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erastovalidia [21]
3 years ago
14

Why would a flare be observed in visible light, when they are so much brighter in X-ray and ultraviolet light?

Physics
1 answer:
raketka [301]3 years ago
3 0

Answer:

Because it radiates a spectre of frequencies, this means that it radiates in a continuous range of frequencies, ones with more intensity (like x-ray and ultraviolet) and others with less intensity (like the visible light). While most of the radiation is not in the visible range, there still is a part of the radiation in the visible spectre. And while this part is not the most intense part, the radiation is so large that we can see a very bright visible light.

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Jim runs off a diving board and lands in the water 3 meters from the end of the board. When he runs at the same speed on a highe
satela [25.4K]

Answer:

C.

Explanation:

If we look at the equation x(final) = v(initial)*t + x(initial), where v(initial in the x dir.) in the same both times, and x(initial) is 0, we can conclude that t is responsible for the change in x(final).

4 0
4 years ago
The latent heat of vaporization of H₂O at body temperature (37°C) is 2.42 x 10⁶ J/kg. To cool the body of a 60.4-kg jogger [aver
Darya [45]

Answer:

<h2>0.094 kg</h2>

Explanation:

        Latent heat of vaporization of H_{2}O at 37°C is 2.42\times10^{6}\text{ }\frac{J}{kg}.

        When the sweat on our body evaporates, it absorbs energy from our body to overcome it's Latent heat of vaporisation. Thus our body cools down when sweat evaporates.

       So, Energy absorbed by sweat to evaporate = Energy lost by body

Specific heat capacity of human body = 3500\text{ }\frac{J}{kg\text{ }C^{o}}. Jogger weights 60.4 kg. Body temperature decreases by 1.08\text{ }C^{o}

       Energy absorbed from body = mS\Delta T=3500\times60.4\times 1.08 =228312\text{ }J

       228312\text{ }J=\text{Energy absorbed by sweat}=mC=m\times2.42\times10^{6}\\m=0.094\text{ }kg

∴ 0.094 kg of sweat has evaporated from the body.

8 0
3 years ago
Can some one help me with this so i can bring my grade up
777dan777 [17]

Answer:

a

Explanation:

3 0
3 years ago
A string of length 1.3 m is oscillating in a standing wave pattern. If the tension in the string is 430 N, the string has a mass
Vlad1618 [11]

To solve this problem we will use the concepts related to the speed of a string which is given by the applied voltage and the linear mass density of it. With the speed value we can find the fundamental frequency that will serve as a step to find the maximum speed through the relation of Amplitude and Angular Speed. So:

v = \sqrt{\frac{T}{\mu_e}}

Where,

T = Tension

\mu_e= Linear mass density

v = \sqrt{\frac{430}{0.023}}

v = 136.7m/s

With this value the fundamental frequency would be

f = \frac{v}{2L}

f = \frac{136.7}{2*1.3}

f = 52.6Hz

Finally the maximum speed is given with the relation between the Amplitude (A) and the Angular frequency, then

V_{max} = A\omega

V_{max} = A(2\pi f)

V_{max} = (2.1*10^{-3})(2\pi 52.6)

V_{max} = 0.69m/s

Therefore the correct answer is B.

6 0
4 years ago
Problem 10: A simple pendulum with mass, m=1 kg and length, L=2.0 m is released by
vova2212 [387]

Answer:

A student is conducting a pendulum experiment. Which of the following pieces of safety equipment would be the most vital to conduct this test?

Explanation:

3 0
3 years ago
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