Answer:
d
Explanation:
3.
What part of your eye is responsible for regulating the amount of light that enters your eye?
Answer:
Iris
Explanation:
The iris seems to be the illuminated portion of the eyes which really covers the pupil. It controls the amount of light reaching the eye. The lens is indeed a translucent layer of the retina that serves to concentrate light and objects on the lens.
Answer:
I hope this helps.
Explanation:
Please help! Will mark brainliest.
Answer:
1122.8
Explanation:
12.73 kg x 9.8 m/s^2 x 9m
=1122.786
Rounded=1122.8
Review Conceptual Example 8 before starting this problem. A block is attached to a horizontal spring and oscillates back and forth on a frictionless horizontal surface at a frequency of 3.96 Hz. The amplitude of the motion is 5.95 x 10-2 m. At the point where the block has its maximum speed, it suddenly splits into two identical parts, only one part remaining attached to the spring. (a) What is the amplitude and (b) the frequency of the simple harmonic motion that exists after the block splits
Answer:
a) A' = 0.345 m, b) f = 2,800 Hz
Explanation:
b) The angular velocity of a simple harmonic motion is
w =[tex]\sqrt{\frac{k}{m} }[/tex]
angular velocity and frequency are related
w = 2π f
we substitute
f = 1 /2π √k/m
indicates that the initial frequency value f = 3.96 Hz
in this case the mass is reduced by half
m ’= m / 2
we substitute
f = 2π [tex]\sqrt{\frac{k}{m} }[/tex]
f = √1/2 (2π √k/m)
f = 1 /√2 3.96
f = 2,800 Hz
a) The amplitude of the movement is defined by the value of the initial depalzamienot before an external force that initiates the movement.
When the block is divided into two parts of equal masses as if it were exploding, for which we can use the conservation of moment
initial instant. Right before the division
p₀ = (m₁ + m₁) v
final instant. Right after the split
p_f = m₁ v '
p₀ = p_f
(2 m₁) v = m₁ v ’
v ’= 2v
At this point we can use conservation of energy for the system with only half the block.
Starting point. Where the block divides
Em₀o = K = ½ m v'²
Final point. Point of maximum elongation
Em_f = Ke = ½ k A²
how energy is conserved
Em₀ = Em_f
½ m’ v’² = ½ k A’²
we substitute the previous expressions
½ m/2 (2v)² = ½ k A’²
A’² = 2 m v² / k (1)
Let's use the conservation of energy with the initial conditions, before dividing the block
½ m v2 = ½ k A2
A² = mv² / k = 5.95 10⁻² m²
we substitute in 1
A'² = 2 A²
A ’²= 2 5.95 10⁻²
A ’²= 11.9 10⁻² m
A' = 0.345 m
Specify whether the boiling point, as determined in the miniscale boiling-point apparatus, is the temperature a.of the liquid at the timebubbles first emerge slowly from the liquid. b.at the vapor-liquid interface above the surface of the boiling liquid while a drop of liquid c.is suspended from the thermometer. d.of the liquid at the timebubbles emerge rapidly from the liquid. e.of the heating source at the timebubbles emerge rapidly from the liquid.
Answer:
a. of liquid at the time bubbles first emerge slowly from the liquid.
Explanation:
Boiling point of liquid happens due to heat energy. This is an exothermic reaction as heat is released in to the environment. The initial boiling vapors slowly move away from the liquid and as the temperature increases the vapors start moving quickly.
2) The track for a racing event was designed so that riders jump off the slope at 37 degrees from a height of 1 m. During a race it was observed that the rider remained in mid air for 1.5 seconds. Determine the speed at which he was traveling off the slope, the horizontal distance he travels before striking the ground and the maximum height he attains. Neglect the size of the bike and rider.
