Answer:
a) θ = 2500 radians
b) α = 200 rad/s²
Explanation:
Using equations of motion,
θ = (w - w₀)t/2
θ = angle turned through = ?
w = final angular velocity = 1420 rad/s
w₀ = initial angular velocity = 420
t = time taken = 5s
θ = (1420 - 420) × 5/2 = 2500 rads
Again,
w = w₀ + αt
α = angular accelaration = ?
1420 = 420 + 5α
α = 1000/5 = 200 rad/s²
To locate a specific target or to determine how close submarines are to the seafloor, they use active and passive sound navigation and ranging (or a SONAR, in simple terms.) It emits pulses of sound waves that travel through the water, reflect off the target and relayed back to the ship. By determining how fast the sound wave travels back, the computers on the sub calculate how far they are from the target.
Hope this helps.
Answer:
2.64N
Explanation:
Force = mass * acceleration
Given
mass = 4kg
distance = 1.9m
Time t = 2.4s
Get the acceleration using the equation of motion
S = ut + 1/2at²
1.9 = 0 + 1/2a(2.4)²
1.9 = 5.76a/2
1.9 = 2.88a
a = 1.9/2.88
a = 0.66m/s²
Get the magnitude of the force
Force = 4 * 0.66
Force = 2.64N
Hence the net force acting on the fish is 2.64N
Answer:
a) Acceleration is zero
, c) Speed is cero
Explanation:
a) the equation that governs the simple harmonic motion is
x = A cos (wt +φφ)
Where A is the amplitude of the movement, w is the angular velocity and φ the initial phase determined by the initial condition
Body acceleration is
a = d²x / dt²
Let's look for the derivatives
dx / dt = - A w sin (wt + φ)
a = d²x / dt² = - A w² cos (wt + φ)
In the instant when it is not stretched x = 0
As the spring is released at maximum elongation, φ = 0
0 = A cos wt
Cos wt = 0 wt = π / 2
Acceleration is valid for this angle
a = -A w² cos π/2 = 0
Acceleration is zero
b)
c) When the spring is compressed x = A
Speed is
v = dx / dt
v = - A w sin wt
We look for time
A = A cos wt
cos wt = 1 wt = 0, π
For this time the speedy vouchers
v = -A w sin 0 = 0
Speed is cero
The outer shell can hold 1 electron
