The speed of the object increases
Explanation:
We can answer this question by applying the work-energy theorem, which states that the work done on an object is equal to the change in kinetic energy of the object. Mathematically:
where
W is the work done on the object
are the final and initial kinetic energy of the object, respectively
m is the mass of the object
v is its final speed
u is its initial speed
In this case, the force does a positive amount of work on the object, so
This also implies that
And so
And therefore
which means that the speed of the object increases.
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A Parallel circuit has certain characteristics and basic rules: A parallel circuit has two or more paths for current to flow through. Voltage is the same across each component of the parallel circuit. The sum of the currents through each path is equal to the total current that flows from the source.
The work W done by the electric field in moving the proton is equal to the difference in electric potential energy of the proton between its initial location and its final location, therefore:
where q is the charge of the proton,
, with
being the elementary charge, and
and
are the initial and final voltage.
Substituting, we get (in electronvolts):
and in Joule:
Answer:it experiences no force
Explanation:
a charge moving in a direction parallel to the magnetic field experience no force.since the angle e is 0,force would also be 0
Answer:
W = 1.06 MJ
Explanation:
- We will use differential calculus to solve this problem.
- Make a differential volume of water in the tank with thickness dx. We see as we traverse up or down the differential volume of water the side length is always constant, hence, its always 8.
- As for the width of the part w we see that it varies as we move up and down the differential element. We will draw a rectangle whose base axis is x and vertical axis is y. we will find the equation of the slant line that comes out to be y = 0.5*x. And the width spans towards both of the sides its going to be 2*y = x.
- Now develop and expression of Force required:
F = p*V*g
F = 1000*(2*0.5*x*8*dx)*g
F = 78480*x*dx
- Now, the work done is given by:
W = F.s
- Where, s is the distance from top of hose to the differential volume:
s = (5 - x)
- We have the work as follows:
dW = 78400*x*(5-x)dx
- Now integrate the following express from 0 to 3 till the tank is empty:
W = 78400*(2.5*x^2 - (1/3)*x^3)
W = 78400*(2.5*3^2 - (1/3)*3^3)
W = 78400*13.5 = 1058400 J
