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Elza [17]
2 years ago
12

starlight and starlite sd hd but micro macro in middle magna that is the white wavelenght in light please for our home at what w

avelenght and no b words the truth in light years please ask nasa and type in box for 15 pts thx rem radio carrys audio and video static like light so 2022 white magna middle special award winning question​
Physics
1 answer:
bogdanovich [222]2 years ago
4 0

Answer:

What is the question I can't understand it

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Which of the following is NOT a symptom associated with hypertension?
zepelin [54]
I would say a or d hope this helps haha ( also if you could do brainliest for this that would be AMASINF bc I really want to rank up, u don’t have to tho)
8 0
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The scale is called an absolute temperature scale, and it's zero point is called absolute sero
Kipish [7]
Absolute zero is from the Kelvin scale.
8 0
3 years ago
A 0.23-f capacitor is desired. What area must the plates have if they are to be separated by a 3.8-mm air gap?
tia_tia [17]

The area of the plates must have is(A)= 9.91×10⁷ m²

<h3 /><h3>How can we calculate the value of a area of a capacitor?</h3>

To calculate the the value of a area of the plates of a capacitor, we are using the formula,

C=\frac{\epsilon_0 A}{d}

Or, A= \frac{C\times d}{\epsilon_0}

Here we are given,

C= The desired capacitance of a capacitor.

= 0.23F

d=distance of separation between the plates.

=3.8mm= 0.0038m.

\epsilon_0= permittivity of the vacuum.  

=8.854×10⁻¹²F/m

We have to calculate the area of the plates must have = A m².

Now we put the known values in the above equation, we can get

A= \frac{C\times d}{\epsilon_0}

Or, A=\frac{0.23\times 0.0038}{8.854\times 10^{-12}}

Or, A= 9.91×10⁷ m²

From the above calculation, we can conclude that the area of the plates must have is(A)= 9.91×10⁷m²

Learn more about Capacitor:

brainly.com/question/13578522

#SPJ4

5 0
2 years ago
Merry-go-rounds are a common ride in park playgrounds. The ride is a horizontal disk that rotates about a vertical axis at their
goldfiish [28.3K]

Answer:

2.55348650884 m/s

2.67400481907 m/s²

10.6960192763 m/s²

Explanation:

d = Diameter of the ride = 16 ft=16\times 0.3048=4.8768\ m

r = Radius = \dfrac{4.8768}{2}=2.4384\ m

t = Time taken = 6 seconds

Velocity is given by

v=\dfrac{2\pi r}{t}\\\Rightarrow v=\dfrac{2\pi 2.4384}{6}\\\Rightarrow v=2.55348650884\ m/s

The speed is 2.55348650884 m/s

Acceleration is given by

a=\dfrac{v^2}{r}\\\Rightarrow a=\dfrac{(\dfrac{2\pi 2.4384}{6})^2}{2.4384}\\\Rightarrow a=2.67400481907\ m/s^2

The acceleration is 2.67400481907 m/s²

When the speed is halved

v=\dfrac{2\pi 2.4384}{\dfrac{6}{2}}=5.10697301768\ m/s

a=\dfrac{v^2}{r}\\\Rightarrow a=\dfrac{(\dfrac{2\pi 2.4384}{\dfrac{6}{2}})^2}{2.4384}\\\Rightarrow a=10.6960192763\ m/s^2

The acceleration is 10.6960192763 m/s²

3 0
4 years ago
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