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astra-53 [7]
2 years ago
9

The record for the highest speed achieved in alaboratory for a

Physics
1 answer:
Ira Lisetskai [31]2 years ago
3 0

Answer:

0.000234 seconds

Explanation:

Since the row is 0.15m, its radius of rotation must be 0.15 / 2 = 0.075 m

We can start by calculating the angular speed of the rod:

\omega = \frac{v}{r} = \frac{2010}{0.075} = 26800 rad/s

Since one revolution equals to 2π rad. The speed in revolution per second must be

26800 / 2π = 4265 revolution/s

The number of seconds per revolution, or period, is the inverse:

1/4265 = 0.000234 seconds

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Waves of wavelength 24 cm and amplitude 3.0 em are used to create a standing wave. The distance from one node tothe next in the
NikAS [45]

Answer:

D. 12 cm

Explanation:

A node is a point on a standing wave that does not vibrate.

The nodes of a standing wave are shown in the following sketch.

The red dots are the nodes of the standing wave.

It is observed that the distance between two adjacent nodes is half the wavelength of the wave.

Therefore, if the wavelength of the wave is 24 cm, then the distance from one node to the net must be 24 / 2 = 12 cm.

Hence, choice D is the correct answer.

7 0
1 year ago
Help me please on this
Norma-Jean [14]
So what u do is 2112393921010
6 0
2 years ago
A. 25%<br> B. 20%<br> C. 10%<br> D. 80%<br> please show work :)
Setler79 [48]

Substitute your values into the formula:

W = Work done = 288

Q_{in} = 360

Solve to find e:

e = 288 ÷ 360 = 0.8

Convert e to a percentage by multiplying by 100.

0.8 × 100 = 80

<h2>D. 80%</h2>
4 0
3 years ago
Question 4 of 10
avanturin [10]

Answer:

C. Cables transmit infrared waves over longer distances.

Explanation:

A.pex

5 0
2 years ago
Read 2 more answers
A machine in an ice factory is capable of exerting 3.31 × 102 N of force to pull large blocks of ice up a slope. The blocks each
tankabanditka [31]

Answer:\theta =1.43 ^{\circ}

Explanation:

Given

Force exerted by machine F=3.31 \times 10^2 N

Weight of Block W=mg=1.32\times 10^4 N

Considering inclination to be \theta

After resolving Forces We get

mg\sin \theta =F

\sin \theta =\frac{F}{mg}

\sin \theta =\frac{3.31 \times 10^2}{1.32\times 10^4}

\sin \theta =2.507\times 10^{-2}

\sin \theta =0.02507

\theta =\sin ^{-1}(0.02507)

\theta =1.43 ^{\circ}

7 0
3 years ago
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