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Zigmanuir [339]
3 years ago
13

Who knows the answers to this’s problems

Physics
1 answer:
MArishka [77]3 years ago
6 0

Answer:1. High Intensity Interval Training

HIIT stands for High Intensity Interval Training. HIIT is essentially a type of exercise, be it cardio or resistance training. HIIT. alternates between periods of high intensity, and low intensity (or recovery

2. flexibility and muscular strength

3. true

4. d

5. c

Explanation:

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The surface of the paper is phosphorescent. When light shines on it, some of the energy is absorbed and re-emitted slowly over t
Ad libitum [116K]

Answer:

No

Explanation:

Recall that the hierarchy of wavelength color from minimum wavelength to maximum wavelength is:

V < I < B < G < Y < O < R; and

E  \ \alpha  \ \dfrac{1}{\lambda}

As a result, blue light has a higher energy level than green and red light.

As a result, the surface glows due to the blue LED. The green LED, on the other hand, would not allow the surface to glow as much as the red LED, which has a lower energy level when compared to the green light. As a result, the red LED would not allow the surface to glow as well.

8 0
3 years ago
Factfile about eclips
MAXImum [283]
Its quite hard to give you facts over this but just type into google facts about the sonar/lunar eclipse.

6 0
3 years ago
The nucleus of a certain type of uranium atom contains
BaLLatris [955]

Answer:

charge = electrons + protons

=92+92

=184

8 0
3 years ago
What is the ratio of the sun’s gravitational pull on Mercury to the sun’s gravitational pull on the earth?
Marta_Voda [28]

Answer:

The answer is \frac{F_{Sun-Mercury} }{F_{Sun-Earth} } =0,3709. Let's learn why.

Explanation:

Newton's law of universal gravitation says;

F_{g} =G.\frac{m_{1}.m_{2}}{r^{2}}

Here G is a universal gravitational <u>constant</u> and is measured experimentally.

Sun's gravitational pull on mercury is:

F_{Sun-Mercury} =G.\frac{m_{sun}.3,30.10^{23}}{(5,79.10^{10})^{2} }

Therefore F_{Sun-Mercury} = Gm_{sun} 98,4366

Sun's gravitational pull on Earth is:

F_{Sun-Earth} =G.\frac{m_{sun} 5,97.10^{24} }{(1,50.10^{11}) ^{2}}

Therefore F_{Sun-Earth} =Gm_{sun} 265,33

As a result;

\frac{F_{Sun-Mercury}}{F_{Sun-Earth} }=\frac{Gm_{sun}98,4366}{Gm_{sun}265,33 } =0,3709

4 0
4 years ago
Send help my teacher is playing call me maybe
guajiro [1.7K]

Answer:

lol

Explanation:

3 0
3 years ago
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