Answer:
The options are
A) isochoric.
B) isothermal.
C) adiabatic.
D) isobaric.
The answer is C. Adiabatic
Adiabatic process involves zero loss or gain of heat in a system. This is usually depicted as Q= 0.
An ideal gas being compressed in a well-insulated chamber using a well-insulated piston involves the use of adiabatic process. The insulated chamber and piston helps to prevent heat loss or gain of heat. This is because insulators are poor conductors of heat and electricity.
The process of compressing an ideal gas in a well-insulated chamber using a well-insulated piston is known as adiabatic process.
When an ideal gas in compressed adiabatically, the heat lost is zero and the work is done it which causes increase its temperature.
Q = 0
The well-insulated chamber and piston helps to prevent heat loss by conduction and convection. The results in a zero heat lost to the surroundings.
Thus, we can conclude that the process of compressing an ideal gas in a well-insulated chamber using a well-insulated piston is known as adiabatic process.
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an rlc series circuit has a 40.0 resistor,a 3.00 mH inductor, and a 5.00 capacitor, (a find the circuits impendance at 60.0 Hz
Answer:
Z = 530.89 ohms
Explanation:
Given that,
Resistor, R = 40 ohms
Inductor, L = 3 mH
Capacitor, C = 5 μF
We need to find the impedance of the circuit at frequency of 60 Hz. For a series LCR circuit, its impedance is given by the formula as follows :
[tex]Z=\sqrt{R^2+(X_L-X_c)^2} \\\\Z=\sqrt{R^2+(2\pi fL-\dfrac{1}{2\pi fC})^2} \\\\Z=\sqrt{(40)^2+(2\pi \times 60\times 3\times 10^{-3}-\dfrac{1}{2\pi \times 60\times 5\times 10^{-6}})^2} \\Z=530.89\ \Omega[/tex]
So, the impedance of the circuit is 530.89 ohms.
A container of gas at 2.7 atm pressure and 133 ∘C is compressed at constant temperature until the volume is halved. It is then further compressed at constant pressure until the volume is halved again.
a) What is the final pressure (atm)?
b) What is the final temperature (C)?
Answer:
5.4atm ,66.5°c
Explanation:
Note that we are given the initial temperature to constant,
Thus PV = constant
so volume halved means pressure doubled from P1v1= p2v2
Hence P2 = 2.7*2atm = 5.4atm
next step we keep pressure constant thenV/T = constant
so volume halved means temperature also halved from V1/T1= v2/T2
Hence
T = 133/2 = 66.5°C
what is the significant contribution of Heraclitus
Explanation:
HERACLITUS ALSO SPELLED HERADEITUS, [ BORN C.540 bce, EPHESUS, ANATOLIA [ NOW SELCUK, TURKEY] - DIED C.486 ] GREEK PHILOSOPHER REMEMBERED FOR HIS COSMOLOGY, IN WHICH FIRE FORMS THE BASIC MATERIAL PRINCIPLE OF AN ORDERLY UNIVERSE . LITTLE IS KNOWN ABOUT HIS LIFE, AND THE ONE BOOK HE APPARENTLY WROTE IS LOST.
HOPE IT HELP.....❤❤A person drives a car around a circular cloverleaf with a radius of 77 m at a uniform speed of 10 m/s. What is the acceleration of the car?
Answer:
770m/s
Explanation:
caculation using one of the newton law of motion
A paper pinwheel is spinning in the wind. Which statement is correct about the forces responsible for the rotation?
A. The perpendicular components of gravity and the force of wind are responsible for the rotation.
B. Only the perpendicular component of wind is responsible for the rotation, because gravity points downward.
C. The components of gravity and the force of wind that point through the pivot are responsible for the rotation.
D. Only the perpendicular component of gravity is responsible for the rotation, because wind points toward the pivot.
Answer:
Only the perpendicular component of gravity is responsible for the rotation because wind points toward the pivot.
