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3 years ago

What would be Kelley's weight be in newtons if her mass was 70 kilograms

Physics

2 answers:

ad-work [718]3 years ago

8 0

Answer:

Kelly's weight would be 688.47 Newtons.

Explanation:

1 Kilogram would be 9.81 Newtons.

xxMikexx [17]3 years ago

5 0

It all depends on where Kelley is at the moment.

-- If she's on Mars, she weighs 229 N.

-- If she's on the Moon, she weighs 113N.

-- If she's on Earth, she weighs 686 N.

-- If she's in a space capsule coasting from one body to another, then there's no instrument that can measure any weight of her.

So it completely depends on where she happens to be at the moment.

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3 years ago

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A +1.0 nC charge is at x = 0 cm, a -1.0 nC charge is at x = 1.0 cm and a 4.0 nC at x= 2 cm. What is the electric potential energ

lesantik [10]

Answer:

- 2.7 x 10^-6 J

Explanation:

q1 = 1 nC at x = 0 cm

q2 = - 1 nC at x = 1 cm

q3 = 4 nC at x = 2 cm

The formula for the potential energy between the two charges is given by

$U=\frac{Kq_{1}q_{2}}{r}$

where r be the distance between the two charges

By use of superposition principle, the total energy of the system is given by

$U = U_{1,2}+U_{2,3}+U_{3,1}$

$U=\frac{Kq_{1}q_{2}}{0.01}+\frac{Kq_{2}q_{3}}{0.01}+\frac{Kq_{3}q_{1}}{0.02}$

$U=-\frac{9\times10^{9}\times 1\times10^{-9}\times 1\times10^{-9}}}{0.01}-\frac{9\times10^{9}\times 1\times10^{-9}\times 4\times10^{-9}}}{0.01}+-\frac{9\times10^{9}\times 1\times10^{-9}\times 4\times10^{-9}}}{0.02}$

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3 years ago

A stereo speaker produces a pure \"c\" tone, with a frequency of 262 hz. what is the period of the sound wave produced by the sp

jasenka [17]

Answer:

3.82 ms

Explanation:

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$T=\frac{1}{f}$

where f is the frequency.

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3 years ago

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3 years ago

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A whistle of frequency 564 Hz moves in a circle of radius 71.2 cm at an angular speed of 17.1 rad/s. What are (a) the lowest and

trapecia [35]

Answer:

a) $f'=544.66 \textup{Hz}$

b) $f'=584.75 \textup{Hz}$

Explanation:

Given:

Frequency of the whistle, f = 564 Hz

Radius of the circle, r = 71.2 cm = 0.712 m

Angular speed, ω = 17.1 rad/s

speed of source, $v_s$ = rω = 0.712 × 17.1 = 12.1752 m/s

speed of sound, v = 343 m/s

Now, applying the Doppler's effect formula, we have

$f'=f\frac{v\pm v_d}{v\pm v_s}$

where,

$v_d$ = relative speed of the detector with respect to medium = 0

a) for lowest frequency, we have the formula as:

$f'=f\frac{v}{v+v_s}$

on substituting the values, we get

$f'=564\times\frac{343}{343+12.1752}$

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b) for maximum frequency, we have the formula as:

$f'=f\frac{v}{v-v_s}$

on substituting the values, we get

$f'=564\times\frac{343}{343-12.1752}$

$f'=584.75 \textup{Hz}$

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3 years ago