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
To solve this problem it is necessary to apply the concepts related to the Period based on gravity and length.
Mathematically this concept can be expressed as
Where,
l = Length
g = Gravitational acceleration
First we will find the period that with the characteristics presented can be given on Mars and then we can find the length of the pendulum at the desired time.
The period on Mars with the given length of 0.99396m and the gravity of the moon (approximately 
will be
For the second question posed, it would be to find the length so that the period is 2 seconds, that is:
Therefore, we can observe also that the shorter distance would be the period compared to the first result given.
Answer:
55 ft/s
Explanation:
A₁ = Area of rectangular cross-section at input side = 1.5 x 11 = 16.5 ft²
A₂ = Area of rectangular cross-section at far end = 1.5 x 6 = 9 ft²
v₁ = speed of water at the input side of channel = 30 ft/s
v₂ = speed of water at the input side of channel = ?
Using equation of continuity
A₁ v₁ = A₂ v₂
(16.5) (30) = (9) v₂
v₂ = 55 ft/s
Answer:
B. IT should have a very broad focus with many variables.
