Answer:
A) a = 73.304 rad/s²
B) Δθ = 3665.2 rad
Explanation:
A) From Newton's first equation of motion, we can say that;
a = (ω - ω_o)/t. We are given that the centrifuge spins at a maximum rate of 7000rpm.
Let's convert to rad/s = 7000 × 2π/60 = 733.04 rad/s
Thus change in angular velocity = (ω - ω_o) = 733.04 - 0 = 733.04 rad/s
We are given; t = 10 s
Thus;
a = 733.04/10
a = 73.304 rad/s²
B) From Newton's third equation of motion, we can say that;
ω² = ω_o² + 2aΔθ
Where Δθ is angular displacement
Making Δθ the subject;
Δθ = (ω² - ω_o²)/2a
At this point, ω = 0 rad/s while ω_o = 733.04 rad/s
Thus;
Δθ = (0² - 733.04²)/(2 × 73.304)
Δθ = -537347.6416/146.608
Δθ = - 3665.2 rad
We will take the absolute value.
Thus, Δθ = 3665.2 rad
The reciprocal of the total resistance is equal to the sum of the reciprocals of the component resistances:
1/(120.7 Ω) = 1/<em>R₁</em> + 1/(221.0 Ω)
1/<em>R₁</em> = 1/(120.7 Ω) - 1/(221.0 Ω)
<em>R₁</em> = 1 / (1/(120.7 Ω) - 1/(221.0 Ω)) ≈ 265.9 Ω
The answer for question 2 i guess it’s c
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
Answer:
U = initial velocity, t = time taken, s = distance covered. Deceleration Formula is used to calculate the deceleration of the given body in motion.
