006 (part 1 of 2) 10.0 points

A bucket of warm water can be considered a closed system for a very short period of time. Which best explains this phenomenon? A

zlopas [31]

The reason is that a closed system has nothing going in or out and water retains energy very well also water it polar so it is attracted to itself. It doesnt like contact with other things.
So it's likely to be C

3 0

3 years ago

A sphere of mass 40kg is attaracted by a second sphere of mass 15 kg when their centers are 20 cm apart with a force of 0.1mg-wt

ser-zykov [4K]

<span>gravitational force = 0.1 mg-wt. = 0.1 * 10^-6 * 9.8 N = Gm1m2/r^2
m1 = 40 kg m2 =15 kg and r = 0.2 m
Put in and find G</span>

3 0

3 years ago

The conductors that carry the current to electrical devices and ? equipment are the heart of all electrical systems. There are a

MariettaO [177]

Answer:

utilization / effects

Explanation:

Utilization equipment are those equipment that makes use of electric energy for the purpose of chemical, electronic, lighting, heating, electro-mechanical or other alike purposes. Hence utilization best suits the first question mark in the question. Secondly, there are associated effects when current flows through a conductor, not responses.

5 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

4 years ago

Which of the following can be a centripetal force?a.

Ahat [919]

The answer is D
Friction acts as a centripetal force in the case of a car turning. The friction creates a tangential force allowing the car to turn.
Tension can act as a centripetal force in the case of circles. Such as swinging a mass on a string, the tension force allows it to move in a curved path.
Gravity acts as a centripetal force in the case of satellites. The gravity keeps the satellite within the orbit.

5 0

4 years ago