Answer:
Your zenith is 43 N of 90 deg (equator)
Thus, your zenith is 90 - 43 = 47 deg
(At the N pole your zenith would be 0 deg from the N pole)
It moves to 56 km per hours
Answer:
P = 40.7kPa
Explanation:
To find the pressure on a surface 6 meter below you use the following formula, which takes into account the heights in which pressures are measured and also the density of the fluid and the gravitational acceleration:
(1)
P2: pressure for a height of -6 m = ?
P1: pressure for a height of -2 m = 1.5kPa = 1500 Pa
ρ: density of water = 1000kg/m^3
g: gravitational acceleration = 9.8 ms^2
y2: -6m
y1: -2m
(the height is measure from the water level, because of that, the heights are negative)
You solve the equation (1) for P1:
(2)
Next, you replace the values of all variables in equation (2):

hence, the pressure on a surface 6 m below the water level is 40.7kPa
Complete Question
The complete question is shown on the first uploaded image
Answer:
a

b

Explanation:
From the question we are told that
The amplitude is 
The period is 
The test weight is 
Generally the radial acceleration is mathematically represented as

at maximum angular acceleration

So

Now
is the angular velocity which is mathematically represented as

Therefore
![a_{max} = [\frac{2 * \pi}{T} ]^2 * A](https://tex.z-dn.net/?f=a_%7Bmax%7D%20%3D%20%20%5B%5Cfrac%7B2%20%2A%20%20%5Cpi%7D%7BT%7D%20%5D%5E2%20%2A%20A)
substituting values
![a_{max} = [\frac{2 * 3.142}{17} ]^2 * 0.018](https://tex.z-dn.net/?f=a_%7Bmax%7D%20%3D%20%20%5B%5Cfrac%7B2%20%2A%20%203.142%7D%7B17%7D%20%5D%5E2%20%2A%200.018)

Generally this test weight is mathematically represented as
Where k is the spring constant
Therefore

substituting values


If we ignore air resistance, as we always do,
there is no horizontal acceleration.
Gravity acts only in the vertical direction, and
no other forces act on the ball after the punt.