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3 years ago

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Which of the following is not a use of electromagnetic waves? a night-vision bremote controls c x-rays d all are used of electro

magnetic wavesbremote controls is not a use of electromagnetic waves?

1 answer:

sergey [27]3 years ago

4 0

Use of electromagnetic because it moves very faster than others for example xrays theynar very very slow so that not It it is d.

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A string with a length of 4.00 m is held under a constant tension. The string has a linear mass density of \mu=0.000600~\text{kg

yulyashka [42]

Answer:

$T=245.76N$

Explanation:

We know that the frequency of the nth harmonic is given by $f_n=nf$ , where $f$ is the fundamental harmonic. Since we have the values of two consecutive frequencies, we can do:

$f_{n+1}-f_n=(n+1)f-nf=nf+f-nf=f$

Which for our values means (we do not need the value of <em>n</em>, that is, which harmonics are the frequencies given):

$f=f_{n+1}-f_n=480Hz-400Hz=80Hz$

Now we turn to the formula for the vibration frequency of a string (for the fundamental harmonic):

$f=\frac{1}{2L} \sqrt{\frac{T}{\mu}}$

So the tension is:

$T=\mu(2Lf)^2$

Which for our values is:

$T=(0.0006kg/m)(2(4m)(80Hz))^2=245.76N$

6 0

2 years ago

ow long must a simple pendulum be if it is to make exactly ten swings per second? (That is, one complete vibration takes exactly

Igoryamba

The period T of a pendulum is given by:
$T=2 \pi \sqrt{ \frac{L}{g} }$
where L is the length of the pendulum while $g=9.81 m/s^2$ is the gravitational acceleration.

In the pendulum of the problem, one complete vibration takes exactly 0.200 s, this means its period is $T=0.200 s$ . Using this data, we can solve the previous formula to find L:
$L=g ( \frac{T}{2\pi} )^2=(9.81 m/s^2)( \frac{0.2 s}{2 \pi} )^2=1 \cdot 10^{-3} m=1 mm$

4 0

2 years ago

Choose true or false for each statement regarding a parallel plate capacitor.. The voltage of a disconnected charged capacitor i

OlgaM077 [116]

Answer:

Explanation:

The voltage of a disconnected charged capacitor increases when the plate area is decreased.

When plate area decreases , capacitance C decreases , but charge Q remains constant .

Q = C V where C is capacitance and V is voltage .

when C decreases , V increases for keeping Q constant .

So the statement is true.

The electric field is dependent on the charge density on the plates.

This statement is true .

The voltage of a connected charged capacitor remains the same when the plate area is decreased .

For a connected capacitor , V or voltage is constant which is equal to voltage of charging battery .

So the statement is true .

3 0

3 years ago

A 50.6 g ball of copper has a net charge of 1.6 µc. what fraction of the copper's electrons have been removed? (each copper atom

Fynjy0 [20]

First figure out how many atoms you have with Avogadro's number. Since there are 63.5 grams/mol and you have 50.6 grams, you have (50.6/63.5)6.022E23=4.7986E23 atoms. Since there are 29 protons per atom, there are also 29 electrons per atom, so you should have a total of
29*4.7986E23=1.3916E25 electrons.
Since there is a positive charge you know some of these electrons are missing. How many are missing can be found by dividing the charge you have by the charge on the electron: 1.6E-6/1.6022E-19 = 9.98627E12 electrons are missing.
Now take the ratio of what is missing to what there should be:
9.98627E12/1.3916E25 = 7.1760873E-13

5 0

2 years ago

A sound wave has a frequency of 247 hz and a wavelength of 1.4 m. what is the speed of the sound wave in air?

Pavel [41]

Speed = wavelength * Frequency
s = 247 /s * 1.4 m
s = 345.8 m/s

In short, Your Answer would be 345.8 m/s

Hope this helps!

7 0

3 years ago