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Rasek [7]
2 years ago
12

Please left home at 8 AM to spend the day at in amusement park. He arrived at the park, which was 150 KM from his house, at 10 A

M. Pace explain that his speed 38 KM/H which was what
Physics
1 answer:
antiseptic1488 [7]2 years ago
3 0

Answer:

v=\frac{150km}{2h}=75\frac{km}{h}

Explanation:

We can use the equation for the speed

v=\frac{x}{t}

where x is the distance and t the time. In this case we know that the time spent was 2 hours and the distance was 150km. By replacing we have

v=\frac{150km}{2h}=75\frac{km}{h}

I hope this useful for you

regards

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an electric current of 285.0 ma flows 674. milliseconds. Calculate the amount of electric charge transported. Be sure your answe
yulyashka [42]

Answer:

<em>The amount of electric charge transported = 0.192 C</em>

Explanation:

Electric Charge: This is defined as the product of electric current and time in an electric circuit, The S.I unit of electric charge is Coulombs (C)

Q = It..................... Equation 1

Where Q = Electric charge, I = electric current, t = time.

<em>Given:</em> I = 285 mA, t = 674 milliseconds.

<em>Conversion: (i) Convert from 285 mA to A = (285/1000) A = 0.285 A</em>

<em>       (ii) convert from 674 milliseconds to seconds = (674/1000) s = 0.674 s          </em>

Substituting these values into equation 1

Q = 0.285 × 0.674

<em>Q = 0.192 C</em>

<em>Therefore the amount of electric charge transported = 0.192 C</em>

<em></em>

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2 years ago
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Answer:

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A particle has a charge of -4.25 nC.
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Answer:

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Explanation:

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

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Electric field is given by

E=\dfrac{kq}{r^2}\\\Rightarrow E=\dfrac{8.99\times 10^9\times -4.25\times 10^{-9}}{0.25^2}\\\Rightarrow E=-611.32\ N/C

The magnitude is 611.32 N/C

The electric field will point straight down as the sign is negative towards the particle.

E=\dfrac{kq}{r^2}\\\Rightarrow r=\sqrt{\dfrac{kq}{E}}\\\Rightarrow r=\sqrt{\dfrac{8.99\times 10^9\times 4.25\times 10^{-9}}{13}}\\\Rightarrow r=1.71436\ m

The distance from the electric field is 1.71436 m

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3 years ago
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lilavasa [31]

Answer:

The fundamental wavelength of the vibrating string is 1.7 m.

Explanation:

We have,

Velocity of wave on a guitar string is 344 m/s

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It is required to find the fundamental wavelength of the vibrating string. The fundamental frequency on the string is given by :

f=\dfrac{v}{2l}\\\\f=\dfrac{344}{2\times 0.85}\\\\f=202.35\ Hz

Now fundamental wavelength is :

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So, the fundamental wavelength of the vibrating string is 1.7 m.

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