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3 years ago

Please help on this one?

Physics

1 answer:

pashok25 [27]3 years ago

7 0

Answer:

B) It is connected in parallel and measures potential differences.

Explanation:

A voltmeter measures electrical potential differences between two points in an electric circuit. It is also connected in parallel to measure a device's voltage, while an ammeter is connected in series to measure a device's current.

Hope this helps! :)

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How does the time difference between the first p-wave and the first s-wave relate to the distance to the epicenter?

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S-waves do, the seismograph will detect P-waves arriving first, and S-waves will follow. The time difference, as recorded on a clock, between when the P-waves and S-waves ... earthquake waves speed up with increasing distance, and the lag time graph

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3 years ago

A motorcycle being driven on a dirt path hits a

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Impulse = mass * change in velocity (change in momentum) = Force * change in time

So, F=(m*change in v)/(change in t)
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4 years ago

The food calorie, equal to 4186 J, is a measure of how much energy is released when food is metabolized by the body. A certain b

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Answer:

The hiker has to climb 941.26 to ''work off'' the calories.

Explanation:

Let's first find out how much energy the hiker gets from 150 food calories.

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Energy = 150 * 4186

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To burn all of this off, we set it equal to the gravitational potential energy, and then solve for the height.

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Gravitational Potential Energy (G.P.E) = 629700

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Thus,

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3 years ago

How can a small human retina detect objects larger than itself?

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The correct answer would be A) the lenses of the eye change the size of images

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3 years ago

A 0.80-μm-diameter oil droplet is observed between two parallel electrodes spaced 11 mm apart. The droplet hangs motionless if t

Arisa [49]

A) $2.4\cdot 10^{-16}kg$

The radius of the oil droplet is half of its diameter:

$r=\frac{d}{2}=\frac{0.80 \mu m}{2}=0.40 \mu m = 0.4\cdot 10^{-6}m$

Assuming the droplet is spherical, its volume is given by

$V=\frac{4}{3}\pi r^3 = \frac{4}{3}\pi (0.4\cdot 10^{-6} m)^3=2.68\cdot 10^{-19} m^3$

The density of the droplet is

$\rho=885 kg/m^3$

Therefore, the mass of the droplet is equal to the product between volume and density:

$m=\rho V=(885 kg/m^3)(2.68\cdot 10^{-19} m^3)=2.4\cdot 10^{-16}kg$

B) $1.5\cdot 10^{-18}C$

The potential difference across the electrodes is

$V=17.8 V$

and the distance between the plates is

$d=11 mm=0.011 m$

So the electric field between the electrodes is

$E=\frac{V}{d}=\frac{17.8 V}{0.011 m}=1618.2 V/m$

The droplet hangs motionless between the electrodes if the electric force on it is equal to the weight of the droplet:

$qE=mg$

So, from this equation, we can find the charge of the droplet:

$q=\frac{mg}{E}=\frac{(2.4\cdot 10^{-16}kg)(9.81 m/s^2)}{1618.2 V/m}=1.5\cdot 10^{-18}C$

C) Surplus of 9 electrons

The droplet is hanging near the upper electrode, which is positive: since unlike charges attract each other, the droplet must be negatively charged. So the real charge on the droplet is

$q=-1.5\cdot 10^{-18}C$

we can think this charge has made of N excess electrons, so the net charge is given by

$q=Ne$

where

$e=-1.6\cdot 10^{-19}C$ is the charge of each electron

Re-arranging the equation for N, we find:

$N=\frac{q}{e}=\frac{-1.5\cdot 10^{-18}C}{-1.6\cdot 10^{-19}C}=9.4 \sim 9$

so, a surplus of 9 electrons.

3 0

4 years ago