That's 299,792,458 meters per second.
Answer:
when a custodian pushes a large crate across the gym floor, forces acting on the crate are the one applied by the custodian, frictional force acting in the opposite direction of motion acting between the surface of crate and floor and reaction force (normal force) acting in upward direction normal to the surface
Explanation:
Answer:
The final velocity of the second player is 6.1 m/s.
Explanation:
The final velocity of the second player can be calculated by conservation of linear momentum (p):
(1)
Where:
: is the mass of the first football player = 110 kg
: is the mass of the second football player = 90 kg
: is the initial velocity of the first football player = 5.0 m/s
: is the initial velocity of the second football player = 0 (he is at rest)
: is the final velocity of the first football player = 0 (he stops after the impact)
: is the final velocity of the second football player =?
By solving equation (1) for we have:
Therefore, the final velocity of the second player is 6.1 m/s.
I hope it helps you!
In order to reduce the amount of energy lost due to heat flow, electricity is delivered to our homes using <u>d. low current</u>.
<u>Explanation</u>:
The presence of flow of electrons is said to be electricity. The electricity flows in one direction. The electricity enables function of the electrical devices and machines.
Heat is generated when the electricity flows from one place to another. So the energy is lost in the form of heat.
So to avoid the loss of energy, the electricity is delivered to home using low current. The strength of an electric current can be measured and represented in the unit of ampere.
Answer:
Therefore,
The potential (in V) near its surface is 186.13 Volt.
Explanation:
Given:
Diameter of sphere,
d= 0.29 cm
Charge ,
To Find:
Electric potential , V = ?
Solution:
Electric Potential at point surface is given as,
Where,
V= Electric potential,
ε0 = permeability free space = 8.85 × 10–12 F/m
Q = Charge
r = Radius
Substituting the values we get
Therefore,
The potential (in V) near its surface is 186.13 Volt.
