Answer:
V_{a} - V_{b} = 89.3
Explanation:
The electric potential is defined by
= - ∫ E .ds
In this case the electric field is in the direction and the points (ds) are also in the direction and therefore the angle is zero and the scalar product is reduced to the algebraic product.
V_{b} - V_{a} = - ∫ E ds
We substitute
V_{b} - V_{a} = - ∫ (α + β/ y²) dy
We integrate
V_{b} - V_{a} = - α y + β / y
We evaluate between the lower limit A 2 cm = 0.02 m and the upper limit B 3 cm = 0.03 m
V_{b} - V_{a} = - α (0.03 - 0.02) + β (1 / 0.03 - 1 / 0.02)
V_{b} - V_{a} = - 600 0.01 + 5 (-16.67) = -6 - 83.33
V_{b} - V_{a} = - 89.3 V
As they ask us the reverse case
V_{b} - V_{a} = - V_{b} - V_{a}
V_{a} - V_{b} = 89.3
The 'formulas' to use are just the definitions of 'power' and 'work':
Power = (work done) / (time to do the work)
and
Work = (force) x (distance) .
Combine these into one. Take the definition of 'Work', and write it in place of 'work' in the definition of power.
Power = (force x distance) / (time)
From the sheet, we know the power, the distance, and the time. So we can use this one formula to find the force.
Power = (force x distance) / (time)
Multiply each side by (time): (Power) x (time) = (force) x (distance)
Divide each side by (distance): Force = (power x time) / (distance).
Look how neat, clean, and simple that is !
Force = (13.3 watts) x (3 seconds) / (4 meters)
Force = (13.3 x 3 / 4) (watt-seconds / meter)
Force = 39.9/4 (joules/meter)
<em>Force = 9.975 Newtons</em>
Is that awesome or what !
Answer:
51.82
Explanation:
First of all, let's convert both vectors to cartesian coordinates:
Va = 36 < 53° = (36*cos(53), 36*sin(53))
Va = (21.67, 28.75)
Vb = 47 < 157° = (47*cos(157), 47*sin(157))
Vb = (-43.26, 18.36)
The sum of both vectors will be:
Va+Vb = (-21.59, 47.11) Now we will calculate the module of this vector:

Answer:
3600joules
Explanation:
formula :W=FS
W=work done (J)
F=force (N)
S=displacement moved in the direction of force (m)
200N×18m
=3600J