Work, Kinetic Energy and Potential Energy
6.1 The Important Stuff 6.1.1 Kinetic Energy
For an object with mass m and speed v, the kinetic energy is defined as K = 1mv2
2
(6.1)
Kinetic energy is a scalar (it has magnitude but no direction); it is always a positive number; and it has SI units of kg · m2/s2. This new combination of the basic SI units is
known as the joule:
As we will see, the joule is also the unit of work W and potential energy U. Other energy
1joule = 1J = 1 kg·m2 (6.2) s2
units often seen are:
6.1.2 Work
1erg=1g·cm2 =10−7J 1eV=1.60×10−19J s2
When an object moves while a force is being exerted on it, then work is being done on the object by the force.
If an object moves through a displacement d while a constant force F is acting on it, the force does an amount of work equal to
W =F·d=Fdcosφ (6.3)
where φ is the angle between d and F.
Answer:
Maximum charge will be 
Explanation:
We have given force ion the moving charge 
Maximum speed of the moving charge v = 30 m /sec
Magnetic field 
We have to fond the charge
Force on moving charge is given by
So 
Maximum charge will be
The charge transported as a result of charging is 43200C .
we can note the following data from the statement :
Current (i) = 3A
Time interval (t) = 4 hrs = 14400secs
As we know ,
Current is defined as the rate of flow of the charges between two points when a potential difference is applied between them.
Mathematically , it can be written as
Current(i) = charge (q)/ time(t)
Q = i × t
= 3 × 14400
= 43200 C
By dividing the value of the current passed through the circuit by the time for which the charge is passed gives out the value of the charge passed .
learn more on charges :
brainly.com/question/15238352
#SPJ4
Answer:
V = (-32t +32) ft/s
The velocity of the bowling ball after t seconds is the time derivative of the position function s(t). To obtain the velocity from s(t) we simply differentiate s(t) with respect to t. That is
V = ds(t)/dt = -16×2t + 32×1 = -32t + 32.
When differentiating, you multiply the coefficient of each term in the equation with the power of the variable and then reduce the power by 1 just like above.
Explanation:
