Mass is directly or indirectly related to inertia?

Which of the following are common sorces of radio waves ?

alexgriva [62]

It would help if there are some options, but some common sources:
Televisions
Microwave Ovens
Cell Phones
Satellites
Wired Computers
CD players
etc.

5 0

3 years ago

Someone let me know if I got these 2 questions right or not pls

Alik [6]

Answer:

both are correct .... ....

5 0

3 years ago

Read 2 more answers

A driver, traveling at 22.0 m/s, slows down her 2.00 x 103 kg car to stop for a red light. What work is done by the friction for

Artemon [7]

Some work will be done on friction between wheels and road but it is negligible compared to work done on friction on breaks.

W = Ek = (m*v^2)/2 = 2000*22^2/2 = 1000*22^2 = 484KJ

Because car is not changing its potential energy, there is no work to be done on while changing it which means that all goes on changing kinetic energy (energy of motion)

6 0

4 years ago

What happens when two continental plates meet ?

dalvyx [7]

Answer:

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Explanation:

8 0

3 years ago

Read 2 more answers

A string with a mass density of 3 * 10^-3 kg/m is under a tension of 380 N and is fixed at both ends. One of its resonance frequ

Delvig [45]

Answer:

(a) the fundamental frequency of this string is 65 Hz

(b) the harmonics of the given frequencies are third and fourth respectively.

(c) the length of the string is 2.74 m

Explanation:

Given;

mass density of the string, μ = 3 x 10⁻³ kg/m

tension of the string, T = 380 N

resonating frequencies, 195 Hz and 260 N

For the given resonant frequencies;

$195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } ---(1)\\\\260 = \frac{n+1}{2l} \sqrt{\frac{T}{\mu} } ---(2)\\\\divide \ (2) \ by (1)\\\\\frac{260}{195} = \frac{n+1 }{n} \\\\260n = 195(n+1)\\\\260 n = 195 n + 195\\\\260n - 195n = 195\\\\65n = 195\\\\n = \frac{195}{65} \\\\n = 3$

(c) From any of the equations, solve for Length of the string (L);

$195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } \\\\195 = \frac{3}{2l}\sqrt{\frac{380}{3\times 10^{-3}} } \\\\l = \frac{3}{2\times 195}\sqrt{\frac{380}{3\times 10^{-3}} }\\\\l = 2.74 \ m$

(a) the fundamental frequency is calculated as;

$f_o = \frac{1}{2l} \sqrt{\frac{T}{\mu} } \\\\f_o = \frac{1}{2\times 2.74} \sqrt{\frac{380}{3\times 10^{-3} } }\\\\f_o = 65 \ Hz$

(b) harmonics of the given frequencies;

the first harmonic (n = 1) = f₀ = 65 Hz

the second harmonic (n = 2) = 2f₀ = 130 Hz

the third harmonic (n = 3) = 3f₀ = 195 Hz

the fourth harmonic (n = 4) = 4f₀ = 260 Hz

Thus, the harmonics of the given frequencies are third and fourth respectively.

7 0

3 years ago