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3 years ago

A wave has an amplitude of 0.2 cm.

1 answer:

3 0

The distance between the resting point and maximum height of the wave is 0.2 cm.

The amplitude is measured from the resting point up to the highest point of the wave.

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A 15.0-kg block is dragged over a rough, horizontal surface by a 70.0-N force acting at 20.0° above the horizontal. The block is

Tatiana [17]

Answer:

Explanation:

from the question we are told that

Load $L=15kg$

Force $F=70N$

Angle of inclination $=20.0 degres$

Displacement $m=5 meters$

coefficient of kinetic friction $\alpha =0.300$

3 0

3 years ago

What instrument can you use to measure the circumference of your waist?

wolverine [178]

<h2><em><u>Answer</u></em>:</h2>

<h3>A <u>measuring tape</u> can be used to measure the circumference of your waist.</h3>

3 0

4 years ago

Read 2 more answers

Th answer and how to do it (please do it on paper and send a picture if you can )

Umnica [9.8K]

If series Rs = R1 + R2

8 0

3 years ago

consider the photoelectric effect experiment. in one experiment yellow light shines on a piece of potassium metal. a current is

kompoz [17]

Explanation:

If the intensity of the yellow light increased, meaning more photons will strike the Potassium metal per unit area. This will cause more ejection of electrons from the metal and hence, the strength of current will also increase as we know that

I = Q/t, as the charge increase , the current will also increase.

4 0

3 years ago

An electric fan is turned off, and its angular velocity decreases uniformly from 550 rev/min to 180 rev/min in a time interval o

Sergio039 [100]

Answer:

(a) $\alpha =-1.43rev/sec^2$

(b) 4.17 rev

Explanation:

We have given the angular velocity of fan decreases from 550 rev/min to 180 rev/min

So initial angular speed $\omega _0=550rev/min=\frac{550}{60}=9.166rev/sec$

Final angular speed $\omega =180rev/min=\frac{180}{60}=3rev/sec$

Time is given as t = 4.3 sec

(a) We know that angular acceleration is given by $\alpha =\frac{\omega -\omega _0}{t}=\frac{3-9.166}{4.3}=-1.43rev/sec^2$

(b) We know the relation \Theta =\omega _0t+\frac{1}{2}\alpha t^2=9.166\times 4.3-\frac{1}{2}\times 1.43\times 4.3^2=26.1935rad=\frac{26.1935}{2\pi }=4.17rev

So time $t=\frac{0-9.166}{-1.43}=6.40sec$

5 0

3 years ago