q, they exert the same force on the test charge; both attractive or repulsive. So, the angle between the two forces is 90° in any case. Now, as we know the magnitude of these forces and that they form a 90° angle, we can use the Pythagorean Theorem to calculate the magnitude of the resultant net force:
Then, it means that the net force acting on the test charge has a magnitude of √2F.
Answer:
Explanation:
We shall apply conservation of mechanical energy
kinetic energy of alpha particle is converted into electric potential energy.
1/2 mv² = k q₁q₂/d , d is closest distance
d = 2kq₁q₂ / mv²
= 2 x 9 x 10⁹ x 79e x 2e / 4mv²
= 1422 x2x (1.6 x 10⁻¹⁹)² x 10⁹ /4x 1.67 x 10⁻²⁷ x (1.5 x 10⁷)²
= 3640.32 x 10⁻²⁹ /2x 3.7575 x 10⁻¹³
= 484.4 x 10⁻¹⁶
=48.4 x 10⁻¹⁵ m
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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