HELP ASAP How does the acceleration due to gravity at the surface of a planet change if the planet's radius is doubled?

The coefficient of performance of a residential heat pump is 1.6. Calculate the heating effect in kJ/s this heat pump will produ

Andre45 [30]

Answer:

$Q_{H}=6.4kJ/s$

Explanation:

Given data

Coefficient of performance of a residential heat pump=1.6

Electrical power P=4kW

Required

Heat Q

Solution

The rate of heat produced is given as

$Q_{H}=COP_{HP}Win\\$

Substitute the given values

So

$Q_{H}=4kW*1.6\\Q_{H}=6.4kJ/s$

6 0

3 years ago

Please help me I will give you 20 points !!

nadezda [96]

Answer:

D. The primary source, because it was written by the researcher

Explanation:

One should learn to trust the primary source more because it is the real work of the researcher.

A primary work defines the work of research from his or her experimental findings.

The primary source presents experimental data and other subordinates ones to reach a conclusion as seen from the view of the researcher.
A secondary source implies someone else documenting their own opinion about the primary experimental set up.
This can get dicey in the sense that results can be twisted to serve other ulterior motives.

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

4 years ago

A particle is leaving the Moon in a direction that is radially outward from both the Moon and Earth.

Misha Larkins [42]

Answer:

$2780m/s$

Explanation:

Essentially, Kinetic energy of the particle must equal the combined potential energies of earth and the moon when the object is on the moon's surface, meaning the full equation is

$M_E$ =Mass of Earth= $5.97*10^2^4$

$M_m$ =Mass of Moon= $7.4*10^2^2kg$

$r_E$ =distance from earth's center to the moon's= $3.84*10^8m$

$r_m$ =radius of moon= $1.738*10^6m$

After some algebra, the equation simplifies to

$v=\sqrt{2G*(\frac{M_E}{r_E+r_m}+\frac{M_m}{r_m})}$

Plugging in the values of G, which is $6.67*10^-^1^1 \frac{m^3}{kg*s^2}$ , should yield the proper answer of 2780m/s.

4 0

3 years ago

I just want this information cuz it will come in handy in science soon. Where and how are Gamma rays found?

vodomira [7]

Gamma rays are found in extremely hot regions and they have the smallest wavelength but the most energy. They can be found by the effects they have on matter and it do few basic things with matter, it can collide with a electron and bounce of it as a billiard ball.

7 0

3 years ago

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