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3 years ago

Crests of an ocean wave pass a pier every 10.0 s. if the waves are moving at 5.6 m/s, what is the wavelength of the ocean waves?

1 answer:

4 0

Every 10.0 seconds, a crest of the wave passes the pier. This means that the period of the wave is exactly 10.0 s:
$T=10.0 s$
which means that the frequency of the wave is
$f= \frac{1}{T}= \frac{1}{10.0 s}=0.1 Hz$

The wavelength of a wave is related to its frequency by the relationship
$\lambda= \frac{v}{f}$
where v is the speed of the wave.
In this problem, v=5.6 m/s; if we use the previous formula, we find the wavelength of the wave:
$\lambda= \frac{5.6 m/s}{0.1 Hz}=56 m$

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Jonathan answered 28 out of 35 questions correctly on his chemistry test. What is his percent score?

pav-90 [236]

Seriously? It’s that hard to use a calculator??

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3 years ago

.89m/s A vehicle that starts to move from the rest gets an acceleration of 5m/s² within 2 seconds. Calculate the velocity and di

GuDViN [60]

<u>Explanation:</u>

s = ? u = 0m/s v = ? a = 5m/s² t = 2s

v = u + at

= 0 + (5 x 2)

= 10 m/s

s = ut + 1/2 at²

= (0)(2) + x $\frac{1}{2}$ x 5 x 2²

= 10 m

Hope this helps!

5 0

1 year ago

. A magnetic field has a magnitude of 0.078 T and is uniform over a circular surface whose radius is 0.10 m. The field is orient

Dmitriy789 [7]

Answer:

The magnetic flux through surface is $2.22 \times 10^{-3}$ Wb

Explanation:

Given :

Magnitude of magnetic field $B = 0.078$ T

Radius of circle $r = 0.10$ m

Angle between field and surface normal $\theta =$ 25°

From the formula of flux,

$\phi = B.A$

$\phi = BA\cos \theta$

Where $\theta =$ angle between magnetic field line and surface normal, $A =$ area of circular surface.

$A = \pi r^{2}$

$A = 3.14 \times (0.10) ^{2}$

$A = 0.0314$ $m^{2}$

Magnetic flux is given by,

$\phi = 0.078 \times 0.0314 \times \cos 25$

$\phi = 2.22 \times 10^{-3}$ Wb

Therefore, the magnetic flux through surface is $2.22 \times 10^{-3}$ Wb

6 0

3 years ago

Consider a hydrogen atom in the n = 1 state. The atom is placed in a uniform B field of magnitude 2.5 T. Calculate the energy di

dlinn [17]

Answer:

$E=29\times 10^{-5}eV$

Explanation:

For n-=1 state hydrogen energy level is split into three componets in the presence of external magnetic field. The energies are,

$E^{+}=E+\mu B$ ,

$E^{-}=E-\mu B$ ,

$E^{0}=E$

Here, E is the energy in the absence of electric field.

And

$E^{+} and E^{-}$ are the highest and the lowest energies.

The difference of these energies

$\Delta E=2\mu B$

$\mu=9.3\times 10^{-24}J/T$ is known as Bohr's magneton.

B=2.5 T,

Therefore,

$\Delta E=2(9.3\times 10^{-24}J/T)\times 2.5 T\\\Delta E=46.5\times 10^{-24}J$

Now,

$Delta E=46.5\times 10^{-24}J(\frac{1eV}{1.6\times 10^{-9}J } )\\Delta E=29.05\times 10^{-5}eV\\Delta E\simeq29\times 10^{-5}eV$

Therefore, the energy difference between highest and lowest energy levels in presence of magnetic field is $E=29\times 10^{-5}eV$

6 0

3 years ago

A 5.00-g bullet is fired into a 900-g block of wood suspended as a ballistic pendulum. The combined mass swings up to a height o

Thepotemich [5.8K]

Answer:

The value is $KE_b =0.710 \ J$

Explanation:

From the question we are told that

The mass of the bullet is $m_b = 5.00 \ g = 0.005 \ kg$

The mass of the wood is $m_w = 900 \ g = 0.90\ kg$

The height attained by the combined mass is $h = 8.0 \ cm = 0.08 \ m$

Generally according to the law of energy conservation

$KE _b = PE_c$

Here $KE_b$ is the kinetic energy of the bullet before collision.

and $PE_c$ is the potential energy of the combined mass of bullet and wood at the height h which is mathematically represented as

$PE_m = [m_b + m_w] * g * h$

$KE_b =PE_c = [0.005 + 0.90] * 9.8 *0.08$

=> $KE_b =0.710 \ J$

3 0

3 years ago