Answer:
In my textbook's words-
Amplitude, in physics, the maximum displacement or distance moved by a point on a vibrating body or wave measured from its equilibrium position. It is equal to one-half the length of the vibration path.
Explanation:
The correct definition of amplitude is that it is a maximum displacement
that occurs on a vibrating body from one point to the other.
The initial point of the wave is regarded as its equilibrium position which is
equal to one-half the length of the vibration path.
Amplitude helps to calculate the peak value of different types of waves
such as water waves and in electrical appliances so as to know the peak
current suitable for it.
Read more on https://brainly.com/question/21632362
An aluminum wire having a cross-sectional area equal to 2.20 10-6 m2 carries a current of 4.50 A. The density of aluminum is 2.70 g/cm3. Assume each aluminum atom supplies one conduction electron per atom. Find the drift speed of the electrons in the wire.
Answer:
The drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.
Explanation:
We can find the drift speed by using the following equation:
[tex] v = \frac{I}{nqA} [/tex]
Where:
I: is the current = 4.50 A
n: is the number of electrons
q: is the modulus of the electron's charge = 1.6x10⁻¹⁹ C
A: is the cross-sectional area = 2.20x10⁻⁶ m²
We need to find the number of electrons:
[tex] n = \frac{6.022\cdot 10^{23} atoms}{1 mol}*\frac{1 mol}{26.982 g}*\frac{2.70 g}{1 cm^{3}}*\frac{(100 cm)^{3}}{1 m^{3}} = 6.03 \cdot 10^{28} atom/m^{3} [/tex]
Now, we can find the drift speed:
[tex]v = \frac{I}{nqA} = \frac{4.50 A}{6.03 \cdot 10^{28} atom/m^{3}*1.6 \cdot 10^{-19} C*2.20 \cdot 10^{-6} m^{2}} = 2.12 \cdot 10^{-4} m/s[/tex]
Therefore, the drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.
I hope it helps you!
Consider the air over a city to be a box that measures 100 km per side that reaches up to an altitude of 1.0 km. Wind (clean air) is blowing into the box along one of its sides with a speed of 4 m/s. An air pollutant is emitted into the box at a rate of 10.0 kg/s; the pollutant degrades with a rate constant k = 0.20/hr. a. Find the steady state concentration of the pollutant (µg/m3 ) in the box if the air is assumed to be completely mixed. b. If the wind speed suddenly drops to 1 m/s, estimate the concentration of the pollutant (µg/m3 ) two hours later.
Answer:
a) ρ = 6.25 10⁵ μg / m³, b) ρ = 1 10⁷ μg / m³
Explanation:
Let's analyze the exercise a little before starting, we must know the amount of pollutant in the box, that the one that enters less the one that degrades and with this value find the density or concentration.
Let's start by finding the volume of air that goes into the box
V = Lh x
Let's find the distance of air that enters per unit of time, as it goes at constant speed
x = v₀ t
we substitute
V₀ = Lh v₀ t
At this same time, a quantity of pollutant is distributed
Q₀ = r t
the contaminant that is entering reaches the entire box, therefore the total amount of contaminant is
Q = Qo t
we substitute
Q = r t²
the net amount of pollutant that remains is that less enters the one that degraded in the same time, as they ask for the steady state
[tex]Q_{net}[/tex]= Q - k t
the pollutant concentration is
ρ = Q_net / V
V = L L h
ρ =[tex]\frac{r \ t^2 - k \ t}{ L^2 h}[/tex]
ρ = [tex](r \frac{ L^2}{v_o^2} - k \frac{L}{v_o} ) \frac{1}{L^2 h}[/tex]
ρ = [tex]\frac{r}{ v_o h} -\frac{k}{v_o L h}[/tex]
let's reduce the magnitudes to the SI system
r = 10 kg / s
L = 100 km = 100 10³ m
h = 1 km = 1 10³ m
k = dq / dt = 0.20 1/h ( 1h/3600 s) = 5.5555 10⁻⁵ 1/s
v₀ = 4 m / s
let's calculate
The volume of the box
V = (100 100 1) 109
V = 1 10¹³ m³
ρ = [tex]\frac{10}{ 4^2 \ 1\ 10^3 } - \frac{5.5556 \ 10^{-5}}{ 4 \ 100 \ 10^3 1 \ 10^3}[/tex]
ρ = [tex]6.25 10^{-4} - 1.389 ^{-13}[/tex]
ρ = 6.25 10⁻⁴ kg / m³
let's reduce to μg / m³
ρ = 6.25 10⁻⁻⁴ kg / m³ (10⁹ μg / 1kg)
ρ = 6.25 10⁵ μg / m³
b) in case the air speed decreases to v₀ = 1 m / s
ρ= \frac{10}{ 1^2 \ 1\ 10^3 } - \frac{5.5556 \ 10^{-5}}{ 1 \ 100 \ 10^3 1 \ 10^3}
ρ = 1 10⁻² - 5.5556 10⁻¹³
ρ = 1 10⁻² kg / m³
ρ = 1 10⁷ μg / m³
Stacy collected the data shown in the table.
Number of Washers Total Mass Total Force
0.6 kg 5.9 N
2
1.2 kg 11.8 N
3
1.8 kg 17.6N
4
2.5 kg 24.5 N
5
3.2 kg 31.4 N
What's the general relationship between mass and gravitational force?
Answer:
1.8 Kg 17.6N
Explanation:
I don't know the explanation hahaha
Answer:
The gravitational force on an object increases as the object’s mass increases.