Explanation:
A proton accelerates from rest in a uniform electric field of 660 N/C. At one later moment, its speed is 1.30 Mm/s (nonrelativistic because v is much less than the speed of light). (a) Find the acceleration of the proton. m/s2 (b) Over what time interval does the proton reach this speed? s (c) How far does it move in this time interval? m (d) What is its kinetic energy at the end of this interval? J
Answer:
a) a = 6.31 10¹⁰ m / s² , b) t = 2.06 10⁻⁵ s , c) x = 13.39 m , d) ΔK = 1.41 10⁻¹⁵ J
Explanation:
a) Since they indicate that the speeds are non-relativistic, we can use the kinematics relations and Newton's second law
F = m a
The force in electrical is
F = qE
qE = m a
a = qE / m
we calculate
a = 1.6 10⁻¹⁹ 660 / 1.673 10⁻²⁷
a = 6.31 10¹⁰ m / s²
b) Let's use the one-dimensional kinematics relation
v = v₀ + a t
as part of rest its initial velocity is zero
v = a t
t = v / a
t = 1.30 10⁶ / 6.31 10¹⁰
t = 2.06 10⁻⁵ s
c) We use the kinematics displacement equation
x = v₀ t + ½ a t²
initial velocity is zero
x = ½ a t²
x = ½ 6.31 10¹⁰ (2.06 10⁻⁵)²
x = 1,339 10¹ m
x = 13.39 m
d) the kinetic energy is
ΔK = Kf -K₀
ΔK = ½ m v² - 0
ΔK = ½ 1.673 10⁻²⁷ (1.30 10⁶) 2
ΔK = 1.41 10⁻¹⁵ J
whos on the 1 dollar bills
Answer:
George Washington
Explanation:
The image of the famous first President of the United States-George Washington (from 1789 to 1797) is found on every $1 bill. George Washington's image began to be found on the $1 bill in 1869.
Where on the physical activity pyramid do sedentary activities belong?
Answer:A
Explanation:
i just did the test
Answer: A.
Explanation: Edge said it was right
Examples of sources of physical entropy for secure random number generation include:________
Answer:
Noise in the environment
Electrons emitted via the photoelectric effect
Explanation:
Physical entropy can be seen as a disorderliness that can be observed in a physical system. The noise in a physical system can be used for secure number generation because of its randomness.
Noise is known to have no defined pattern and hence, can be used to creates secure random key generations.
Electrons emitted via the photoelectric effect: Electrons are emitted in a random order whenever a beam of light is incident on a photoelectric material. These electrons emitted can be used for random key generations
A poundal is the force required to accelerate a mass of 1 lbm at a rate of 1 ft/s2 , and a slug is the mass of an object that will accelerate at a rate of 1 ft/s2 when subjected to a force of 1 lbf. (a) Calculate the mass in slugs and the weight in poundals of a 135 lbm woman (i) on earth and (ii) on the moon, where the acceleration of gravity is one-sixth of its value on earth. (b) A force of 405 poundals is exerted on a 35.0-slug object. At what rate (m/s2 ) does the object accelerate?
Answer:
We are given that
The
Mass of person = 135lbm
Weight of person on earth will be
= mass x gravity constant
= 135lbm x 32.174 ft/s2
=4343.49 lbm-ft/s2 x 1 poundal-s2/lbm-ft
= 4343.49 poundal
Again we are given that
Mass of person = 135 lbm
But here remember Mass remains the same
So
Weight of person on moon will be
mass x gravity constant on moon
= 135 lbm x 32.174 ft/s2 x 1/6
= 723.9 lbm-ft/s2 x 1 poundal-s2/lbm-ft
= 723.9poundal
But we know that
1 ft slug / lbf-s2 = 32.174 ft lbm/ lbf-s2
So
1 slug = 32.174 lbm
So then the Mass of person
= 135lbm x 1slug/32.174 lbm
=4.2slugs
So finally
Weight = 405 poundals is same as
405 lbm-ft/s2
So
Mass = 35 slug x 32.174 lbm/slug = 1126.09 lbm
Acceleration rate = weight /mass
= (405 lbm-ft/s2) / (1126.09 lbm)
= 0.3597ft/s2 x 0.305m/ft
= 1.179 m/s²
if abus travelling at 20m/s is subject to steady decceleration of 5m/s².how long will it take yo come to rest.
