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
Answer:
2014.44 N
Explanation:
mass of spacecraft, m = 1850 kg
distance r = 3 x R
where r be the radius of earth.
g be the acceleration due to gravity on the surface of earth and g' be the acceleration due to gravity at height


g' = g / 9
g' = 9.8 / 9 = 1.089 m/s²
Force of gravity on the space craft
F = m g' = 1850 x 1.089
F = 2014.44 N
Thus, the force of gravity on the space craft at height is 2014.44 N.
Answer:
= 3289.8 m / s
Explanation:
This exercise can be solved using the definition of momentum
I = ∫ F dt
Let's replace and calculate
I = ∫ (at - bt²) dt
We integrate
I = a t² / 2 - b t³ / 3
We evaluate between the lower limits I=0 for t = 0 s and higher I=I for t = 2.74 ms
I = a (2,74² / 2- 0) - b (2,74³ / 3 -0)
I = a 3,754 - b 6,857
We substitute the values of a and b
I = 1500 3,754 - 20 6,857
I = 5,631 - 137.14
I = 5493.9 N s
Now let's use the relationship between momentum and momentum
I = Δp = m
- m v₀o
I = m
- 0
= I / m
= 5493.9 /1.67
= 3289.8 m / s
I'm not quite sure what happens to Fay so I didn't finish but hope it helps