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3 years ago

What is addictive about vaping?

Physics

1 answer:

AnnZ [28]3 years ago

4 0

Answer:
The last one, nicotine from tobacco

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so classmate states that continental drift could not be possible because it would take far too much force to move tectonic plate

Juliette [100K]

Any student who states that tectonic plates are too massive to move doesn't understand how much heat exists in the center of the earth. This heat is capable of moving the most massive of plates.

5 0

3 years ago

Physical Science, please explain!

DedPeter [7]

Answer:

Explanation:

because the net force is greater in the right direction

3 0

3 years ago

Read 2 more answers

On a hot summer day, 3.50 ✕ 106 J of heat transfer into a parked car takes place, increasing its temperature from 36.5°C to 44.4

anygoal [31]

Answer:

a) $\Delta s=443037.9747\ J.K^{-1}$

b) $\Delta s=31868131.8681\ J.K^{-1}$

Explanation:

Given:

amount of heat transfer occurred, $dQ=3.5\times 10^6\ J$

initial temperature of car, $T_i=36.5+273=309.5\ K$

final temperature of car, $T_f=44.4+273=317.4\ K$

We know that the change in entropy is given by:

$\Delta s=\frac{dQ}{T_f-T_i}$

$\Delta s=\frac{3.5\times 10^6}{(44.4-36.5)}$ (heat is transferred into the system of car)

$\Delta s=443037.9747\ J.K^{-1}$

amount of heat transfer form the system of house, $dQ=5.8\times 10^8\ J$

initial temperature of house, $T_i=23.5+273=296.5\ K$

final temperature of house, $T_f=5.3+273=278.3\ K$

$\Delta s=\frac{dQ}{T_f-T_i}$

$\Delta s=\frac{5.8\times 10^8}{278.3-296.5}$

$\Delta s=31868131.8681\ J.K^{-1}$

6 0

3 years ago

A circle has an initial radius of 50ft when the radius begins decreasing at a rate of 2ft/s. what is the rate of change of the a

valkas [14]

The area of the circle with radius r is
A = πr²

The rate of change of area with respect to time is
$\frac{dA}{dt} = \frac{dA}{dr} . \frac{dr}{dt} =2 \pi r. \frac{dr}{dt}$

The rate of change of the radius is given as
$\frac{dr}{dt} =-2 \, \frac{ft}{s}$
Therefore
$\frac{dA}{dt} =-4 \pi r \, \frac{ft^{2}}{s}$

When r = 10 ft, obtain
$\frac{dA}{dt}|_{r=10 \, ft} = -40 \pi \, \frac{ft^{2}}{s}$

Answer: - 40π ft²/s (or - 127.5 ft²/s)

7 0

4 years ago

A 72 kg skydiver can be modeled as a rectangular "box" with dimensions 21 cm × 41 cm × 170 cm . if he falls feet first, his drag

LuckyWell [14K]

Formula for terminal velocity is:

Vt = √(2mg/ρACd)
Vt = terminal velocity = ?
m = mass of the falling object = 72 kg
g = gravitational acceleration = 9.81 m/s^2
Cd = drag coefficient = 0.80
ρ = density of the fluid/gas = 1.2 kg/m^3
A = projected area of the object (feet first) = 0.21 m * 0.41 m = 0.0861 m^2

Therefore:

Vt = √(2 * 72 * 9.81 / 1.2 * 0.0861 * 0.80)

Vt = 130.73 m/s

4 0

3 years ago