All categories

4 years ago

Heat from the sun moves through space by the process of

2 answers:

7 0

I don't know what your options are, but heat moves from the sun moves in 3 ways: radiation, conduction, and convection. I hope this helps at least a little

tino4ka555 [31]4 years ago

5 0

Answer: Radiation

Explanation:

There are three modes of heat transfer:

<u>Conduction: </u>Heat transfers from hotter body to colder one via conduction when they are placed in contact with each other. The molecules vibrate and transfer heat.

<u>Convection: </u>Heat transfers in fluids via convection. The heat is carried by bulk motion of fluid.

Radiation: Heat transfer through space takes place via radiation. The electromagnetic waves carry the energy from the Sun and moves through space. This mode does not depend on the medium. The electromagnetic waves carry the energy and travel at the speed of light.

You might be interested in

Assuming the earth is a uniform sphere of mass M and radius R, show that the acceleration of free fall at the earth's surface is

zimovet [89]

use Newton's gravitational law and 2nd law .....

8 0

2 years ago

Use free fall in a sentance

Softa [21]

Free fall means rapid fall or a downward motion due to gravity. Sentence example: When Joe was accidentally bumped by Sarah, he was sent towards a free fall down the escalator, leading to a serious injury on his arm and two legs.

5 0

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

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

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

What is the gravitational force that two objects would feel if they are 3.5 meters apart? Object 1 has a mass of 10x10^5 kg and

polet [3.4K]

Answer:

1.63 N

Explanation:

F = GMm/r^2

= (6.67x10^-11)(10x10^5)(3x10^5) / 3.5^2

= 1.63 N ( 3 sig. fig.)

6 0

3 years ago

Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its te

Zarrin [17]

Answer:

a) (dT/dt) = -0.3 [T - 70]

b) (dT/dt) = -0.3 {T - [66 cos ((π/30)t)]}

c) (dT/dt) = -18 {T - [66 cos (2πt)]}

with t in hours

d) (dT/dt) = -32.4 [T - 57.6 - 118.8 cos (2πt)]

with T in Fahrenheit and t in hours

Explanation:

The Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its temperature and that of its surroundings.

If the temperature of the object = T

Temperature of the surroundings = Ambient temperature = TA(t)

(dT/dt) ∝[T - TA(t)]

Introducing the constant of proportionality, k

(dT/dt) = k [T - TA(t)]

Temperature is in degree Celsius and time is in minutes.

Because the temperature of the body is decreasing, we introduce a minus sign

(dT/dt) = -k [T - TA(t)]

a) If TA(t) = 70°C, k = 0.3

(dT/dt) = -0.3 [T - 70]

b) The ambient temp TA(t) = 66 cos ((π/30)t) degrees Celsius (time measured in minutes).

(dT/dt) = -k [T - TA(t)]

(dT/dt) = -k {T - [66 cos ((π/30)t)]}

(dT/dt) = -0.3 {T - [66 cos ((π/30)t)]}

c) If we measure time in hours the differential equation in part (b) changes.

1 hour = 60 mins

If t is now expressed in hours,

t hours = (60t) mins

(dT/dt) = -k {T - [66 cos ((π/30)t)]}

dT = -k {T - [66 cos ((π/30)t)]} dt

dT = -k {T - [66 cos ((π/30)60t)]} d(60t)

(dT) = -60k {T - [66 cos ((π/30)60t)]} dt

(dT/dt) = -60k {T - [66 cos (2πt)]}

with t in hours, k = 0.3, 60k = 18

(dT/dt) = -18 {T - [66 cos (2πt)]}

d) If we measure time in hours and we also measure temperature in degrees Fahrenheit, the differential equation in part (c) changes even more.

If T is in degree Fahrenheit

T°F = (5/9)(T°F - 32) degrees Celsius

T°F = [(5T/9) - 17.78] degrees Celsius

(dT/dt) = -60k {T - [66 cos (2πt)]}

time already converted to hours.

dT = -60k {T - [66 cos (2πt)]} dt

66 cos (2πt) degrees Celsius = {(9/5) [66 cos (2πt)] + 32} degrees Fahrenheit = {[118.8 cos (2πt)] + 57.6} degrees Fahrenheit

d[(5T/9) - 17.78] = -60k {T - [118.8 cos (2πt) + 57.6]} dt

(5/9) dT = -60k [T - 57.6 - 118.8 cos (2πt)] dt

(5/9) (dT/dt) = -60k [T - 57.6 - 118.8 cos (2πt)]

(dT/dt) = -108k [T - 57.6 - 118.8 cos (2πt)]

k = 0.3, 108k = 32.4

(dT/dt) = -32.4 [T - 57.6 - 118.8 cos (2πt)]

with T in Fahrenheit and t in hours

Hope this Helps!!!

7 0

4 years ago