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ruslelena [56]
3 years ago
12

What are the frequencies of EM spectrum

Physics
1 answer:
kap26 [50]3 years ago
3 0

The EM spectrum has no limits.  Any frequency you can imagine
is the frequency of some electromagnetic radiation somewhere.


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Please answer the following question
aleksklad [387]

Answer:

1. Light Bulb

2. A switch

3. A battery

4. Ammeter

5. Resistor

6. Voltmeter

4 0
3 years ago
Change in speed over a given period of time is
Dvinal [7]

Explanation:

Acceleration is the change in speed over a given time period

8 0
3 years ago
Read 2 more answers
Imagine a system where a block rests on an inclined plane. The block is then given an initial push so that it starts sliding dow
Helen [10]

Answer:

statement - 'The work done by friction is equal to the sum of the work done by the gravity and the initial push' is correct.

Explanation:

The statement ''The work done by friction is equal to the sum of the work done by the gravity and the initial push" is correct.

The above statement is correct because, the initial push will tend to slide down the block thus the work done by the initial push will be in the downward direction. Also, the gravity always acts in the downward direction. thus, the work done done by the gravity will also be in the downward direction

here, the downward direction signifies the downward motion parallel to the inclined plane.

Now we know that the work done by the friction is against the direction of motion. Thus, the friction force will tend to move the block up parallel to the inclined plane.

Hence, for the block to stop sliding the the above statement should be true.

6 0
3 years ago
The distance between adjacent nodes in a standing wave pattern in a length of string is 25.0 cm:A. What is the wavelength of wav
mina [271]

A) 50 cm

B) 10000 cm/s

Explanation

Step 1

A)

If you know the distance between nodes and antinodes then use this equation:

\begin{gathered} \frac{\lambda}{2}=D \\ \text{where}\lambda\text{ is the wavelength} \\ D\text{ is the distance betw}een\text{ nodes} \end{gathered}

then, let

D=\text{ 25 cm }

now, replace to find the wavelength

\begin{gathered} \frac{\lambda}{2}=25 \\ \text{Multiply both sides by 2} \\ \frac{\lambda}{2}\cdot2=25\cdot2 \\ \lambda=50\text{ Cm} \end{gathered}

so, the wavelength is

A) 50 cm

Step 2

The speed of a wave can be found using the equation

v=\lambda f

or velocity = wavelength x frequency,

then,let

\begin{gathered} \lambda=50\text{ cm} \\ f=200\text{ Hz} \end{gathered}

replace and evaluate

\begin{gathered} v=\lambda f \\ v=50\text{ cm }\cdot200\text{ HZ} \\ v=10000\text{ }\frac{\text{cm}}{s} \end{gathered}

so

B) 10000 cm/s

I hope this helps you

6 0
1 year ago
Assume the following values: d1 = 0.880 m , d2 = 1.11 m , d3 = 0.560 m , d4 = 2.08 m , F1 = 510 N , F2 = 306 N , F3 = 501 N , F4
dsp73

Answer:

= 2630.6 N.m

Explanation:

(FR)x = ΣFx = -F4 = -407 N

(FR)y = ΣFy =-F1-F2 -F3 = -510 - 306 - 501 = -1317 N

(MR)B =ΣM + Σ(±Fd)

= MA + F1(d1 +d2) + F2d2 - F4d3

= 1504 + 510(0.880+1.11) +306(1.11) - 407(0.560)

= 2630.64 N.m (counterclockwise)

6 0
3 years ago
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