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1 year ago

Select all the wave properties that can change how we perceive light waves.amplitudeamplitudecrestscreststimetimefrequency

Physics

1 answer:

ASHA 777 [7]1 year ago

6 0

The properties that change how we perceive light waves are the following:

The amplitude of the light wave changes the brightness of light relative to other light waves of the same wavelenghth.

The frequency of the light wave changes the color and the type of the light wave.

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A 0.200 H inductor is connected in series with a 88.0 Ω resistor and an ac source. The voltage across the inductor is vL=−(12.0V

Step2247 [10]

Answer:

a. (VL)R/ωL[1 - cos[ωt]] = (10.84 V)[1 - cos[(487rad/s)t]]

b. 1.084 mV

Explanation:

a. Since it is a series circuit, the current in the inductor is the same as the current in the resistor.

Now, the voltage across the inductor vL = -Ldi/dt.

So, the current, i = -1/L∫vLdt.

Now, vL = −(12.0V)sin[(487rad/s)t] and L = 0.200 H

Substituting these into i, we have

i = -1/L∫vLdt

= -1/0.200H∫[−(12.0V)sin[(487rad/s)t]]dt.

= -[−(12.0V)]/0.200H∫[sin[(487rad/s)t]]dt.

= 60V/H∫[sin[(487rad/s)t]]dt

Integrating i, we have

i = 60V/H ÷ [(487rad/s)[-cos[(487rad/s)t]] + C

at t = 0, i(0) = 0

0 = 60V/H ÷ [(487rad/s)[-cos[(487rad/s)× 0]] + C

0 = 60V/H ÷ [(487rad/s)[-cos[0]] + C

0 = 60V/H ÷ [(487rad/s)[-1]+ C

C = 60V/H ÷ [(487rad/s)

So, i = 60V/H ÷ [(487rad/s)[-cos[(487rad/s)t]] + 60V/H ÷ [(487rad/s)

i = 60V/H ÷ [(487rad/s)[1 - cos[(487rad/s)t]]

i = (0.123A)[1 - cos[(487rad/s)t]] = VL/ωL[1 - cos[ωt]] where ω = 487rad/s and VL = 12.0 V and L = 0.200 H

So, the voltage across the resistor vR = iR where R = resistance of resistor = 88.0 Ω

So, vR = iR = VL/ωL[1 - cos[ωt]] × R = (VL)R/ωL[1 - cos[ωt]]

= (0.123A)[1 - cos[(487rad/s)t]] × 88.0 Ω

= (10.84 V)[1 - cos[(487rad/s)t]]

b. vR at t = 2.00 ms = 0.002 s

So, vR = (10.84 V)[1 - cos[(487rad/s)(0.002)]]

= (10.84 V)[1 - cos[0.974]]

= (10.84 V)[1 - 0.9999]

= (10.84 V)(0.0001)

= 0.001084

= 1.084 mV

3 0

3 years ago

Although the evidence is weak, there has been a concern in recent years over possible health effects from the magnetic fields ge

sergey [27]

Answer:

The magnetic field strength is $B = 2 \mu T$

Explanation:

From the question we are told that

The length line above the ground is $R = 20m$

The current of the line is $I = 200A$

The voltage of the line is $V = 110kV$

Generally magnetic field strength is mathematically represented as

$B = \frac{\mu_o I}{2 \pi R}$

Where $\mu_0$ is the permeability of free space $= 4\pi * 10^{-7} N/A^2$

$B = \frac{(4\pi * 10^{-7} N/A^2) *200}{2 \pi *20}$

$= (2.0*10^{-7})[\frac{200}{20} ]$

$= 2*10^{-6}T$

$= 2 \mu T$

Earths magnetic field is approximately given as $50 \mu T$

So the percentage would be

$= \frac{Magnetic \ Field \ Intensity \ Of \ Line}{Earth's \ Magnetic \ Field} * 100$

$= \frac{2 \mu T }{50 \mu T } * 100$

$=4$ %

7 0

2 years ago

Medical cyclotrons need efficient sources of protons to inject into their center. In one kind of ion source, hydrogen atoms (i.e

antoniya [11.8K]

Answer:

$r=5.278\times 10^{-4}\ m$

Explanation:

Given that:

magnetic field intensity, $B=0.07\ T$

kinetic energy of electron, $KE=1.2\ eV= 1.2\times 1.6\times 10^{-19}\ J= 1.92\times 10^{-19}\ J$

we have mass of electron, $m=9.1\times 10^{-31}\ kg$

<em>Now, form the mathematical expression of Kinetic Energy:</em>

$KE= \frac{1}{2} m.v^2$

$1.92\times 10^{-19}=0.5\times 9.1\times 10^{-31}\times v^2$

$v^2=4.2198\times 10^{11}$

$v=6.496\times 10^6\ m.s^{-1}$

<u>from the relation of magnetic and centripetal forces we have the radius as:</u>

$r=\frac{m.v}{q.B}$

$r=\frac{9.1\times 10^{-31}\times 6.496\times 10^6 }{1.6\times 10^{-19}\times 0.07}$

$r=5.278\times 10^{-4}\ m$

6 0

3 years ago

1. On this graph, what is the change in velocity between 5 seconds and 8 seconds?

alukav5142 [94]

Can I see the graph so I can help you

8 0

2 years ago

The drag force that resists the motion of a car traveling at 80 km h^- 1 is 300 N.

kobusy [5.1K]

The power require to keep the car traveling is 6,666 W.

The power of the engine at the given efficiency is 3,999.6 W.

<h3>What is Instantaneous power?</h3>

This the product of force and velocity of the given object.

The power require to keep the car traveling is calculated as follows;

P = Fv

$P = 300\ N \ \times \ \frac{80 \ kmh^{-1}}{3.6 \ km h^{-1}/m/s} \\\\
P = 300 \ N \times 22.22 \ m/s\\\\
P = 6,666 \ W$

The power of the engine at the given efficiency is calculated as follows;

$E = \frac{P_{out}}{P _{in}} \times 100\%\\\\
60\% = \frac{P_{out}}{6,666} \times 100\%\\\\
0.6 = \frac{P_{out}}{6,666} \\\\
P_{out} = 3,999.6 \ W$

Learn more about efficiency here: brainly.com/question/15418098

8 0

2 years ago