Answer:
4 seconds
Explanation:
Given:
v₀ = 20 m/s
v = 0 m/s
a = -5 m/s²
Find: t
v = at + v₀
0 m/s = (-5 m/s²) t + 20 m/s
t = 4 s
The tidal lung volume of human breathing, representing the amount of air inhaled and exhaled in a normal breath, is 500 cm3. (Assume atmospheric pressure.)(a) What is the number of molecules of air inhaled with each human breath when the air temperature is 18.0°C?________molecules(b) If the molar mass of air is 28.96 g/mol, what is the mass of air molecules inhaled with each breath? (Assume the air temperature is 18.0°C.)(c) It has been calculated that all of the air in Earth's atmosphere could be collected into a sphere of diameter 1,999 km at a pressure of 1.00 atm. What is the mass of the air in Earth's atmosphere? (Assume the density of air used in this calculation was 1.225 kg/m3.)(d) If all 7 billion humans on Earth inhaled simultaneously, what percentage of the atmosphere would be inhaled during this process? (Assume the air temperature is 18.0°C everywhere on Earth.)
Answer:
Explanation:
Temperature of air = 18°C = 273 + 18 = 291 K .
volume = 500 cc = 0 .5 litre .
pressure = one atmosphere ( atm) .
From gas equation , we can calculate this volume at NTP as follows.
volume = .5 x ( 273 / 291 ) litre
= 0.469 litre .
In any gas at NTP , 22.4 litre contains 6.02 x 10²³ molecules
.469 litre will contain 6.02 x 10²³ x .469 / 22.4 molecules
= 126 x 10²⁰ molecules .
b )
one mole = 6.02 x 10²³ molecules
6.02 x 10²³ molecules has weight of 28.96 grams
126 x 10²⁰ molecules has weight of 28.96 x 126 x 10²⁰ / 6.02 x 10²³ grams
= .606 gram .
c )
volume of all the air in the atmosphere = volume of sphere
= 4 / 3 x π x R³
= ( 4 / 3) x 3.14 x (999.5 x 10³ )³ m³
= 4.18 x 10¹⁸ m³
density of air = 1.225 kg / m³
mass of air = 1.225 x 4.18 x 10¹⁸ kg
= 5.12 x 10¹⁸ kg
d )
volume of air inhaled by 7 billion people
= . 5 x 7 x 10⁹ litre
= 3.5 x 10⁶ m³ .
Total volume of air in atmosphere = 4.18 x 10¹⁸ m³
required percentage
= 3.5 x 10⁶ x 100 / 4.18 x 10¹⁸
= .8373 x 10⁻¹⁰ % .
An optical fiber made of glass with an index of refraction 1.50 that is coated with a material with index of refraction 1.30 has a critical angle of
Answer:
A critical angle of 60.1°
Explanation:
Let's say
n1 Refractive index of rarer medium
n2 Refractive index of denser medium
So using the relation
စc= Sin^ -1(n1/n2)
So
စc = Sin^-1(1.3/1.5) = 60.1°
5. What occurs when an enzyme and a substrate interact at an
active site?
a. Activation energy is reduced.
b. The products are bound irreversibly.
c. The enzyme is changed by the reaction.
d. Activation energy is increased.
Answer:
B
Explanation:
The enzime's active site bindes to the substrate.... when an enzime binds to a substrate it forms a enzime-substrate complex
When an enzyme and a substrate interact at an active site, the products are bound irreversibly. Therefore, option B is correct.
What are enzymes?Enzymes can be described as proteins that act as biological catalysts by accelerating reactions. The enzymes may act on the molecules called substrates, and the enzyme changes the substrates into different molecules known as products.
All metabolic processes require enzyme catalysis to occur at rates fast enough to sustain life. Metabolic pathways will depend upon enzymes to catalyze each step.
Enzymes catalyze more than 5,000 biochemical reaction types. Enzymes enhance reaction rates by lowering their activation energy. Some enzymes make their conversion of substrate to product occurs.
Enzymes are much larger than substrates, they can have 62 amino acid residues to over 2,500 residues in the animal fatty acid synthase. The catalytic site is located next to binding sites where residues are oriented to the substrates. The catalytic site and binding site compose the active site.
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PHYSICS HW HELP PLS!! explain how you got it too thank you! :)
1.) F= Gm1m2/d2 for m2
2.) R= V/I for I
3.) = xd/L for d
4.) ac= v^2/r for v
5.) A= πr^2 for r
6.) vf^2= vi^2+2ax for x
Answer:
Explanation:
Can you please place the square roots in a proper manner so I may assist you?
