Answer:
The sum of the initial and final velocity is divided by 2 to find the average. The average velocity calculator uses the formula that shows the average velocity (v) equals the sum of the final velocity (v) and the initial velocity (u), divided by 2.
Answer:
t=17.838s
Explanation:
The displacement is divided in two sections, the first is a section with constant acceleration, and the second one with constant velocity. Let's consider the first:
The acceleration is, by definition:
So, the velocity can be obtained by integrating this expression:
The velocity is, by definition: 
, so
.
Do x=11 in order to find the time spent.
At this time the velocity is: 
This velocity remains constant in the section 2, so for that section the movement equation is:
The left distance is 89 meters, and the velocity is 
, so:
So, the total time is 14.303+3.5355s=17.838s
Answer:
The distance between the ships changing at 6PM is 21.29Km/h
Explanation:
Ship A is sailing east at 35Km/h and ship B is sailing West at 25Km/h
Given
dx/dt= 35
dy/dt= 25
dv/dt= ???? at t= 6PM - 2PM= 4
Therefore t=4
We know ship A travels at 150km in the x-direction and Ship A at t=4 travels at 4.35 Which is 140 also in x-direction
So, we use:
;
But ship B travels at t=4, at 4.25 =100 in the y-direction
so, let's use the equation:
Lets use 2DD' = 2xx' + 2yy'
Differentiating with respect to t we have:
D•d(D)/dt = -(10)•dx/dt + 100•dy/dt
=100.5 d(D)/dt = (-10)•35 + (100)•25
When t=4, we have x=(140-150) =10 and y=100
=100.5
= 100.5 dD/dt = 10.35 +100.25
= dD/dt = 21.29km/h
-- "Work" is the product of (force exerted on the object) x (distance the object moves in the direction of the force).
Since the orbit is a circle, the gravitational force toward the center is always perpendicular to the orbit. The object never moves in the direction of the force. If it did, it wouldn't be 'R' away from the center of the circle.
So the product of (gravitational force) x (distance in the direction of the force) is always zero.
-- Even if the orbit ISN't a circle . . . there are some parts of the orbit that aren't quite perpendicular to the gravitational force. If the satellite is traveling through one of those parts AND getting closer to the central body, then gravity is doing positive work on the satellite. If the satellite is traveling through one of those parts and getting FARTHER from the central body, then the satellite is the one doing positive work, and gravity is doing 'negative work'. The work done by gravity ... and the work done by the satellite ... is zero over a complete revolution, although not zero at every point.
This is exactly the definition of a "Conservative Force" ... a force that does zero work through one trip around any CLOSED path. Gravity is a conservative force, and so is the Electrostatic force.
Force = mass × acceleration
mass of an electron =9.1×10^-31 kg
given acceleration...
find force
