The potential on the surface of the charged sphere is 3.9 × 10⁷ volts.
Given:
The diameter of the sphere is 2.30 m and the charge of 1.10 mc.
The potential on the surface of a charged sphere is given by,

Where V is the potential on the surface, Q is the charge, R is the radius and k is the Coulomb's constant.
V = 8.99×10⁹×5×10⁻³/(2.30÷2)
V = 3.9 × 10⁷ volts
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<span>Anion are associated with exothermic process.</span>
This question involves the concept of the conservation of energy.
(a) The child's speed halfway will be "12.13 m/s".
(b) The child's speed three-fourth way will be "14.86 m/s".
(a)
Using the <em>law of conservation of energy</em>:
Potential Energy Lost = Kinetic Energy Gained

where,
v = speed = ?
g = acceleration due to gravity = 9.81 m/s²
h = height lost = halfway = height/2 = 15 m/2 = 7.5 m
Therefore,

<u>v = 12.13 m/s</u>
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(b)
Using the <em>law of conservation of energy</em>:
Potential Energy Lost = Kinetic Energy Gained

where,
v = speed = ?
g = acceleration due to gravity = 9.81 m/s²
h = height lost = three-fourth way down = 
Therefore,

<u>v = 14.86 m/s</u>
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Learn more about the <em>law of conservation</em> of energy here:
brainly.com/question/20971995?referrer=searchResults
The attached picture shows the <em>law of conservation of energy</em>.
Answer:
Explanation:
True
It is correct as the total rotational energy is the sum of transnational kinetic energy of each small piece of the object.
Consider a rigid body made of small particles each of mass dm have same angular velocity ω and the radius of the rotational path is r.
The velocity of each particle is v = r ω
So, the kinetic energy of each particle is
dK = 0.5 dm x v² = 0.5 x dm x r² x ω²
The total kinetic energy
K = 0.5 x I ω²
where, I is the moment of inertia
Answer:
Motors are the most common application of magnetic force on current-carrying wires. Motors have loops of wire in a magnetic field. When current is passed through the loops, the magnetic field exerts torque on the loops, which rotates a shaft. Electrical energy is converted to mechanical work in the process
Explanation:
hope that helps!