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4 days ago

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Early investigators (including Thomas Young) measured the thickness of wool fibers using diffraction. One early instrument used

a collimated beam of 560 nm light to produce a diffraction pattern on a screen placed 30 cm from a single wool fiber.
part a: If the fiber's diameter was 18 μm, what was the width of the central maximum on the screen?

1 answer:

ValentinkaMS [1.1K]4 days ago

8 0

Answer:

$w= 1.867\times10^{-2}$

Explanation:

Provided:

fiber diameter d= 18 μm

screen distance D= 30 cm

wavelength λ= 560 nm

from this, we can determine the fringe width

$w=\frac{2\lambda D}{d}$

substituting the values yield

$w=\frac{2\times560\times10^{-9}\times0.3}{18\times10^{-6}$

$w= 1.867\times10^{-2}$

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A spinning wheel is slowed down by a brake, giving it a constant angular acceleration of 25.60 rad/s2. During a 4.20-s time inte

Yuliya22 [1153]

We will use the equations of rotational kinematics,

$\theta =\theta _{0} + \omega_{0} t+ \frac{1}{2}\alpha t^2$ (A)

$\omega^2= \omega^2_{0} +2\alpha\theta$ (B)

Here, $\theta$ and $\theta _{0}$ denote the final and initial angular displacements, respectively, whereas $\omega$ and $\omega_{0}$ represent final and initial angular velocities, and $\alpha$ is the angular acceleration.

We are provided with $\alpha = - 25.60 \ rad/s^2, \theta = 62.4 \ rad$ and $t = 4.20 \ s$ .

By substituting these values into equation (A), we have

$62.4 \ rad = 0 + \omega_{0} 4.20 \ s + \frac{1}{2} (- 25.60 \ rad/s^2) ( 4.20)^2 \\\\ \omega_{0} = \frac{220.5+ 62.4 }{4.20} =67.4 \ rad/s$

Now, using equation (B),

$\omega^2=(67.4 \ rad/s)^2 + 2 (- 25.60 \ rad/s^2)62.4 \ rad \\\\\ \omega = 36.7 \ rad/s$

This indicates that the wheel's angular speed at the 4.20-second mark is 36.7 rad/s.

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8 days ago

Jade and her roommate Jari commute to work each morning, traveling west on I-10. One morning Jade left for work at 6:45 A.M., bu

Maru [1053]

Answer:

Jari

Explanation:

To determine who is traveling faster, we need to evaluate their gradients. A steeper slope indicates a higher speed.

For Jari's path, starting point is (0, 0) and (6, 7) is another point.

The gradient is the difference in y divided by the difference in x:

Change in y=7-0=7

Change in x=6-0=6

Thus, the slope equals 7/6.

For Jade, her first point is (0, 10) and another is (6, 16).

Change in y=16-10=6

Change in x=6-0=6

Thus, the slope equals 6/6=1.

It's evident that 7/6 exceeds 6/6 or 1, proving Jari is quicker than Jade.

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4 days ago

A person is rowing across the river with a velocity of 4.5 km/hr northward. The river is flowing eastward at 3.5 km/hr (Figure 4

Yuliya22 [1153]

Answer: Her velocity magnitude (v) relative to the shore is 5.70 km/h.

Explanation:

Let Q be the speed of the boat, and P be the speed of the river flow.

R represents the resultant velocity combining boat velocity and river current.

According to vector addition using the law of triangles:

$R=\sqrt{P^2+Q^2+2PQCos\theta}$

From the diagram:

P = 3.5 km/h, Q = 4.5 km/h

$\theta= 90^o$

$R=\sqrt{P^2+Q^2+2PQCos\theta}=\sqrt{(3.5)^2+(4.5)^2+3.5\times 4.5\times cos90^o}=5.70$

$(Cos90^o=0),(sin 90^o=1)$

$\alpha =tan^{-1}\frac{Qsin\theta}{P+Qcos\theta}=tan^{-1}\frac{4.5 sin 90^o}{3.5+4.5 cos90^o}=tan^{-1}\frac{4.5}{3.5}=52.12^o$

Therefore, her velocity magnitude relative to the shore is 5.70 km/h.

8 0

17 days ago

A student solving a physics problem for the range of a projectile has obtained the expression r= v20sin(2θ)g where v0=37.2meter/

ValentinkaMS [1144]

The formula for range is:

$R = \frac{v_o^2 sin2\theta}{g}$

Given values are:

$v_0=37.2m/s$

where θ equals 14.1 degrees

$g=9.80m/s^2$

Using the equation above,

$R = \frac{37.2^2 sin2*14.1}{9.80}$

The calculated range is 66.7 meters.

Therefore, the range is approximately 66.1 meters.

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16 days ago

Read 2 more answers

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el

Sav [1095]

Complete Question

An aluminum "12 gauge" wire measures a diameter of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field E in the wire varies over time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is recorded in seconds.

At time 5 seconds, I = 1.2 A.

We need to find the charge Q traveling through a cross-section of the conductor from time 0 to time 5 seconds.

Answer:

The charge is $Q =2.094 C$

Explanation:

The question indicates that

The wire’s diameter is $d = 0.205cm = 0.00205 \ m$

The radius of the wire is $r = \frac{0.00205}{2} = 0.001025 \ m$

Aluminum's resistivity is $2.75*10^{-8} \ ohm-meters.$

The electric field variation is described as

$E (t) = 0.0004t^2 - 0.0001 +0.0004$

The charge is effectively given by the equation

$Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt$

Where A is the area expressed as

$A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2$

Thus,

$\frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega$

Therefore

$Q = 120 \int\limits^{t}_{0} { E(t) } \, dt$

By substituting values

$Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt$

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.$

The question states that t = 5 seconds

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.$

$Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }$

$Q =2.094 C$

5 0

10 days ago