1. An object is traveling at a constant velocity of 30 m/s when it experiences a constant

A jet starts at rest at the end of a runway and reaches a speed of 80 m/s in 20 s. What is its acceleration?

ahrayia [7]

Answer:

Acceleration=velocity/time.

=80/2=40m/s^2.

4 0

3 years ago

Monochromatic light of variable wavelength is incident normally on a thin sheet of plastic film in air. The reflected light is a

BartSMP [9]

Answer:

thickness t = 528.433 nm

Explanation:

given data

wavelength λ1 = 477.1 nm

wavelength λ2 = 668.0 nm

n = 1.58

solution

we know for constructive interference condition will be

2 × t × μ = (m1+0.5) × λ1 ....................1

2 × t × μ = (m2+0.5) × λ2 ....................2

so we can say from equation 1 and 2

(m1+0.5) × λ1 = (m2+0.5) × λ2

so

$\frac{\lambda 2}{\lambda 1} = \frac{m1+0.5}{m2+0.5}$ ..............3

put here value and we get

$\frac{668.0}{477.1} = \frac{m1+0.5}{m2+0.5}$

$\frac{m1+0.5}{m2+0.5}$ = 1.4

$\frac{m1+0.5}{m2+0.5} = \frac{7}{5}$ ...................4

so we here from equation 4

m1+0.5 = 7

m1 = 3 .................5

m2+0.5 = 4

m2 = 2 .................6

so now put value in equation 1

2 × t × μ = (m1+0.5) × λ1

2 × t × 1.58 = (3+0.5) × 477.1

solve it we get

thickness t = 528.433 nm

4 0

3 years ago

Interactive Solution 8.29 offers a model for this problem. The drive propeller of a ship starts from rest and accelerates at 2.3

MAXImum [283]

Answer:

Δθ = 15747.37 rad.

Explanation:

The total angular displacement is the sum of three partial displacements: one while accelerating from rest to a certain angular speed, a second one rotating at this same angular speed, and a third one while decelerating to a final angular speed.
Applying the definition of angular acceleration, we can find the final angular speed for this first part as follows:

$\omega_{f1} = \alpha * \Delta t = 2.38*e-3rad/s2*2.04e3s = 4.9 rad/sec (1)$

Since the angular acceleration is constant, and the propeller starts from rest, we can use the following kinematic equation in order to find the first angular displacement θ₁:

$\omega_{f1}^{2} = 2* \alpha *\Delta\theta (2)$

Solving for Δθ in (2):

$\theta_{1} = \frac{\omega_{f1}^{2}}{2*\alpha } = \frac{(4.9rad/sec)^{2}}{2*2.38*e-3rad/sec2} = 5044.12 rad (3)$

The second displacement θ₂, (since along it the propeller rotates at a constant angular speed equal to (1), can be found just applying the definition of average angular velocity, as follows:

$\theta_{2} =\omega_{f1} * \Delta_{t2} = 4.9 rad/s * 1.48*e3 s = 7252 rad (4)$

Finally we can find the third displacement θ₃, applying the same kinematic equation as in (2), taking into account that the angular initial speed is not zero anymore:

$\omega_{f2}^{2} - \omega_{o2}^{2} = 2* \alpha *\Delta\theta (5)$

Replacing by the givens (α, ωf₂) and ω₀₂ from (1) we can solve for Δθ as follows:

$\theta_{3} = \frac{(\omega_{f2})^{2}- (\omega_{f1}) ^{2} }{2*\alpha } = \frac{(2.42rad/s^{2}) -(4.9rad/sec)^{2}}{2*(-2.63*e-3rad/sec2)} = 3451.25 rad (6)$