Explanation:
This is the answer on Edmentum. :)
A tank initially holds 100 gallons of salt solution in which 50 lbs of salt has been dissolved. A pipe fills the tank with brine at the rate of 3 gpm, containing 2 lbs of dissolved salt per gallon. Assuming that the mixture is kept uniform by stirring, a drain pipe draws out of the tank the mixture at 2 gpm. Find the amount of salt in the tank at the end of 30 minutes.
A. 171.24 lbs
B. 124.11 lbs
C. 143.25 lbs
D. 105.12 lbs
Answer:
A. 171.24 Ibs
Explanation:
To find the amount of salt in the tank,
Let Q = Amount of salt in the mixture
And let 100 + (3-2)t = 100 + t be the volume of mixture at anytime t.
Rate of gain - Rate of loss = dQ / dt
Concentration of salt = Q / (100+t)
For the linear differential equation,
dQ / dt = 3(2) - 2 [Q/ (100 + t)]
dQ /dt + Q [2 / (100 + t)] = 6
The general solution of the linear differential equation is:
Q (i.f) = ∫ A(t) (i.f) dt + C
Therefore,
i.f = e ^ ∫ P(t) dt
And P(t) = 2 / (100 + t)
i.f = e ^ ∫ 2 / (100 + t)
= e ^ 2㏑ (100 + t)
= e ^ ㏑ (100 + t) ^2 = (100 + t) ^2
Q(100 + t) ^ 2 = ∫6 (100 + t) ^2 dt + C
Q(100 + t) ^2 = 2(100 + t) ^ 3 + C
When t = 0, Q = 50
Therefore,
50( 100) ^2 = 2(100) ^3 + C
C = -1.5 * 10 ^6
therefore, when t = 30,
Q (100 + 30) ^2 = 2(100 + 30) ^3 - 1.5 * 10 ^6
Q (400) ^2 = 2(130) ^3 - 1.5 * 10 ^6
Q = 171.24 Ibs
The amount of salt in the tank at the end of 30 minutes is 171.24 lbs.
The given parameters:
Initial volume of the tank, i = 100 gallonsRate of gain of salt = 3 gpmRate of loss of salt = 2 gpmThe linear differential equation of the salt solution is calculated as follows;
[tex]\frac{dx}{dt} = Gain - loss[/tex]
where;
x is the salt concentrationThe salt concentration at time t, is calculated as follows;
[tex]\frac{dx}{dt} = 2(3) - 2(\frac{X}{100 + t} )\\\\\frac{dx}{dt} = 6 - 2(\frac{X}{100 + t} )\\\\\frac{dx}{dt} +2(\frac{X}{100 + t} ) =6[/tex]
Apply the general solution of linear differential equation as follows;
[tex]X(f) = \int\limits {At} \, dt \ + C\\\\f = e^{\int\limits {At} \, dt}\\\\ f = e^{\int\limits {\frac{2}{100 + t} } \, dt}\\\\f = e^{2 ln(100 + t)}\\\\f = (100 + t)^2[/tex]
[tex]X(100 + t)^2 = \int\limits {6(100 + t)^2} \, dt \ + \ C\\\\ X(100 + t)^2 = 2(100 + t)^3 + C[/tex]
When t = 0 and X = 50
[tex]50(100 + 0)^2 = 2(100+ 0)^3 + C\\\\C = -1.5 \times 10^6[/tex]
When t = 30 min, the concentration is calculated as;
[tex]X (100 + 30)^2 = 2(100 + 30)^3- 1.5 \times 10^6\\\\X(130)^2 = 2(130)^3 - 1.5\times 10^6\\\\X(130)^2 = 2894000\\\\X = \frac{2894000}{130^2} \\\\X = 171.24 \ lbs[/tex]
Learn more about solution of Linear differential equation here: https://brainly.com/question/5508539
the ____ is a particle with one unit of positive change
a. proton
b. positron
c. electron
d. nucleus
Answer:
a proton because it has a positive charge
Answer:
The answer is
B)
Which two statements help explain why digital storage of data is so reliable?
A. Memory chips are sturdy.
U B. Digital data usually deteriorate over time.
C. It is usually possible to recover data from a memory chip even
when the device containing it is broken.
D. Digital data are easier to copy than analog data are, making them
more accessible to thieves.
Answer:
A. Memory chips are sturdy.
C. It is usually possible to recover data from a memory chip even when the device containing it is broken.
Explanation:
Digital storage of data refers to the process which typically involves saving computer files or documents on magnetic storage devices usually having flash memory. Some examples of digital storage devices are hard drives, memory stick or cards, optical discs, cloud storage, etc.
A reliable storage ensures that computer files or documents are easily accessible and could be retrieved in the event of a loss.
The two statements which help explain why digital storage of data is so reliable are;
A. Memory chips are sturdy: they are designed in such a way that they are compact and firm.
C. It is usually possible to recover data from a memory chip even when the device containing it is broken.
Answer:
A and C
Explanation:
got it right on a p e x
Which of the following is a part of both geocentric model and heliocentric model
Answer:
These planets rotate around the sun in a circular path. Likewise in a heliocentric model it is believed that the sun is at the center of the universe and the planet earth along with all other planet move around it. Thus in both geocentric model and heliocentric model bodies in space move in circular orbits.
Answer:
The bodies in space move in circular orbits
Explanation:
I got it right on my test
