Subtract the following and give the answer using significant figures 3.400-0.7591

Antifreeze is a "must have" in an Albertan winter because it helps keep your vehicle engine from freezing in cold temperatures.

Natasha_Volkova [10]

Products such as antifreeze are composed of organic compounds that are classified as alcohols. (a)

Maybe those other classes of chemicals also lower the freezing temperature of water, just like alcohol does. I don't know. But alcohol is what's used to make anti-freeze. I'm guessing alcohol must be cheaper, less toxic, and less corrosive inside the engines' cooling systems than any of that other stuff is.

4 0

2 years ago

The ink drops have a mass m = 1.00×10^−11 kg each and leave the nozzle and travel horizontally toward the paper at velocity v =

luda_lava [24]

Answer:

9.98 × 10⁻⁹ C

Explanation:

mass, m = 1.00 × 10⁻¹¹ kg

Velocity, v = 23.0 m/s

Length of plates D₀ = 1.80 cm = 0.018 m

Magnitude of electric field, E = 8.20 × 10⁴ N/C

drop is to be deflected a distance d = 0.290 mm = 0.290 × 10⁻³ m

density of the ink drop = 1000 kg/m^3

Now,

Time = $\frac{\textup{Distance}}{\textup{Velocity}}$

or

Time = $\frac{\textup{0.016}}{\textup{23}}$

or

Time = 6.9 × 10⁻⁴ s

Now, force due to the electric field, F = q × E

where, q is the charge

Also, Force = Mass × acceleration

q × E = 1.00 × 10⁻¹¹ × a

or

a = $\frac{q\times8.20\times10^4}{1\times10^{-11}}$

Now from the Newton's equation of motion

$d=ut+\frac{1}{2}at^2$

where,

d is the distance

u is the initial speed

a is the acceleration

t is the time

or

$0.290\times10^{-3}=0\times(6.9\times10^{-4})+\frac{1}{2}\times(\frac{q\times8.20\times10^4}{1\times10^{-11}})\times(6.9\times10^{-4})^2$

or

q = 9.98 × 10⁻⁹ C

4 0

2 years ago

an electric current of 285.0 ma flows 674. milliseconds. Calculate the amount of electric charge transported. Be sure your answe

yulyashka [42]

Answer:

The amount of electric charge transported = 0.192 C

Explanation:

Electric Charge: This is defined as the product of electric current and time in an electric circuit, The S.I unit of electric charge is Coulombs (C)

Q = It..................... Equation 1

Where Q = Electric charge, I = electric current, t = time.

Given: I = 285 mA, t = 674 milliseconds.

Conversion: (i) Convert from 285 mA to A = (285/1000) A = 0.285 A

(ii) convert from 674 milliseconds to seconds = (674/1000) s = 0.674 s

Substituting these values into equation 1

Q = 0.285 × 0.674

Q = 0.192 C

Therefore the amount of electric charge transported = 0.192 C

5 0

2 years ago

The Nucleus of the Atom is in the center of the Atom, not in the outer rings & orbitals.

chubhunter [2.5K]

Answer:

true

Explanation:

this the nucleus is located at the centre and contains protons and neutrons

3 0

2 years ago

A 0.272-kg volleyball approaches a player horizontally with a speed of 12.6 m/s. The player strikes the ball with her fist and c

Nady [450]

(a) +9.30 kg m/s

The impulse exerted on an object is equal to its change in momentum:

$I= \Delta p = m \Delta v = m (v-u)$

where

m is the mass of the object

$\Delta v$ is the change in velocity of the object, with

v = final velocity

u = initial velocity

For the volleyball in this problem:

m = 0.272 kg

u = -12.6 m/s

v = +21.6 m/s

So the impulse is

$I=(0.272 kg)(21.6 m/s - (-12.6 m/s)=+9.30 kg m/s$

(b) 155 N

The impulse can also be rewritten as

$I=F \Delta t$

where

F is the force exerted on the volleyball (which is equal and opposite to the force exerted by the volleyball on the fist of the player, according to Newton's third law)

$\Delta t$ is the duration of the collision

In this situation, we have

$\Delta t = 0.06 s$

So we can re-arrange the equation to find the magnitude of the average force:

$F=\frac{I}{\Delta t}=\frac{9.30 kg m/s}{0.06 s}=155 N$

6 0

2 years ago