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Murljashka [212]

1 year ago

13

Light from a monochromatic source shines through a double slit onto a screen 5.00 m away. The slits are 0.180 mm apart. The dark

bands on the screen are measured to be 1.70 cm apart. What is the wavelength of the incident light? (A) 612 nm (B) 457 nm (C) 306 nm (D) 392 nm

1 answer:

Nitella [24]1 year ago

4 0

Answer:

Wavelength of incident light, $\lambda = 612 nm$

Given:

Distance between slit and screen, x = 5.00 m

slit width, d = 0.180 mm

width of the fringe, $\beta = 1.70 cm = 0.017 m$

Solution:

To calculate the wavelength of the incident light, $\lambda$ :

$\beta = \frac{x\lambda }{d}$

$\lambda = \frac{\beta d}{x}$

$\lambda = \frac{0.017\times 0.180\times 10^{- 3}}{5} = 6.12\times 10^{- 7}m = 612 nm$

$\lambda = 612 nm$

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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

Tresset [83]

Complete Question

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 electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is $Q =2.094 C$

Explanation:

From the question we are told that

The diameter of the wire is $d = 0.205cm = 0.00205 \ m$

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

The resistivity of aluminum is $2.75*10^{-8} \ ohm-meters.$

The electric field change is mathematically defied as

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

Generally the charge is mathematically represented as

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

Where A is the area which is mathematically represented as

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

So

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

Therefore

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

substituting values

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

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

From the question we are told that t = 5 sec

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

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

$Q =2.094 C$

5 0

2 years ago

1. A2 .7-kg copper block is given an initial speed of 4.0m/s on a rough horizontal surface. Because of friction, the block final

tatyana61 [14]

Answer:

A. Increase in temperature is 0.0176 degree Celsius. b. the remaining energy will be lost.

Explanation:

The mass of copper block = 7kg

Initial speed = 4.0 m/s

Specific heat of copper = 0.385 j/g degree Celcius.

a. The increase in temperature is calculated below:

$\text{Change in kinetic energy} = \frac{1}{2}mv^{2} \\= \frac{1}{2} \times 2.7 \times 4^{2} = 21.6 J \\$

85% of energy is converted into internal energy.

$mc\DeltaT = 21.6 \times 0.85 \\2.7 \times 385 \times \Delta T = 21.6 \times 0.85 \\\Delta T = 0.0176 degree \ celsius$

b. The remaining 15 per cent of kinetic energy will be lost and it will be changed into other forms.

7 0

1 year ago

Suggest reasons why poaching for subsistence is likely to be less damaging to the biodiversity of an area than poaching for prof

dlinn [17]

An example for ruining a biodiversity is fishing. The two factors that have contributed to increased fishing in deep ocean waters in recent years are the human population growth and decreased fishing opportunities inshore. Increase population growth increases the demand for food which also leads to increase in fish demand. Because the fish demand is high, inshore fishing opportunities decrease that is why deep ocean waters is the new venue for fishing. This may sound absurd but poaching for subsistence is likely to be less damaging to he biodiversity <span>of an area than poaching for profit. Because the people do not care anymore to the biodiversity that they interrupted just to get back more profit. They do not care what must be taken from it like getting bigger fishes and leaving the smaller ones behind to maintain productivity.</span>

5 0

1 year ago

(Another tomato/skyscraper problem.) You are looking out your window in a skyscraper, and again your window is at a height of 45

Ivan

Answer:

1027.2 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 32.2 ft/s

$s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{450-\frac{1}{2}\times 32.2\times 2^2}{2}\\\Rightarrow u=192.8\ ft/s$

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{192.8^2-0^2}{2\times 32.2}\\\Rightarrow s=577.20\ m$

The height the tomato would fall is 450+577.2 = 1027.2 m

6 0

2 years ago

An insurance company hired your group to help investigate an insurance claim following a car accident. In the accident, two cars

Romashka-Z-Leto [24]

Answer:

v=8m/s

Explanation:

To solve this problem we have to take into account, that the work done by the friction force, after the collision must equal the kinetic energy of both two cars just after the collision. Hence we have

$W_{f}=E_{k}\\W_{f}=\mu N=\mu(m_1+m_1)g\\E_{k}=\frac{1}{2}[m_1+m_2]v^2$

where

mu: coefficient of kinetic friction

g: gravitational acceleration

We can calculate the speed of the cars after the collision by using

$W_f=(0.7)(1650kg+1900kg)(9.8\frac{m}{s^2})=24353J\\24353J=\frac{1}{2}(1650kg+1900kg)v^2\\v=\sqrt{\frac{24353J}{1775kg}}=3.70\frac{m}{s}$

Now , we can compute the speed of the second car by taking into account the conservation of the momentum

$P_b=P_a\\m_1v_1+m_2v_2=(m_1+m_2)v\\\\v_2=\frac{(m_1+m_2)v-m_1v_1}{m_1}\\\\v_2=\frac{(1650kg+1900kg)(3.7\frac{m}{s})-(1650kg)(16\frac{m}{s})}{(1650kg)}\\\\v_2=8\frac{m}{s}$

the car did not exceed the speed limit

Hope this helps!!

7 0

1 year ago

Read 2 more answers