Answer:
Copper would experience induced magnestism mostly easily.
Explanation:
Permalloy would experience induced magnetism most easily.
I don't know the options but that is correct too!!
In higher mass stars, repeating cycles of fusion will create heavier elements in layers
until which element is created at the center of the core?
hydrogen
iron
uranium
helium
Write the relation connecting Celsius scale and Fahrenheit scale of temperature
Answer: See explanation
Explanation:
Celsius and Fahrenheit are the scales that are used in the measurement of temperature. Celcius is also refered to as centigrade. The relation that exist between Celsius and Fahrenheit is typically proportional.
The conversion from Celsius to Fahrenheit is expressed as:
F = (9/5 × C) + 32.
The conversion from Fahrenheit to Celsius is expressed as:
C = 5/9(F - 32)
For example to convert 100°C to Fahrenheit will be:
F = (9/5 × C) + 32.
F = (9/5 × 100) + 32
F = 180 + 32
F = 212°F
the velocity of a ship in the unit of m/s moving initially along west is given by V(t) = 5-2t. It’s acceleration at t=1 s is given by:
1. 0 m/s^2
2. 2m/s^2 along west
3. 2m/s^2 along east
4. None
Whis one is correct?
Answer:
4. None
Explanation:
Applying,
a(t) = dV(t)/dt
Where a(t) = Acceleration of the ship at a given time.
From the question,
Given: V(t) = 5-2t
Therefore,
a(t) = dV(t)/dt = 5 m/s²
Hence it's acceleration is 5 m/s²
From the question,
The right option is 4. None
how many continents do have in africa
Answer:
There was 7 continents in africa
Testing shows that a sample of wood from an artifact contains 50% of the original amount of carbon-14. Given that the half-life of carbon-14 is 5730
years, how old is the artifact?
A. 22,920 years
B. 11,460 years
C. 5730 years
D. 2865 years
Answer:
C. 5730 years
Explanation:
N(t) = N(0)e^-kt
The half-life is T = 5730 years,
e^-kT = 1/2
→ k = - ln(1/2) / T
→ - ln(1/2) / 5730
→ 1.209681 x 10^-4 years^-1
The amount present dropped to 50%.
Then one half-life has elapsed, so the age is 5730 years.
PLEASE HELP ME WITH THIS ONE QUESTION
The half-life of Barium-139 is 4.96 x 10^3 seconds. A sample contains 3.21 x 10^17 nuclei. How much of the sample is left after 1.98 x 10^4 seconds?
A) 8.03 x 10^16 nuclei
B) 4.01 x 10^16 nuclei
C) 2.02 x 10^16 nuclei
D) 1.61 x 10^17 nuclei
OPTION C is the correct answer.
The radioactive decay follows the first order kinetics. The number of atoms decaying at any time is proportional to the number of atoms present at that instant. The amount of sample left is 2.02 x 10¹⁶nuclei. The correct option is C.
What is half-life?The time required for the decay of one half of the amount of the species is defined as the half-life period of a radionuclide. The half-life period is a characteristic of a radionuclide. The half lives can vary from seconds to billions of years.
The isotope decay of an atom is given by the equation:
ln [A] = -kt + ln [A]₀
The rate constant, k is:
k = ln 2 / Half life
k = ln 2 / 4.96 x 10³
k = 1.40 × 10⁻⁴ s⁻¹
t = 1.98 x 10⁴
[A]₀ = 3.21 x 10¹⁷
ln [A] = -1.40 × 10⁻⁴ × 1.98 x 10⁴ + ln [3.21 x 10¹⁷] = 37.538
[A] = 2.02 x 10¹⁶ nuclei
Thus the correct option is C.
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need in hurry please
How can magnetic levitation be improved?
Assume that the energy lost was entirely due to friction and that the total length of the PVC pipe is 1 meter. Use this length to compute the average force of friction (for this calculation, you may neglect uncertainties).
The question is incomplete. The complete question is :
Assume that the energy lost was entirely due to friction and that the total length of the PVC pipe is 1 meter. Use this length to compute the average force of friction (for this calculation, you may neglect uncertainties).
