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andreyandreev [35.5K]
3 years ago
11

Define the problem. Look around your house or outside. Is there anything that you're wondering about or have questions about?

Physics
1 answer:
bekas [8.4K]3 years ago
6 0

Answer:

Water

Explanation:

How do we have water at our disposal, like that's crazy.

You might be interested in
A 0.25-kg ball attached to a string is rotating in a horizontal circle of radius 0.5 m twice every second, what is the tension i
Svetradugi [14.3K]

Answer:

T = 19.75 N

Explanation:

given,

mass of ball = 0.25 Kg

radius = 0.5 m

frequency = 2 s⁻¹

tension in the string = ?

angular velocity

ω = 2 π f

ω = 2 π x 2

ω = 12.57 rad/s

tension on the string is equal to the centripetal force

T = m ω² r

T = 0.25 x 12.57² x 0.5

T = 19.75 N

Tension in the string is equal to T = 19.75 N

3 0
3 years ago
The tangential acceleration of a cart moving at a constant speed in a horizontal circle is:
qaws [65]

Answer:

a=0                                  ∵ \alpha=0\ rad.s^{-1}

Explanation:

The tangential acceleration of a cart moving at a constant speed in a circle is:

The angular velocity is constant when the circular speed is constant.

We know that the (instantaneous) tangential velocity of such object is given by:

v=r.\omega

Now for angular acceleration we have a constant angular speed:

\alpha=0\ rad.s^{-1}

And angular acceleration is related to tangential acceleration as:

a=r.\alpha

\Rightarrow a=0

7 0
3 years ago
____________ are used to calculate the distance a continent has moved in a year. a. Space satellites c. Laser beams b. Mirrors d
Law Incorporation [45]
<span> Space satellites, laser beams, mirrors</span> are used to calculate the distance a continent has moved in a year.

Therefore, your correct answer would be "all of the above".
4 0
2 years ago
A generator with �# ' = 300 V and Zg = 50 Ω is connected to a load ZL = 75 Ω through a 50-Ω lossless line of length l = 0.15λ. (
ki77a [65]

Answer:

a. Zin = 41.25 - j 16.35 Ω

b. V₁ = 143. 6 e⁻ ¹¹ ⁴⁶

c.  Pin = 216 w

d. PL = Pin = 216 w

e. Pg = 478.4 w , Pzg = 262.4 w

Explanation:

a.

Zin = Zo * [ ZL + j Zo Tan (βl) ] / [ Zo + j ZL Tan (βl) ]  

βl = 2π / λ * 0.15 λ = 54 °

Zin = 50 * [ 75 + j 50 Tan (54) ] / [ 50 + j 75 Tan (54) ]

Zin = 41.25 - j 16.35 Ω

b.

I₁ = Vg / Zg + Zin ⇒ I₁ = 300 / 41.25 - j 16.35 = 3.24 e ¹⁰ ¹⁶

V₁ = I₁ * Zin = 3.24 e ¹⁰ ¹⁶ * ( 41.25 - j 16.35)

V₁ = 143. 6 e⁻ ¹¹ ⁴⁶

c.

Pin = ¹/₂ * Re * [V₁ * I₁]

Pin = ¹/₂ * 143.6 ⁻¹¹ ⁴⁶ * 3.24 e ⁻ ¹⁰ ¹⁶ = 143.6 * 3.24 / 2 * cos (21.62)

Pin = 216 w

d.

The power PL and Pin are the same as the line is lossless input to the line ends up in the load so

PL = Pin

PL = 216 w

e.

Pg Generator

Pg = ¹/₂ * Re * [ V₁ * I₁ ] = 486 * cos (10.16)

Pg = 478.4 w

Pzg dissipated

Pzg = ¹/₂ * I² * Zg = ¹/₂ * 3.24² * 50

Pzg = 262.4 w

4 0
3 years ago
You are trying to overhear a juicy conversation, but from your distance of 23.0 m , it sounds like only an average whisper of 40
Len [333]
If its asking the distance for the 65 db then use a proportion, if otherwise pleas clarify. It sounds like a pretty juicy conversation.
4 0
3 years ago
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