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andriy [413]
3 years ago
5

A turntable reaches an angular speed of "45 rev/min" in "4.10 s" after being turned on. What is its angular acceleration?

Physics
1 answer:
Marina86 [1]3 years ago
3 0

Answer:

1.15 rad/s²

Explanation:

given,

angular speed of turntable = 45 rpm

                       =45\times \dfrac{2\pi}{60}

                       =4.71\ rad/s

time, t = 4.10 s

initial angular speed = 0 rad/s

angular acceleration.

\alpha = \dfrac{\omega_f-\omega_0}{t}

\alpha = \dfrac{4.71-0}{4.10}

\alpha = 1.15\ rad/s^2

Hence, the angular acceleration of the turntable is 1.15 rad/s²

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You are tasked with calibrating the springs for a pinball machine. Your method of testing the springs is by attaching masses ont
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Answer:

490.5\ \text{N/m}

Explanation:

m = Mass attached to spring = 14 kg

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

x = Displacement of spring = 78-50=28\ \text{cm}

k = Spring constant

The force balance of the system is given by

kx=mg\\\Rightarrow k=\dfrac{mg}{x}\\\Rightarrow k=\dfrac{14\times 9.81}{0.28}\\\Rightarrow k=490.5\ \text{N/m}

The spring constant for that spring is 490.5\ \text{N/m}.

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3 years ago
si comparamos la gota de agua cayendo en una gotera, y el sonido del agua hirviendo,ambos con la misma intensidad ¿cuál sonido t
mrs_skeptik [129]

Answer:

ambas

Explanation:

Tendrían el mismo sonido ya que la intensidad es la misma.

8 0
3 years ago
The colour of star depend on its temperature, why?​
taurus [48]

<em>Another key factor that determines a star's colour is its temperature. As stars become hotter, the overall radiated energy increases, and the peak of the curve changes to shorter wavelengths. To put it another way, when a star heats up, the light it produces moves toward the blue end of the spectrum.</em>

4 0
2 years ago
Electricity costs 6 cents per kilowatt hour. In one month one home uses one megawatt hour of electricity. How much will the elec
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Answer:

Cost of 1000  kilowatt hour = 6000 cents

Explanation:

Given that

Electricity cost is 6 cents per kilowatt hour.

And we have to found out the cost for one megawatt hour

We know that

1 kilowatt = 1000 watt

1 megawatt = = 1000000 watt

1 megawatt = 1000 kilowatt

1 megawatt hour = 1000 kilowatt hour  

Given that cost of 1 kilowatt hour = 6 cents

So the cost of 1000  kilowatt hour = 6 x 1000 cents

Cost of 1000  kilowatt hour = 6000 cents

3 0
3 years ago
A microwave oven operating at 1.22 × 108 nm is used to heat 165 mL of water (roughly the volume of a teacup) from 23.0°C to 100.
ANTONII [103]

<u>Answer:</u> The number of photons are 3.7\times 10^8

<u>Explanation:</u>

We are given:

Wavelength of microwave = 1.22\times 10^8nm=0.122m    (Conversion factor:  1m=10^9nm  )

  • To calculate the energy of one photon, we use Planck's equation, which is:

E=\frac{hc}{\lambda}

where,

h = Planck's constant = 6.625\times 10^{-34}J.s

c = speed of light = 3\times 10^8m/s

\lambda = wavelength = 0.122 m

Putting values in above equation, we get:

E=\frac{6.625\times 10^{-34}J.s\times 3\times 10^8m/s}{0.122m}\\\\E=1.63\times 10^{-24}J

Now, calculating the energy of the photon with 88.3 % efficiency, we get:

E=1.63\times 10^{-24}\times \frac{88.3}{100}=1.44\times 10^{-24}J

  • To calculate the mass of water, we use the equation:

Density=\frac{Mass}{Volume}

Density of water = 1 g/mL

Volume of water = 165 mL

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{165mL}\\\\\text{Mass of water}=165g

  • To calculate the amount of energy of photons to raise the temperature from 23°C to 100°C, we use the equation:

q=mc\Delta T

where,

m = mass of water = 165 g

c = specific heat capacity of water = 4.184 J/g.°C

\Delta T = change in temperature = T_2-T_1=100^oC-23^oC=77^oC

Putting values in above equation, we get:

q=165g\times 4.184J/g.^oC\times 77^oC\\\\q=53157.72J

This energy is the amount of energy for 'n' number of photons.

  • To calculate the number of photons, we divide the total energy by energy of one photon, we get:

n=\frac{q}{E}

q = 53127.72 J

E = 1.44\times 10^{-24}J

Putting values in above equation, we get:

n=\frac{53157.72J}{1.44\times 10^{-24}J}=3.7\times 10^{28}

Hence, the number of photons are 3.7\times 10^8

4 0
3 years ago
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