Mass of the ball : 16.3 g
Predicted range : 0.3503 m
Actual range : 1.09 m
Solution :
Given that :
The predicted range is 0.3503 m
Time of the fall is :
[tex]$t=\sqrt{\frac{2H}{g}}$[/tex]
[tex]v_1t= 0.35[/tex] ...........(i)
[tex]v_0t= 1.09[/tex] ...........(ii)
Dividing the equation (ii) by (i)
[tex]$\frac{v_0t}{v_1t}=\frac{1.09}{035} = 3.11$[/tex]
∴ [tex]v_0=3.11 \ v_1[/tex]
Now loss of energy = change in the kinetic energy
[tex]$W=\frac{1}{2} m [v_0^2-v_1^2]$[/tex]
[tex]$W=\frac{1}{2} \times (16.3 \times 10^{-3}) \times [v_0^2-\left(\frac{v_0}{3.11}\right)^2]$[/tex]
[tex]$W=7.307\times 10^{-3} \ v_0^2$[/tex]
If f is average friction force, then
(f)(L) = W
(f) (1) = [tex]$7.307\times 10^{-3} \ v_0^2$[/tex]
(f) = [tex]$7.307\times 10^{-3} \ v_0^2$[/tex]
The Average force of friction is ( F ) = 7.307 * 10⁻³ v₀²
Given data:
Predicted range ( v₁t ) = 0.3503 m
Actual range ( v₀t ) = 1.09 m
mass = 16.3 g
First step : Determine the value of V₀
[tex]t = \sqrt{\frac{2H}{g} }[/tex] , v₁t = 0.3503 , ( v₀t ) = 1.09 m
To obtain the value of V₀
Divide ( v₀t ) by ( v₁t ) = 1.09 / 0.3503 = 3.11 v₁
∴ V₀ = 3.11 v₁
Next step : Determine the average force of friction ( f )
given that loss of energy results in a change in kinetic energy
W = [tex]\frac{1}{2} m ( vo^{2} - v1^{2} )[/tex]
= 1/2 * 16.3 * 10⁻³ * [ v₀² - [tex](\frac{v_{0} }{3.11} )^{2}[/tex] ]
∴ W = 7.307 * 10⁻³ v₀²
Average force of friction = W / Actual length
= 7.307 * 10⁻³ v₀² / 1
∴ Average force of friction ( F ) = 7.307 * 10⁻³ v₀²
Hence we can conclude that the average force of friction is 7.307 * 10⁻³ v₀²
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Your question has some missing data below are the missing data related to your question
Mass of the ball : 16.3 g
Predicted range : 0.3503 m
Actual range : 1.09 m
In a certain region of space near earth's surface, a uniform horizontal magnetic field of magnitude B exists above a level defined to be y = 0. Below y = 0 , the field abruptly becomes zero (seethe figure). A vertical square wire loop has resistivity rho mass density rhom, diameter d, and side length l. It is initially at rest with its lower horizontal side at y = 0 and is then allowed to fall under gravity, with its plane perpendicular to the direction of the magnetic field.
a) While the loop is still partially immersed in the magnetic field (as it fallsinto the zero-field region), determine the magnetic "drag" forcethat acts on it at the moment when its speed is v.
b) Assume that the loop achieves a terminal velocity vt before its upper horizontal side exits the field. Determine a formulafor vt
c) If the loop is made of copper and B = 0.80 T find vt
Answer:
a) F = [tex]\frac{\pi d^2B^2lv}{16p}[/tex]
b) attached below
c) 0.037 m/s
Explanation:
a) Determine the magnetic "drag" force acting at the moment
speed = v
first step: determine current in the loop
I = [tex]\frac{\pi d^2}{16pl} B lv[/tex] ----- ( 1 )
given that the current will induce force on the three sides of the loop found in the magnetic field
forces on vertical sides = + opposite
we will cancel out
hence equation 1 becomes
F = [tex]\frac{\pi d^2B^2lv}{16p}[/tex] ( according to Lenz law we can say that the direction of force is upwards and this force will slow down the decrease in flux )
b) Determine the formula for Vt
attached below
c) Find Vt
given :
B = 0.80 T
density of copper = 8.9 * 10^3 kg/m^3
resistivity of copper = 1.68 * 10^-8 Ωm
∴ Vt = 16 ( 8.9 * 10^3 kg/m^3 ) ( 1.68 * 10^-8 Ωm ) ( 9.8 m/s^2 ) / ( 0.08 T)^2
= 0.037 m/s
Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.57 A out of the junction. The current of wire 2 is 0.65 A out of the junction.