Answer:
3.277 m
Explanation:
Given :
Maximum Height (Hmax) = (u²sin²θ) / 2g
Xv = Xh + Uv * t + 0.5gt²
Xv and Xh are vertical and horizontal distances
-1 = 0 + sin37 * 1.5 Uv + 0.5*-9.8*1.5^2
-1 = 0 + 0.903Uv - 11.025
-1 + 11.025 = 0.903Uv
10.025 = 0.903Uv
Uv = 10.025 / 0.903
Uv = 11.10 m/s
Hmax = 1 + (u²sin²θ) / 2g
= (11.10^2 * (sin37)^2) / 2*9.8
= 44.624360 / 19.6
= 2.277
Hmax = 1 + 2.277
Hmax = 3.277 m
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
A plastic cup weighing 100 g floats on water so that 1/4 of the volume of the cup is immersed in water. How much volume of oil can be poured into the glass to keep it still sinking? The density of the oil is 900 kg / m
Answer:
Any floating object displaces a volume of water equal in weight to the object's MASS. ... If you place water and an ice cube in a cup so that the cup is entirely full to the ... If you take a one pound bottle of water and freeze it, it will still weigh one ... Fresh, liquid water has a density of 1 gram per cubic centimeter (1g = 1cm^3, ...
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)
What is the correct definition of amplitude
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
. [30%] We first showed that The electric field for a point charge radiating in 3-dimensions has a distance dependence of 1/r 2 (see Equation 1). In Problem 1 you showed that the electric field for a point charge radiating in 2-dimensions has a distance dependence of 1/r . Consider again the 2-dimensional case described in Problem 1. What distance dependence do you expect for the electric potential
Answer:
Answer is explained in the explanation section below.
Explanation:
Note: This question is incomplete and lacks necessary data to solve. As it mentioned the reference of problem number 1, which is missing in this question. However, I have found that question on the internet and will be solving the question accordingly.
Solution:
The relation between electric field and the electric potential is:
E = [tex]\frac{dV}{dr}[/tex]
So, making dV the subject, we have:
dV = E x dr
Integrating the above equation, we get.
V = [tex]\int\limits^_ {} \,[/tex]E x dr Equation 1
Now, in 2-D
E is inversely proportional to the radius r.
E ∝ 1/r
So, we can write: replacing E ∝ 1/r in the equation 1
V ∝ [tex]\int\limits^_ {} \,[/tex][tex]\frac{1}{r}[/tex] x dr
Which implies that,
V ∝ log (r)
Hence, distance dependence expected for the electric potential = ln (r)
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!
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
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
Two train whistles have identical frequencies of 180 Hz. When one train is at rest in the station and the other is moving nearby, a commuter standing on the train platform hears beats with a frequency of 2.00 beats/s when the whistles sound at the same time. What are the two possible speeds and directions that the moving train can have?
Actual answers :3.85 m/s away from the station and 3.77 m/s towards the station from the book. I just need to know how to get to the answers.
Answer:
-3.77 m/s
3.85 m/s
Explanation:
given that
Frequency at stationary = 180 Hz
Beat frequency = 2 Hz
Using Doppler effect, we know that
f' = f[(v ± v0) / (v ± vs)], where
v = speed of sound, 343 m/s
v0 = speed of the observer, 0
vs = speed of light, ?
f = stationary frequency, 180 Hz
f' = stationary ± beat frequency, 180 ± 2
Applying the formula, we have
f' = f[(v ± v0) / (v ± vs)]
182 = 180 [(343 + 0) / (343 + vs)]
182/180 = 343 / 343 + vs
343 + vs = 343 * 180/182
343 + vs = 339.23
vs = 339.23 - 343
vs = -3.77 m/s
Again, using
f' = f[(v ± v0) / (v ± vs)]
178 = 180 [(343 + 0) / (343 + vs)]
178/180 = 343 / 343 + vs
343 + vs = 343 * 180/178
343+ vs = 346.85
vs = 346.85 - 343
vs = 3.85 m/s
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. :)
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