A right triangle has a hypotenuse of 37 and a leg of 10. What is the second leg of the triangle?
Answer:
The second leg of the triangle is 35.62
Explanation:
Given that
Hypotenuse = 37
A leg = 10
Based on the above information
The second leg of the triangle is
According to the Pythagorean theorem
a^2 + b^2 = c^2
where,
a and b = legs
And, the c = hypotenuse.
Now put these values to the above formula
So,
10^2 + b^2 = 37^2
b^2 = 37^2 - 10^2
b^2 = 1269
b = sqrt(1269)
b = 35 .62
Hence, the second leg of the triangle is 35.62
The other leg is 35 mm.
Joey drives his Skidoo 13 kilometres north. He stops for lunch and then drives 10kilometres south. What distance did he cover? What was his displacement?
Answer:
Total distance covered (scalar quantity) = 23 km
Displacement (vector quantity) = 3 km north from the original starting point
Explanation:
Since he drove 13 km north and then 10 km south, the total distance he cover in his drive was: 13 km + 10 km = 23 km
On the other hand, his displacement was 3 km north from where he started.
If the charge remains the same but the radius of the sphere is doubled, the electric flux coming out of it will be
Answer:
Explanation:
We shall apply Gauss's theorem for electric flux to solve the problem . According to this theorem , total electric flux coming out of a charge q can be given by the following relation .
∫ E ds = q / ε
Here q is assumed to be enclosed in a closed surface , E is electric intensity on the surface so
∫ E ds represents total electric flux passing through the closed surface due to charge q enclosed in the surface .
This also represents total flux coming out of the charge q on all sides .
This is equal to q / ε where ε is a constant called permittivity which depends upon the medium enclosing the charge . For air , its value is 8.85 x 10⁻¹² .
If charge remains the same but radius of the sphere enclosing the charge is doubled , the flux coming out of charge will remain the same .
It is so because flux coming out of charge q is q / ε . It does not depend upon surface area enclosing the charge . It depends upon two factors
1 ) charge q and
2 ) the permittivity of medium ε around .
in a young's double-slit experiment the center of a bright fringe occurs wherever waves from the slits differ in the distance they travel by a multiple of
Answer:
The path difference will be zero.
Explanation:
Given that,
In a young's double-slit experiment the center of a bright fringe occurs wherever waves from the slits differ in the distance.
Suppose, we find the multiple of wavelength
In young's double slit experiment,
We need to calculate the multiple of wavelength
Using formula of path difference
[tex]\Delta x=d\sin\theta[/tex]....(I)
[tex]d\sin\theta=m\lambda[/tex]
Where, m = fringe number
[tex]\lambda[/tex] = wavelength
For center of a bright fringe ,
m = 0
Put the value into the formula
[tex]d\sin\theta=0\times\lambda[/tex]
[tex]d\sin\theta=0[/tex]
Put the value in equation (I)
[tex]\Delta x=0[/tex]
Hence, The path difference will be zero.
Add the following vectors using head-to-tail method and verify your results using the component method.
1. Vector A has a magnitude of 4.0 cms at an angle of 30 degrees with the positive X-axis; vector B has a magnitude of 3.0 cms at 90 degrees with + X-axis; vector C has a magnitude of 5.0 cms at 120 degrees with + X-axis.
2. In addition to vectors A, B, and C in problem # 1, there is a vector D with a magnitude of 6.0 cms at an angle of 210 degrees with +X-axis.
Note: Use a graph paper to show your work for problem number 1 and 2 using the head-to-tail method.
Do the component method analysis of both of these problems using the component method of adding vectors. Show your work step by step by taking the x and y components of all the vectors; adding all the x and y components together and then finding the resultant vector magnitude and direction knowing x and y are perpendicular to each other.
Compare the results of the graphical analysis as well as the component method and find the percent error using the component method as standard result
Answer:
See attached image for the requested graphs.
Part 1 : Vector sum is about 9.4 cm long in the graph, from components: about 9.35 cm long. Percent difference = 0.5%
Part 2: Vector sum is about 7.6 cm long in the graph and from components: 7.57 cm long. Percent difference = 0.4%
Explanation:
Part 1.