Required:
a. How many electrons per second move past a point in wire 3?
b. In which direction do the electrons move -- into or out of the junction?
Answer:
a. 1.56 × 10¹⁸ electrons per second
b. The electrons in wire 3 flow into the junction.
Explanation:
Here is the complete question
Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.65 A out of the junction. (a) How many electrons per second move past a point in wire 3? (b) In which direction do the electrons move in wire 3 -- into or out of the junction?
Solution
(a) How many electrons per second move past a point in wire 3?
Using Kirchhoff's current law, at the junction, i₁ + i₂ + i₃ = 0 where i₁ = current in wire 1 = 0.40 A, i₂ = current in wire 2 = 0.65 A and i₃ = = current in wire 3,
So, i₃ = -(i₁ + i₂)
taking current flowing into the junction as positive and those leaving as negative, i₁ = + 0.40 A and i₂ = -0.65 A
So, i₃ = -(i₁ + i₂)
i₃ = -(0.40 A + (-0.65 A))
i₃ = -(0.40 A - 0.65 A)
i₃ = -(-0.25 A)
i₃ = 0.25 A
Since i₃ = 0.25 C/s and we have e = 1.602 × 10⁻¹⁹ C per electron, then the number of electrons flowing in wire 3 per second is i₃/e = 0.25 C/s ÷ 1.602 × 10⁻¹⁹ C per electron = 0.1561 × 10¹⁹ electrons per second = 1.561 × 10¹⁸ electrons per second ≅ 1.56 × 10¹⁸ electrons per second
(b) In which direction do the electrons move -- into or out of the junction?
Given that i₃ = + 0.25 A and that positive flows into the junction, thus, the electrons in wire 3 flow into the junction.
Can someone help me
Answer:
Explanation:before the phase change the substance is a particle.
A Michelson interferometer operating at a 400 nm wavelength has a 3.70-cm-long glass cell in one arm. To begin, the air is pumped out of the cell and mirror M2 is adjusted to produce a bright spot at the center of the interference pattern. Then a valve is opened and air is slowly admitted into the cell. The index of refraction of air at 1.00 atm pressure is 1.00028.
How many bright-dark-bright fringe shifts are observed as the cell fills with air?
Answer:
[tex]m=42\ fringes[/tex]
Explanation:
From the question we are told that:
Wavelength [tex]\lambda=400nm[/tex]
Length of cell arm [tex]h=3.70cm[/tex]
Refraction of air at at 1.00 atm pressure [tex]n=1.00028.[/tex]
Generally the equation for Number of shifts is mathematically given by
[tex]m=N-N_o[/tex]
Since
[tex]N_0=\frac{2t}{\lambda_0}[/tex]
Therefore
[tex]m=\frac{2t}{\lambda_0/n}-\frac{2t}{\lambda_0}[/tex]
[tex]m=\frac{2t}{\lambda_0} n-1[/tex]
[tex]m=\frac{2(3.7*10^{-2})}{400*10^{-9}}*(1.00028-1)[/tex]
[tex]m=51.8[/tex]
[tex]m=42\ fringes[/tex]
how do you use the coefficient to calculate the number of atoms in each molecule?
Answer:
To find out the number of atoms: MULTIPLY all the SUBSCRIPTS in the molecule by the COEFFICIENT. (This will give you the number of atoms of each element.)
Explanation:
A laser beam enters one of the sloping faces of the equilateral glass prism (n=1.42) and refracts through the prism. Within the prism the light travels horizontally. What is the angle between the direction of the incident ray and the direction of the outgoing ray?
