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vladimir1956 [14]

3 years ago

15

A wave of infrared light has a speed of 6m/s and a wave length of 12 m. What is the frequency of this wave?

1 answer:

erastovalidia [21]3 years ago

6 0

Frequency = speed / wavelength

(6 m/s) / (12 m) = 0.5 Hz.

That's not infrared light.
Infrared light waves move about 50 million times faster than that, and they're only about 0.00000007 as long as that.

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A metal disk of radius 6.0 cm is mounted on a frictionless axle. Current can flow through the axle out along the disk, to a slid

Galina-37 [17]

Answer:

0.09 N

Explanation:

We are given that

Radius of disk,r=6 cm= $\frac{6}{100}=0.06 m$

1 m=100 cm

B=1 T

Current,I=3 A

We have to find the frictional force at the rim between the stationary electrical contact and the rotating rim.

$dF=IBdr$

$dF=IBdr$

$\tau=rdF=IBrdr$

$\tau=\int_{0}^{R}IBr dr$

$\tau=IB(\frac{R^2}{2}$

Torque due to friction

$\tau=R\times F$

Where friction force=F

$R\times F=\frac{IBR^2}{2}$

$F=\frac{IBR}{2}$

Substitute the values

$F=\frac{3\times 1\times 0.06}{2}$

$F=0.09 N$

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2 years ago

The engine in your car is sometimes called: A. A 2-stroke engine

Juli2301 [7.4K]

Answer:A

Explanation:

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There are process in 4 stroke engine

Intake: Intake of air
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2 years ago

Dams hold collected water back from flowing at their normal rate. What type of energy is in the collected water above the dam? c

Andreyy89

It would be mechanical potential energy.

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2 years ago

Read 2 more answers

A 80 W light bulb (normally run at 120 V) is attached to a transformer. The voltage source in the transformer is 65 V and Np = 3

Marina CMI [18]

67.8 turns needed by the secondary coil to run the bulb.

<u>Explanation</u>:

We know that,

$\text { Electric power }(p)=\frac{V^{2}}{R}$

$\text { Hence, } \frac{P_{1}}{P_{2}}=\frac{V_{1}^{2} / R}{V_{2}^{2} / R}$

$\frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}$

For calculating number of turns

$\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}$

Given that,

$80 \mathrm{W}\left(P_{1}\right) \text { bulb with voltage } 120 \mathrm{V}\left(V_{1}\right) \text { is connected to a transformer. }$

$\text { The source voltage of a transformer is }\left(V_{P}\right) \text { is } 65 \mathrm{V}$

$\text { The number of turns in primary winding of transformer is }\left(N_{P}\right) \text { is } 30 .$

We need to find the number of turns in the secondary winding $\left(N_{S}\right)$ to run the bulb at 120W $\left(P_{2}\right)$

Firstly find the secondary voltage in the transformer use, $\frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}$

$\frac{80}{120}=\frac{120^{2}}{V_{2}^{2}}$

$V_{2}^{2}=\frac{120^{2} \times 120}{80}$

$V_{2}^{2}=\frac{1728000}{80}$

$V_{2}^{2}=21600$

$V_{2}=\sqrt{21600}$

$V_{2}=146.9 \mathrm{V}=V_{S}$

Now, finding the number of turns in secondary coil. Use, $\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}$

$\frac{30}{N_{S}}=\frac{65}{146.9}$

$N_{S}=\frac{30 \times 146.9}{65}$

$N_{S}=\frac{4407}{65}$ $N_{S}=67.8$

The number of turns in the secondary winding are 67.8 turns.

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Answer:

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Explanation:

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2 years ago