The Basketball with the greatest gravitational potential energy is : Basketball 4 feet above the ground ( c )
<u>Given data :</u>
masses = constant
H₁ = 2 ft
H₂ = 3 ft
H₃ = 4 ft
g = 9.81 m/s²
<h3><u>Procedure to determine the basketball with the greatest gravitational P.E </u></h3>
we will apply the equation below
Gravitational potential energy ( U ) = mgh
U = 2 * 9.81 = 19.62
U = 3 * 9.81 = 29.43
U = 4 * 9.81 = 39.24
From the calculations the basketball with the greatest gravitational potential energy is the basketball at 4ft above the ground
Learn more about gravitational potential energy : brainly.com/question/15896499
Answer:
Explanation:
When an moving electric charge passes through a uniform magnetic field
its motion becomes circular .
If m be the mass v be the velocity , q be the charge on the mass B be the magnetic field and R be the radius of circular path
force on the moving charge created by magnetic field
= B q v
Centripetal force required for circular motion
= m v² / R
For balancing
B q v = m v² / R
v = B q R / m
Time period of rotation
T = 2π R / v
= 2 π R m / B q R
= 2 π m / B q
For first particle
T₁ = 2 π m₁ / B q₁
For second particle
T₂ = 2 π m₂ / B q₂
q₁ = q₂ and 10 m₁ = m₂ ( given )
Putting the values in second equation
T₂ = 2 π 10 m₁ / B q₁
= 10 x 2 π m₁ / B q₁
= 10 T₁
Given T₁ = T
T₂ = 10 T
Answer: 109.89 Nm
Explanation:
The maximum torque will be calculated as the force multiplied by the perpendicular distance. This will be:
Torque = force × perpendicular distance
torque = 333 × 0.33
= 109.89 Nm