Answer:
30.5°
Explanation:
Since the light travels horizontally through the prism, it undergoes minimum deviation. So, the angle between the direction of the incident ray and that of the outgoing ray D is gotten from
n = [sin(D + α)/2]/sin(α/2) where n = refractive index of prism = 1.42 and α = angle of prism = 60° (since it is a n equilateral glass prism).
Making D subject of the formula, we have
n = [sin(D + α)/2]/sin(α/2)
nsin(α/2) = [sin(D + α)/2]
(D + α)/2 = sin⁻¹[nsin(α/2)]
D + α = 2sin⁻¹[nsin(α/2)]
D = 2sin⁻¹[nsin(α/2)] - α
So, substituting the values of the variables into the equation, we have
D = 2sin⁻¹[nsin(α/2)] - α
D = 2sin⁻¹[1.42sin(60°/2)] - 60°
D = 2sin⁻¹[1.42sin(30°)] - 60°
D = 2sin⁻¹[1.42 × 0.5] - 60°
D = 2sin⁻¹[0.71] - 60°
D = 2(45.23°) - 60°
D = 90.46° - 60°
D = 30.46°
D ≅ 30.5°
A capacitor is constructed of two large, identical, parallel metal plates separated by a small distance d. A battery fully charges the capacitor and is then disconnected. The plate separation is now increased to a distance of 2d. What would be the change, if any, of the voltage across the capacitor, the electric field between the plates, and the energy stored in the capacitor?
Answer:
The answer is "Option D".
Explanation:
Please find the complete question in the attached file.
As plate separation increased to 2d the capacitance get halred but the change remain same
[tex]\therefore V=\frac{Q}{C}[/tex]
The voltage doubles are now electric field remain same because both the distance and voltage get doubled.
[tex]\to E=\frac{v}{d}\ = \frac{2v}{2d}\\\[/tex]
So,
[tex]energy=\frac{1}{2}\ \frac{Q^2}{C}\\\\c'=\frac{C}{2}\\\\E'=2E[/tex]
What effect does increased blood flow have on the body when performing exercises? A. delivers more sugar to organs B. delivers more energy to muscles C. delivers more oxygen to the body D. delivers more protein to muscles Please select the best answer from the choices provided. O A . OB ос OD Next Submit Save and Exit Mark this and return
The bus travelled at velocity 15 ms-l for 5 minutes before it came to a stop. By using suitable linear equation, calculate the distance the bus has travelled.
Answer:
ans: 2250 meters
Explanation:
initial velocity (U)= 15 m/s
final velocity (V) = 0m/s , since need to come in rest
total time taken (T) = 5 min= 300 seconds
total distance covered (S)= UT + 1/2 aT^2 ,
a= acceleration
S= 15 × 300 + 0.5 ×(0 - 15) × 300
since a = (V - U)/ T
S = 4500 - 2250
S= 2250 m
A magnetic field of 0.276 T exists in the region enclosed by a solenoid that has 517 turns and a diameter of 10.5 cm. Within what period of time must the field be reduced to zero if the average magnitude of the induced emf within the coil during this time interval is to be 12.6 kV
Answer:
The period the field must be reduced to zero is 9.81 x 10⁻⁵ s
Explanation:
Given;
initial value of the magnetic field, B₁ = 0.276 T
number of turns of the solenoid, N = 517 turns
diameter of the solenoid, d = 10.5 cm = 0.105 m
induced emf, = 12.6 kV = 12,600 V
when the field becomes zero, then the final magnetic field value, B₂ = 0
The induced emf is given by Faraday's law;
[tex]emf = -\frac{NA\Delta B}{t} \\\\emf = -\frac{NA (B_2 -B_1)}{t} \\\\t = -\frac{NA (B_2 -B_1)}{emf}\\\\t = \frac{NA (B_1 -B_2)}{emf}\\\\where;\\\\t \ is \ the \ time \ when \ B = 0 \ \ (i.e\ B_2 = 0)\\\\A \ is \ the \ area \ of \ the \ coil\\\\A = \frac{\pi d^2}{4} = \frac{\pi (0.105)^2}{4} = 0.00866 \ m^2\\\\t= \frac{(517) \times (0.00866)\times (0.276 -0)}{12,600}\\\\t = 9.81 \times 10^{-5} \ s[/tex]
Therefore, the period the field must be reduced to zero is 9.81 x 10⁻⁵ s
For a different power plant the power input (not output) is 3570 MW. The coal has an energy content of 28,000 kJ per kg. How much coal must be put into the plant each day
Answer:
31475404.8 kg/day
Explanation:
From the information given:
The power plant capacity W = 3570 MW
Energy content of the coal = 28000 kJ/kg
let assume that the thermal efficiency = 35%
Recall that:
1 kw = 3600 kJ/hr provided that the energy conversion is 100% efficient
But assuming the thermal efficiency = 35%.
