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3 years ago

If a star’s radiation peaks in the ultraviolet region of the spectrum, what could you conclude?

Physics

2 answers:

Alex_Xolod [135]3 years ago

4 0

Answer:

It is a very intense star

Explanation:

As we know by Electromagnetic radiations the energy of spectrum in decreasing order is given as

1) Gamma Rays

2) X rays

3) Ultraviolet Rays

4) Visible Light

5) Infrared waves

6) Microwaves

7) Radio waves

So here we can say that ultraviolet rays are at peak intensity for the spectrum in the radiation which means major proportion of the spectrum is consisting this part of the electromagnetic waves

So here the star would be very intense due to large number of this spectrum.

PSYCHO15rus [73]3 years ago

3 0

The short answer is "it is a very intense star."

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A wave has a frequency of 15,500 Hz and a wavelength of 0.20 m. What is the

Hoochie [10]

Answer:

3100 m/s

Explanation:

The relationship between frequency and wavelength of a wave is given by the wave equation:

$v=f\lambda$

where

v is the speed of the wave

f is its frequency

$\lambda$ is the wavelength

For the wave in this problem,

f = 15,500 Hz

$\lambda=0.20 m$

Therefore, the wave speed is

$v=(15500)(0.20)=3100 m/s$

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3 years ago

Erica throws a tennis ball against a wall, and it bounces back. Which force is responsible for sending the ball back to Erica?

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Kinetic energy is responsible for this

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3 years ago

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a fan acquires a speed of 180 rpm in 4s, starting from rest. calculate the speed of the fan at the end of the 5th second startin

KengaRu [80]

Answer:

225 rpm

Explanation:

The angular acceleration of the fan is given by:

$\alpha = \frac{\omega_f - \omega_i}{\Delta t}$

where

$\omega_f$ is the final angular speed

$\omega_i$ is the initial angular speed

$\Delta t$ is the time interval

For the fan in this problem,

$\omega_i = 0\\\omega_f = 180 rpm\\\Delta t=4 s$

Substituting,

$\alpha = \frac{180-0}{4}=45 rpm/s$

Now we can find the angular speed of the fan at the end of the 5th second, so after t = 5 s. It is given by:

$\omega' = \omega_i + \alpha t$

where

$\omega_i = 0\\\alpha = 45 rpm/s\\t = 5 s$

Substituting,

$\omega' = 0 + (45)(5)=225 rpm$

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3 years ago

The temperature of a smelting furnace is found to be 2000 degree Celsius.find the temperature on Fahrenheit scale

Dima020 [189]

Answer:

3632.

Explanation:

Based on the formula:

2000×(9/5)+32=3632

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2 years ago

A small block with a mass of 0.0600 kg is attached to a cord passing through a hole in a frictionless, horizontal surface (Figur

pogonyaev

Answer with Explanation:

We are given that mass of block=0.0600 kg

Initial speed of block=0.63 m/s

Distance of block from the hole when the block is revolved=0.47 m

Final speed=3.29 m/s

Distance of block from the hole when the block is revolved= $9\times 10^{-2}m$

a.We have to find the tension in the cord in the original situation when the block has speed = $v_0=0.63 m/s$

$T=\frac{mv^2}{r}$

Because tension is equal to centripetal force

Substitute the values

$T=\frac{0.06\times (0.63)^2}{0.47}=0.05 N$

b. $v=3.29 m/s$

$T=\frac{mv^2}{r}=\frac{0.06\times (3.29)^2}{0.09}=7.2 N$

c.Work don=Final K.E-Initial K.E

$W=\frac{1}{2}m(v^2-v^2_0)$

$W=\frac{1}{2}(0.06)((3.29)^2-(0.63)^2)$

$W=0.31 J$

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3 years ago

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