A submarine dives from rest a 100-m distance beneath the surface of an ocean. Initially, the submarine moves at a constant rate

(6-16)/4.0=-2.5 m/s²
The car's acceleration is -2.5 m/s²

For motion in a circle.

Centripetal acceleration is calculated as mv²/r = mω²r

where v represents linear velocity, r equals radius which is diameter/2 equating to 1/2 or 0.5m

. Here, m is the mass of the object, which is 175g or 0.175kg.

The angular speed, ω, is derived from Angle covered / time

                         = 2 revolutions per 1 second

                         = 2 * 2π  radians for each second

                         = 4π  radians per second

Thus, Centripetal Acceleration = mω²r = 0.175*(4π)² * 0.5. Utilize a calculator

                                                         ≈13.817  m/s²

. The acceleration's magnitude is approximately 13.817  m/s² and it is oriented towards the center of the circular path.

The tension in the string equates to m*a

                                   = 0.175*13.817

                                   = 2.418 N

Response:

The intensity of light 18 feet underwater is about 0.02%

Clarification:

Employing Lambert's law

Let dI / dt = kI, where k is a proportionality factor, I represents the intensity of incident light, and t indicates the thickness of the medium

Then dI / I = kdt

Taking logarithms,

ln(I) = kt + ln C

I = Ce^kt

At t=0, I=I(0) implies C=I(0)

I = I(0)e^kt

At t=3 & I=0.25I(0), we find 0.25=e^3k

Solving for k gives k = ln(0.25)/3

k = -1.386/3

k = -0.4621

I = I(0)e^(-0.4621t)

I(18) = I(0)e^(-0.4621*18)

I(18) = 0.00024413I(0)

The intensity of light 18 feet underwater is about 0.2%

Consider the diagram below.

m₁ = 8.5 x 10⁶ kg, the starship's mass
m₂ = 10⁴ kg, the shuttlecraft's mass
a₁ =  acceleration of the starship
a₂ = the acceleration of the shuttle
F = 4 x 10⁴ N, the force exerted

Let y represent the distance covered by the starship
Let x denote the distance covered by the shuttlecraft
If t indicates the travel time, then
y = 0.5a₁t²                  (1)
x = 0.5a₂t²                  (2)

F = m₁a₁ = m₂a₂         (3)
Additionally,
x + y = 14000 m          (4)

From (2), we derive
a₁ = (4 x 10⁴ N)/(8.5 x 10⁶ kg) = 4.706 x 10⁻³ m/s²
a₂ = (4 x 10⁴ N)/(10⁴ kg) = 4 m/s²

From (1), (2) and (4), we find
0.5*(t s)²*(4 + 4.706 x 10⁻³ m/s²) = 14000 m
2.002353t² = 14000
t² = 6991.774 s²
t = 83.617 s

Thus
x = 0.5*4*6991.774 = 13984 m = 13.984 km
y = 0.5*4.706 x 10⁻³*6991.774 = 16.452 m

The starship moves roughly 16.5 meters while towing the shuttlecraft by 13.98 kilometers.

Result: The starship shifts by 16.5 m (to the nearest tenth)

B = µo*N*I/2r  

Thus, B = 4πx10^-7*150*1.6/2*3.5 = 4.31x10^-5T