The graphical addition of the three vectors A, B and C gives a vector sum of approximately 9.4 cm at an angle of about 84 degrees
The component form addition is
Ax + Bx + Cx = 3.5 + 0 + (-2.5) = 1
Ay + By + Cy = 2.0 + 3 + 4.3 = 9.3
Magnitude: [tex]\sqrt{1^2+9.3^2} \approx 9.35\,\,cm[/tex]
The percent difference is: (9.4-9.35) * 100 /9.35 = 0.5%
Part 2.
The graphical addition of the four vectors (A, B, C, and D) measures approximately 7.6 cm at an angle of about 124 degrees.
The component form addition is
Ax + Bx + Cx + Dx = 3.5 + 0 + (-2.5) + (-5.2) = -4.2
Ay + By + Cy + Dy = 2.0 + 3 + 4.3 + (-3) = 6.3
Magnitude = [tex]\sqrt{(-4.2)^2+6.3^2} \approx 7.57\,\,cm[/tex]
The percent difference is: (7.6-7.57) * 100 /7.57 = 0.4%
1: A generator produces electricity at about 21,600 V. This voltage needs to be increased to about 345,000 V via a transformer before being transmitted through the power grid. What is the ratio of windings between the secondary coil and the primary coil needed to do this
Answer:
16:1
Explanation:
From the question,
Since the voltage of the transformer is increased, then the winding in the secondary coil is greater than the winding in the primary coil
Applying transformer's equation
Vs/Vp = Ns/Np............. Equation 1
Where Vs = Secondary voltage, Vp = primary Voltage, Ns/Np = Ratio of the winding between the secondary coil to the winding in the primary coil.
Given: Vs = 345000 V, Vp = 21600 V
Substitute these values into equation 1
Ns/Np = 345000/21600
Ns/Np = 16:1
A small metal ball is given a negative charge, then brought near to end A of an insulating rod. What happens to end A of the rod when the ball approaches it closely this first time
Answer:
Weak attraction will be observed
Explanation:
This is because the charge induction is less than the insulating rod, and a postive charge is induced when a a negative sphere is placed closed to the end of the rod
Draw the velocity vectors starting at the black dots and the acceleration vectors including those equal to zero. The orientation of the vectors will be graded. The location and length of the velocity vectors will be graded. The location and length of the nonzero acceleration vectors will not be graded. To draw a zero vector click at the point of its location.
Answer:
hello your question is incomplete attached below is the remaining part and the solution
Explanation:
The accelerations at points 1,2,3 is zero and this is because the velocities at points 1,2,3 are constants. attached below is the velocity vector diagram and the acceleration vectors
A wheel rotating about a fixed axis has a constant angular acceleration of 4.0 rad/s2. In a 4.0-s interval the wheel turns through an angle of 80 radians. Assuming the wheel started from rest, how long had it been in motion at the start of the 4.0-s interval?
a. 3 s.
b. 6 s.
c. 9 s.
d. 12 s.
Answer:
a. 3 s.
Explanation:
Given;
angular acceleration of the wheel, α = 4 rad/s²
time of wheel rotation, t = 4 s
angle of rotation, θ = 80 radians
Apply the kinematic equation below,
[tex]\theta = \omega_1 t \ + \ \frac{1}{2} \alpha t^2\\\\80 = 4\omega_1 + \frac{1}{2}*4*4^2\\\\80 = 4\omega_1 + 32\\\\ 4\omega_1 = 48\\\\ \omega_1 = \frac{48}{4}\\\\ \omega_1 = 12 \ rad/s[/tex]
Given initial angular velocity, ω₀ = 0
Apply the kinematic equation below;
[tex]\omega_1 = \omega_o + \alpha t_1\\\\12 = 0 + 4t\\\\4t = 12\\\\t = \frac{12}{4}\\\\t = 3 \ s[/tex]
Therefore, the wheel had been in motion for 3 seconds.
a. 3 s.
Acceleration: A water rocket can reach a speed of 75 m/s in 0.050 seconds from launch. What is its average acceleration
Answer:
1500 m/s squared
Hope this helps :)
Explanation:
A string is waved up and down to create a wave pattern with a wavelength of 0.5 m. If the waves are generated with a frequency of 2 Hz, what is the speed of the wave that travels through the string to the other end?
