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Gelneren [198K]
3 years ago
13

Need help on all three pizza?

Physics
1 answer:
ahrayia [7]3 years ago
7 0
1. 12m/s and here’s how. The equation for calculating velocity of a wave is lambda x frequency. So 2m x 6Hz. Hz can be 1/s so the only unit of measurement we get is meters and seconds. So the answer will be 12m/s.

2. Frequency=2Hz period= 0.5 seconds
Equation for frequency is velocity/wavelength. 10m/s divides by 5m = 2Hz.
Equation for period is 1 / frequency. 1 divided by 2 = 0.5 seconds

3. The answer is the picture of you can’t read my hand writing comment and say “ I need the info about the wave because I can’t read your hand writing “ thank you and I hoped I helped

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An object of irregular shape has a characteristic length of L = 0.5 m and is maintained at a uniform surface temperature of Ts =
goblinko [34]

Answer:

The value of the average convection coefficient is 20 W/Km².

Explanation:

Given that,

For first object,

Characteristic length = 0.5 m

Surface temperature = 400 K

Atmospheric temperature = 300 K

Velocity = 25 m/s

Air velocity = 5 m/s

Characteristic length of second object = 2.5 m

We have same shape and density of both objects so the reynold number will be same,

We need to calculate the value of the average convection coefficient

Using formula of  reynold number for both objects

R_{1}=R_{2}

\dfrac{u_{1}L_{1}}{\eta_{1}}=\dfrac{u_{2}L_{2}}{\eta_{2}}

\dfrac{h_{1}L_{1}}{k_{1}}=\dfrac{h_{2}L_{2}}{k_{2}}

Here, k_{1}=k_{2}

h_{2}=h_{1}\times\dfrac{L_{1}}{L_{2}}

h_{2}=\dfrac{q}{T_{2}-T_{1}}\times\dfrac{L_{1}}{L_{2}}

Put the value into the formula

h_{2}=\dfrac{10000}{400-300}\times\dfrac{0.5}{2.5}

h_{2}=20\ W/Km^2

Hence, The value of the average convection coefficient is 20 W/Km².

7 0
4 years ago
Why are coal,nuclear,oil,and natural gas determined to be non-renewable types of energy resources ​
IrinaVladis [17]

Answer:

Once used as a energy source they cannot be charged/used again.

Explanation:

7 0
3 years ago
A roller coaster car of mass m= 300 kg is released from rest at the top of a 60 m high hill (position A), and rolls with a negli
Andrews [41]

Answer: The principle of conservation of energy, angular speed and centripetal force

Explanation:

At point A, the car experienced maximum of potential energy

As it moves down the hill, the potential energy decreases while the kinetic energy increases.

The maximum kinetic energy of the car is needed for the attainment of enough centripetal force to help the car move through the loop without falling .

4 0
3 years ago
3. A ray of light consisting of blue light (wavelength 480 nm) and red light (wavelength 670 nm) is incident on a thick piece of
Alex Ar [27]

Answer:

The angular separation between the refracted red and refracted blue beams while they are in the glass is 42.555 - 42.283 = 0.272 degrees.

Explanation:

Given that,

The respective indices of refraction for the blue light and the red light are 1.4636 and 1.4561.

A ray of light consisting of blue light (wavelength 480 nm) and red light (wavelength 670 nm) is incident on a thick piece of glass at 80 degrees.

We need to find the angular separation between the refracted red and refracted blue beams while they are in the glass.

Using Snell's law for red light as :

n_1\sin\theta_1=n_2\sin\theta_2\\\\\theta_2=\sin^{-1}((\dfrac{n_2}{n_1})\sin\theta_1)\\\\\theta_2=\sin^{-1}((\dfrac{1}{1.4561})\sin(80))\\\\\theta_2=42.555

Again using Snell's law for blue light as :

n_1\sin\theta_1=n_2\sin\theta'_2\\\\\theta'_2=\sin^{-1}((\dfrac{n_2}{n_1})\sin\theta_1)\\\\\theta'_2=\sin^{-1}((\dfrac{1}{1.4636 })\sin(80))\\\\\theta'_2=42.283

The angular separation between the refracted red and refracted blue beams while they are in the glass is 42.555 - 42.283 = 0.272 degrees.

7 0
4 years ago
The images output from your new color laser printer seem to be a little too blue. What can you do to fix this?
Svetradugi [14.3K]

Answer:

The images output from your new color laser printer seem to be a little too blue. to fix this problem we need to calibrate the printer.

Explanation:

This can be done by opening the toolbox, clicking in the device setting folder their you get print quality page click on it. Under the print quality option click on the calibrate next to calibrate now. Then click OK unless when the 'your request has been sent to the device' appears on the screen. When the calibration ends again try printing. calibrating is useful for managing the proper alignment of the inkjet cartridge nozzle to the paper and each other, without proper calibration the print quality deteriorates.

3 0
3 years ago
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