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BigorU [14]
3 years ago
13

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!! I CANNOT RETAKE THIS AND I NEED ALL CORRECT ANSWERS ONLY!!!

Physics
2 answers:
valentinak56 [21]3 years ago
6 0

Answer:

Electric Current

Explanation:

The flow (or free movement) of these electrons through a wire.

Pretty sure :)

wolverine [178]3 years ago
6 0

I think it is electric current

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If the velocity of blood flow in the aorta is normally about 0.32 m/s, what beat frequency would you expect if 4.40-MHz ultrasou
dusya [7]

Answer:

The beat frequency is 0.0019 MHz.

Explanation:

Given that,

Velocity = 0.32 m/s

Frequency = 4.40 MHz

Speed of wave = 1540 m/s

We need to calculate the frequency

Case (I),

Observer is moving away from the source

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f'=\dfrac{v-v'}{v}f

Where, v' = speed of observer

Put the value into the formula

f'=\dfrac{1540-0.32}{1540}\times4.40

f'=4.399\ MHz

Case (II),

Cell is as the source of sound of frequency f' and it moving away from the observer.

Using formula of frequency

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f''=\dfrac{1540-0.32}{1540+0.32}\times4.399

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\Delta f= f'-f''

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4 0
3 years ago
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3 years ago
Particles 1 and 2 of charge q1 = q2 = +3.20 × 10−19 C are on a y axis at distance d = 17.0 cm from the origin. Particle 3 of cha
mel-nik [20]

Answer:

(a) 0.17 m

(b) 5.003 m

(c) 6.38 × 10^{-26} N

(d) 7.37 ×10^{-29} N

Explanation:

(a) The minimum value of x will occur when q3 = 0 m or at origin and q1, q2 are at 0.17 m so the distance between q3 and q1, q2 is 0.17 m, therefore the <em>minimum value of x= 0.17 m</em>.

(b) The maximum value of x will occur when q3 = 5 m because it is said in the question that 5 is the maximum distance travelled by q3. To find the hypotenuse i.e. the distance between q3 and q1,q2, we use Pythagoras theorem.

h^{2} = b^{2} + p^{2}

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<em>Hence, the maximum distance is 5.002 m</em>

(c) For minimum magnitude we use the minimum distance calculated in (a)

Minimum Distance = 0.17 m

For electrostatic force=     F=\frac{kq1q2}{x^{2} }

F=\frac{9 x 10^{9} x3.2x10^{-19}x 6.4x10^{-19}  }{0.17^{2} }

F= 6.38×10^{-26} N

(d) For maximum magnitude, we use the maximum distance calculated in (b)

Maximum Distance = 5.002 m

Using the formula for electrostatic force again:

F =  \frac{9x10^{9}x3.2x10^{-19}x6.4x10^{-19}   }{5.002^{2} } }

F= 7.37×10^{-29 N

4 0
3 years ago
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