Wave speed = (wavelength) x (frequency)
Wave speed = (0.5 m) x (2 /sec)
Wave speed = 1 m/s
a train it speed steading from 10 m/s to 20m/s in 1 minutes average speed during this time in m/s
how far does it travel while increase its speed
Explanation:
Given:
v₀ = 10 m/s
v = 20 m/s
t = 60 s
Find: v_avg and Δx
v_avg = ½ (v + v₀)
v_avg = ½ (20 m/s + 10 m/s)
v_avg = 15 m/s
Δx = v_avg t
Δx = (15 m/s) (60 s)
Δx = 900 m
A resistance of 4 ohm is offered by a conductor when a potential difference of 6 V l'd applied across it. Calculate the current through it?.
Answer:
1.5 Ampere
Explanation:
The formula is V=IR
so now we can derive out the formula which is
I =V
R
I = 6
4
I = 1.5 Ampere
A 1700 W laser emits light with a wavelength of 700 nm into a 3.0 mm diameter beam. What force does the laser beam exert on a completely absorbing target?
Answer:
F=5.7×10⁻⁶
Explanation:
Not knowing a formula outright, I decided to follow the units of some relationships I did know. Radiation pressure is defined as force per area and also intensity divided by velocity (the speed of light here of course). Breaking intensity down into power per area and isolating force gave me the relationship F=(Power/Velocity),where power is given and the velocity is a constant.
My work is in the attachment, where I double checked the units too, comment with any questions.
The force exerted by the beam on a completely absorbing target is of 0.0120 N.
Given data:
The power of laser beam is, P = 1700 W.
The wavelength of light emitted from the laser is, [tex]\lambda = 700 \;\rm nm[/tex].
The diameter of the beam is, [tex]d =3.00 \;\rm mm =3.00 \times 10^{-3} \;\rm m[/tex].
When the laser beam strikes the target, then the force exerted by the laser beam on the target is equal to the product of the laser power and the area of projection of beam.
Then the expression is,
[tex]P = \dfrac{F}{A}\\\\F = P \times \dfrac{\pi}{4} d^{2}[/tex]
Solving as,
[tex]F = 1700 \times \dfrac{\pi}{4} \times (3.00 \times 10^{-3})^{2}[/tex]
F = 0.0120 N
Thus, we can conclude that the force exerted by the beam on a completely absorbing target is of 0.0120 N.
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(A) What is the maximum tension possible in a 1.00-millimeter-diameter nylon tennis racket string?
(B) If you want tighter strings, what do you do to prevent breakage: use thinner or thicker strings? Why? What causes strings to break when they are hit by the ball?
Complete Question
(A) What is the maximum tension possible in a 1.00- millimeter-diameter nylon tennis racket string?
(B) If you want tighter strings, what do you do to prevent breakage: use thinner or thicker strings? Why? What causes strings to break when they are hit by the ball?
The tensile strength of the nylon string is [tex]600*10^{6} \ N/m^2[/tex]
Answer:
A
T = 471.3 \ N
B
To prevent breakage the thickness of the string is increased
String breakage when the racket hit the ball is as a result of the string not being thick enough to withstand the increase in tension
Explanation:
From the question we are told that
The diameter is [tex]d = 1.00 \ mm = 0.001 \ m[/tex]
The tensile strength of the nylon string is [tex]\sigma = 600 *10^{6} \ N/m^2[/tex]
Generally the radius is mathematically evaluated as
[tex]r= \frac{d}{2}[/tex]
=> [tex]r = \frac{0.001}{2}[/tex]
=> [tex]r = 0.0005 \ m[/tex]
The cross sectional area is mathematically represented as
[tex]A = \pi r^2[/tex]
=> [tex]A = 3.142 * (0.005)^2[/tex]
=> [tex]A = 7.855*10^{-7}\ m^2[/tex]
Generally the tensile strength of nylon is mathematically represented as
[tex]\sigma = \frac{T}{ A }[/tex]
Where T is the tension on the maximum tension on the string
So
[tex]T = \sigma * A[/tex]
=> [tex]T = 600*10^{6} * 7.855*10^{-7}[/tex]
=> [tex]T = 471.3 \ N[/tex]
Form the equation above we see that
[tex]T \ \alpha \ A[/tex]
So if the tension is increased to prevent breakage the thickness of the string is increased(i. e the cross-sectional area )
String breakage when the racket hit the ball is as a result of the string not being thick enough to withstand the increase in tension