Answer:
A= 148.92 m/s²
Explanation:
Given that
U(x,y) = (6.00 )x² - (3.75 )y ³
m= 0.04 kg
Now force in the x-direction
Fx= - dU/dx
U(x,y) = (6.00 )x² - (3.75 )y ³
dU/dx= 12 x
When x=0.4 m
dU/dx= 12 x 0.4 = 4.8
So we can say that
Fx= - 4.8 N
From Newtons law
F= m a
- 4.8 = 0.04 x a
a = -120 m/s²
Acceleration in x direction ,a = -120 m/s²
In y -direction
F= - dU/dy
U(x,y) = (6.00 )x² - (3.75 )y ³
dU/dy = 0 - 3.75 x 3 y²
When y = 0.56 m
dU/dy = - 3.75 x 3 x 0.56 x 0.56
dU/dy = - 3.52
So we can say that force in y -direction
F= 3.52 N
F= m a'
3.52 = 0.04 x a'
a'=88.2 m/s²
acceleration in y direction is 88.2 m/s²
The resultant acceleration
A= 148.92 m/s²
Answer:
F = ⅔ F₀
Explanation:
For this exercise we use Coulomb's law
F = k q₁q₂ / r²
let's use the subscript "o" for the initial conditions
F₀ = k q² / r²
now the charge changes q₁ = q₂ = 2q and the new distance is r = 3 r
we substitute
F = k 4q² / 9 r²
F = k q² r² 4/9
F = ⅔ F₀
Jack------------ force of 92.5 n eastward-------Fjack(X)=92.5 n Fjack(Y)=0
<span>jill ------------------------------- force of 89.9 n northeast
Fjill(X)=cos45*89.9=63.57 n
</span>Fjill(Y)=sin45*89.9=63.57 n<span>
</span>jane -----------------------------force of 163 n southeast
Fjane(X)=cos45*163=115.26 n
Fjane(X)=-sin45*163=-115.26 n
Ftotal (X)=92.5+63.57+115.26=271.33 n
Ftotal (Y)=0+63.57-115.26=-51.69 n
Fotal=((271.33)^2+(-115.26)^2) ^0.5=294.80 n southeast
the magnitude of the net force the people exert on the donkey. is 294.80 n southeast
Force = change of momentum / time taken
Force = (90x3)/0.6
The K.E. at the maximum height is zero, no matter what that height is.
