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3 years ago

Convergent faults _____.

Physics

1 answer:

marissa [1.9K]3 years ago

8 0

Answer:

Move towards each other and create mountains

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A -0.06 charge that moves downward is in a uniform electric field with a strengthened of 200 N/C. What is the magnitude and dire

Nookie1986 [14]

Answer:

The magnitude of the force is 12 N Upwards

Explanation:

The force on a positive charge will be in the same direction as the field, but the force on a negative charge will be in the opposite direction to the field. Thus the direction of the force is upward.

Given;

magnitude of charge, q = 0.06 C

magnitude of electric field, E = 200 N/C

The magnitude of the force is given by;

F = qE

F = 0.06 x 200 N/C

F = 12 N Upwards

Therefore, the magnitude of the force is 12 N Upwards

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3 years ago

Which element is a metalloid?

Ann [662]

Answer:

The metalloids are located on the right side of the periodic table in a "step-like" arrangement.

All of the possible metalloids are:

boron (B), silicon (Si), germanium (Ge), arsenic (As), antimony (Sb), tellurium (Te), and polonium (Po)

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1 year ago

How could you compare the energy carried in two different longitudinal waves

jenyasd209 [6]

<span>U could compare them using the intensity technique when bending waves are negligible in comparison with quasi-longitudinal waves.</span>

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3 years ago

Read 2 more answers

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

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