The items in the following list are all units of matter. Which is the smallest unit that retains the properties of the matter?

Convert 1215 m to km.<br> Answer:

IgorC [24]

Answer:

it is 12.15 km

Explanation:

1215 km ÷1000

=12.15

please mark as brainliest

8 0

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

3 years ago

Suppose you wanted to use the Earth's magnetic field to make an AC generator at a location where the magnitude of the field is 0

Likurg_2 [28]

Answer:

The angular velocity I would have to rotate it in order to generate an emf of amplitude 1.0 V is 254.65 rad/s

Explanation:

given information:

B = 0.5 mT = 0.0005 T

N = 1000

r = 5 cm = 0.05 m

emf, ε = 1 V

according to Faraday's law

ε = -N dΦ/dt, Φ = B A

= - N d( B A)/dt

= - N d( B A cos ωt)/dt

= - N B A d(cos ωt)/dt

= N B A ω sin ωt

A = πr², so

ε = N B πr² ω sin ωt

where

ε = emf

N = number of coil turn

B = magnetic field

r = radius

ω = angular velocity

Φ = magnetic flux

emf maximum, sin ωt = 1. So,

ε = N B πr² ω

ω = ε/N B πr²

= 1/[(1000) (0.0005) π (0.05)²

= 254.65 rad/s

6 0

3 years ago

Your name is Galileo Galilei and you toss a weight upward at 20 feet per second from the top of the Leaning Tower of Pisa (heigh

Veseljchak [2.6K]

Answer:

a) v(t) = -32.2 ft/s² · t + 20 ft/s

b) h(t) = 184 ft + 20 ft/s · t - 16.1 ft/s² · t²

c) The weight will reach its maximum height after 0.62 s. The maximum height will be 190 feet.

Explanation:

Hi there!

a) Since the only force that acts on the weight is the gravity force, the object is under a constant downward acceleration g = -32.2 ft/s² (it is negative because we consider the upward direction as positive). The acceleration is the variation of the velocity over time (dv/dt). Then:

dv/dt = g

Separating variables:

dv = g dt

Integrating from the initial velocity, v0, to v and from t = 0 to t, we obtain:

v - v0 = g t

v = g t + v0

Then:

v(t) = -32.2 ft/s² · t + 20 ft/s

b) The velocity of the weight is the variation of the height over time:

dh/dt = v(t)

dh/dt = g t + v0

Separating varibles:

dh = g t dt + v0 dt

Integrating from initial height, h0, to h and from t = 0 to t:

h - h0 = 1/2 · g · t² + v0 · t

h = h0 + v0 · t + 1/2 · g · t²

Then:

h(t) = 184 ft + 20 ft/s · t - 1/2 · 32.2 ft/s² · t²

h(t) = 184 ft + 20 ft/s · t - 16.1 ft/s² · t²

c) When the weight reaches its maximum height, its velocity will be zero. Then, using the equation of velocity we can obtain the time at which the weight is at the maximum height:

v(t) = -32.2 ft/s² · t + 20 ft/s

0 = -32.2 ft/s² · t + 20 ft/s

-20 ft/s/ -32.2 ft/s² = t

t = 0.62 s

The weight will reach its maximum height after 0.62 s.

The maximum height will be h(0.62 s):

h(t) = 184 ft + 20 ft/s · t - 16.1 ft/s² · t²

h(0.62 s) = 184 ft + 20 ft/s · (0.62 s) - 16.1 ft/s² · (0.62 s)²

h(0.62 s) = 190 ft

The maximum height will be 190 feet.

3 0

3 years ago

Read 2 more answers

What happens if the value of VOq doubles

Crazy boy [7]

If Voq doubles, abN gets multiplied by 16 .

If Voq=3, abN=81 .

8 0

3 years ago