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3 years ago

Can anyone help?6th question.free brainliest answer...

Physics

1 answer:

DIA [1.3K]3 years ago

6 0

There's a crest and a trough in each complete wave. So the question is describing 10 complete waves.

After that, the question becomes somewhat murky. It goes on to say "its time period is 0.2 seconds".

-- The "time period" of a wave is usually defined as the time for one complete wave. If that's what the phrase means, then ...

Frequency = ( 1/0.2sec )

Frequency = 5 Hz.

= = = = = = = = = =

BUT ... Is the question awkwardly trying to tell us that the 10 waves take 0.2 seconds ? If that's what it's saying, then ...

Frequency = (10) / (0.2 sec)

Frequency = 50 Hz .

You might be interested in

Glass is transparent to visibile light under normal conditions; however, at extremely high intensities, glass will absorb most o

8_murik_8 [283]

Answer:

3 photons

Explanation:

The energy of a photon E can be calculated using this formula:

$E=\frac{hc}{\lambda}$

Where $h$ corresponds to Plank constant (6.626070x10^-34Js), $c$ is the speed of light in the vacuum (299792458m/s) and $\lambda$ is the wavelength of the photon(in this case 800nm).

$E=\frac{hc}{\lambda}=\frac{(6.626070\times10^{-34})(299792458)}{800\times10^{-9}}=\frac{1.986445812\times10^-25}{800}=2.483057265\times10^{-19}J$

Tranform the units

$1eV=1.602176634\times10^{-19}J\\2.483057265\times10^{-19}J(\frac{1eV}{1.602176634\times10^{-19}J})=1.549802445eV$

The band Gap is 4eV, divide the band gap between the energy of the photon:

$\frac{4ev}{1.549802445eV}=2.508974118$

Rounding to the next integrer: 3.

Three photons are the minimum to equal or exceed the band gap.

4 0

3 years ago

Hello i need help with this!!

mr Goodwill [35]

Answer:

Yes

Explanation:

If lamp A burnt out there would still be a wire above it that connects lamp B and C to the power source

3 0

1 year ago

An ideal refrigerator does 130. 0 j of work to remove 780. 0 j of heat from its cold compartment during each cycle. what is the

zheka24 [161]

The refrigerator's coefficient of performance is 6.

The heat extracted from the cold reservoir Q cold (i.e., inside a refrigerator) divided by the work W required to remove the heat is known as the coefficient of performance, or COP, of a refrigerator (i.e., the work done by the compressor). The required inside temperature and the outside temperature have a significant impact on the COP.

As the inside temperature of the refrigerator decreases, its coefficient of performance decreases. The coefficient of performance (COP) of refrigeration is always more than 1.

The heat produced in the cold compartment, H = 780.0 J

Work done in ideal refrigerator, W = 130.0 J

Refrigerator's coefficient of performance = H/W

= 780/130

= 6

Therefore, the refrigerator's coefficient of performance is 6.

Energy conservation requires the exhaust heat to be = 780 + 130

= 910 J

Learn more about coefficient here:

brainly.com/question/18915846

#SPJ4

5 0

1 year ago

A telephone line has a signal-to-noise ratio of 1000 and a bandwidth of 4 KHz. What is the maximum data rate supported by this l

ivolga24 [154]

Answer:

The maximum data rate supported by this line is 39900 bps

Explanation:

The maximum data rate supported by this line can be obtained using the formula below

c = W*log2(S/N+1)

where;

c is the maximum data rate supported by the line

W is the bandwidth = 4kHz

S/N+1 is the signal to noise ratio = 1001

c = 4*log2(1001)

c = 39868.9 ≅ 39900 bps

Therefore, the maximum data rate supported by this line is 39900 bps

5 0

3 years ago

(I) In a ballistic pendulum experiment, projectile 1 results in a maximum height h of the pendulum equal to 2.6 cm. A second pro

Kipish [7]

Answer:

The second projectile was 1.41 times faster than the first.

Explanation:

In the ballistic pendulum experiment, the speed (v) of the projectile is given by:

$v = \frac{m + M}{m} \cdot \sqrt{2gh}$

where m: is the mass of the projectile, M: is the mass of the pendulum, g: is the gravitational constant and h: is the maximum height of the pendulum.

To know how many times faster was the second projectile than the first, we need to take the ratio for the velocities for the projectiles 2 and 1:

$\frac{v_{2}}{v_{1}} = \frac{\frac{m_{2} + M}{m_{2}} \cdot \sqrt{2gh_{2}}}{\frac{m_{1} + M}{m_{1}} \cdot \sqrt{2gh_{1}}}$ (1)

where m₁ and m₂ are the masses of the projectiles 1 and 2, respectively, and h₁ and h₂ are the maximum height reached by the pendulum by the projectiles 1 and 2, respectively.

Since the projectile 1 has the same mass that the projectile 2, we can simplify equation (1):

$\frac{v_{2}}{v_{1}} = \frac{\sqrt{h_{2}}}{\sqrt{h_{1}}}$

$\frac{v_{2}}{v_{1}} = \frac{\sqrt{5.2 cm}}{\sqrt{2.6 cm}}$

$\frac{v_{2}}{v_{1}} = 1.41$

Therefore, the second projectile was 1.41 times faster than the first.

I hope it helps you!

8 0

2 years ago