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3 years ago

Can you make sound you can see?

Physics

1 answer:

navik [9.2K]3 years ago

5 0

No, sound waves are invisible to our eyes

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47. the beam is supported by two rods ab and cd that have cross-sectional areas of 12mm^2 and 8mm^2, respectively. determine the

Ugo [173]

Let the beam is of length L

Now the stress on both the end is same

now we can say that torque on the beam due to two forces must be zero

$N_1* x = N_2* (L - x)$

also we know that stress at both ends are same

$\frac{N_1}{12} = \frac{N_2}{8}$

$2*N_1 = 3*N_2$

Now from two equations we have

$\frac{3}{2}N_2*x = N_2* (L - x)$

solving above equation we have

$x = \frac{2}{5}L$

<em>so the load is placed at distance 0.4L from the end of 12 mm^2 area</em>

8 0

3 years ago

Please help and if you get this and answer the other one on my thing i’ll give u brainliest

quester [9]

Answer:

The higher the energy, the shorter the wavelength.

Explanation:

3 0

3 years ago

A radioactive element undergoes decay via the loss of two alpha particles to form a stable daughter isotope. Following the decay

jeyben [28]

Answer:

Atomic number of the daughter isotope = (Atomic number of the parent isotope) - 4

Explanation:

An alpha particle has atomic number of 2 and a mass number of 4

Losing two alpha particles would lead to a reduction of 4 in atomic number and 8 in mass number for the radioactive isotope, to give the stable daughter isotope.

8 0

3 years ago

A magnetic field directed along the x-axis changes with time according to B (0.06t2+2.25) T, where t is in seconds. The field is

Gemiola [76]

Answer:

$E = 2 \times 10^{-3} V$

Explanation:

As we know that rate of change in flux will induce EMF

So here we can

$EMF = \frac{d\phi}{dt}$

now we have

$EMF = \pi r^2\frac{dB}{dt}$

now we also know that induced EMF is given by

$\int E. dL = \pi r^2\frac{dB}[dt}$

$E (2\pi r) = \pi r^2\frac{dB}{dt}$

$E = \frac{r}{2}(\frac{dB}{dt})$

now plug in all values in it

$E = \frac{0.0133}{2}(0.12 t)$

$E = 8 \times 10^{-4} (2.50) = 2 \times 10^{-3} V/m$

5 0

4 years ago

Two tiny conducting spheres are identical and carry charges of -18.8 µC and +46.5 µC. They are separated by a distance of 2.47 c

sergiy2304 [10]

Answer:

$F=-12896N$ , attractive.

Explanation:

For calculating this force we use the Coulomb Law:

$F=\frac{kq_1q_1}{r^2}$

Where $k=9\times10^9Nm^2/C^{-2}$ is the Coulomb's constant, $q_1$ and $q_2$ the values of each charge and r the distance between them.

Since the Coulomb's constant as I wrote it is in S.I. we have to write all the magnitudes in that system of units, and substitute:

$F=\frac{(9\times10^9Nm^2/C^{-2})(-18.8\times10^{-6}C)(46.5\times10^{-6}C)}{(0.0247m)^2}=-12896N$

This force is attractive since both charges are of opposite sign.

3 0

3 years ago

Can you make sound you can see?​

Can you make sound you can see?