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kiruha [24]
3 years ago
7

In this wave, there is a vertical line which represents what part of the wave?

Physics
2 answers:
Leya [2.2K]3 years ago
5 0

Answer:

The amplitude of a wave, other things being equal, is the measure of its intensity. Thus, the louder a sound the greater is the amplitude of the system of waves to which it is due. The same applies to ether waves, whether they are perceived in the electro-magnetic, light,...

Explanation:

ANSWER IS AMPLITUDE HOPE THIS HELPS

nydimaria [60]3 years ago
4 0

Answer:Height

Explanation:

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Olivia put a glass of water in the freezer. She left it there for three hours. When she returned, the water had turned to ice. W
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A. freezing, when water turns to ice the water is turning from a liquid to a solid.
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3 years ago
What is the is the 94th element of the periodic table
shutvik [7]

Answer:

plutonium

Explanation:

6 0
4 years ago
How you use physics in your<br> everyday life.
stepan [7]

Answer:

Explanation:

Physics gets involved in your daily life right from you wake up in the morning. The buzzing sound of an alarm clock helps you wake up in the morning as per your schedule. The sound is something that you can't see, but hear or experience. Physics studies the origin, propagation, and properties of sound

5 0
3 years ago
When a body of mass 0.25 kg is attached to a vertical massless spring, it is extended 5.0 cm from its unstretched length of 4.0
lora16 [44]

Answer:

d=0.165m

Explanation:

Given

m=0.25kg,x_{1}=5cm*\frac{1m}{100cm}=0.05m,x_{2}=4cm*\frac{1m}{100cm}=0.04m,v=2\frac{rev}{s}

The tension of the spring is

F_{k}=K*x_{1}=m*g

K=\frac{m*g}{x_{1}}

K=\frac{0.25kg*9.8m/s^2}{0.05m}=49N/m

The force in the spring is equal to centripetal force so

F_{c}=\frac{m*v^2}{r}

v=w*r=2\pi*r

But Fc is also

Fc=KxΔr

F_{c}=K*(r-x_{2})

Replacing

m*4\pi^2*r=K*(r-x_{2})

0.25kg*4\pi^2*r=49*(r-0.04m)

r=0.205m

total distance is

d=0.205-0.04=0.165m

3 0
3 years ago
T=2pi square root 1/g solve for g.<br> Explanation would be really helpful.
Natalija [7]

I added individual steps for clarity. Note that g must be positive if the solution is to be real.

T=2\pi \sqrt{\frac{1}{g}}=2\pi g^{-\frac{1}{2}}\\g^{-\frac{1}{2}} = \frac{T}{2\pi}\\(g^{-\frac{1}{2}})^{-2} = (\frac{T}{2\pi})^{-2}\\g = \frac{4\pi^2}{T^2}\,\,\,, g>0}

Let me know if you have any questions.

7 0
3 years ago
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