Then:
Heat input = 3600/0.35 = 10286 kJ/kw.hr
Now, for producing 1 kw.hr, the quantity of the required coal = 10286/28000
= 0.36736 kg
For 3570 MW, the amount of coal that must be input is:
= 0.36736 kg × 3570000
= 1311475.2 kg/hr
= 1311475.2 × 24 kg/day
= 31475404.8 kg/day
The following 1H NMR absorptions were obtained on a spectrometer operating at 200 MHz and are given in Hz downfield from TMS. Convert the absorptions to δ units. a) 416 Hz = δ b) 1.97×103 Hz = δ c) 1.50×103 Hz = δ
Answer:
For (a): The chemical shift is [tex]2.08\delta[/tex]
For (b): The chemical shift is [tex]9.85\delta[/tex]
For (c): The chemical shift is [tex]7.5\delta[/tex]
Explanation:
To calculate the chemical shift, we use the equation:
[tex]\text{Chemical shift in ppm}=\frac{\text{Peak position (in Hz)}}{\text{Spectrometer frequency (in MHz)}}[/tex]
Given value of spectrometer frequency = 200 MHz
For (a):Given peak position = 416 Hz
Putting values in above equation, we get:
[tex]\text{Chemical shift in ppm}=\frac{416Hz}{200MHz}\\\\\text{Chemical shift in ppm}=2.08\delta[/tex]
For (b):Given peak position = [tex]1.97\times 10^3 Hz[/tex]
Putting values in above equation, we get:
[tex]\text{Chemical shift in ppm}=\frac{1.97\times 10^3Hz}{200MHz}\\\\\text{Chemical shift in ppm}=9.85\delta[/tex]
For (c):Given peak position = [tex]1.50\times 10^3 Hz[/tex]
Putting values in above equation, we get:
[tex]\text{Chemical shift in ppm}=\frac{1.50\times 10^3Hz}{200MHz}\\\\\text{Chemical shift in ppm}=7.5\delta[/tex]
please help very easy 5th grade work giving brainliest
Answer:
the answer is option B because opposit sides of the magnets attract each other
a car is moving at 56 miles per hour the kinetic energy of that car is how much energy does that same car have when it moves at
Complete Question
A car is moving at 56 miles per hour. The kinetic energy of that car is 5 × 105 J. How much energy does the same car have when it moves at 90 miles per hour? Answer in units of J
Answer:
[tex]K.E_2=640000J[/tex]
Explanation:
From the question we are told that:
Initial Velocity [tex]v_1=56 mil/h=>25.0342m/s[/tex]
[tex]K.E_1=5*10^5J[/tex]
Final Velocity [tex]V_2=90mil/h=>40.2336m/s[/tex]
Generally the equation for Kinetic Energy is mathematically given by
[tex]K.E=0.5mv^2[/tex]
Therefore
[tex]K.E_1=0.5mv_1^2[/tex]
[tex]5*10^5J=m(25.0342)^2[/tex]
[tex]m=797.82kg[/tex]
Therefore The Final K.E_2 is
[tex]K.E_2=0.5mv_2^2[/tex]
[tex]K.E_2=0.5*(797.82kg)(40.2336m/s)^2[/tex]
[tex]K.E_2=640000J[/tex]
A dog accelerates at 1.50 m/s2 to reach a velocity of 13.5 m/s while covering a distance of 49.3 m. What was his initial velocity?
Let v be the dog's initial velocity. Then
(13.5 m/s)^2 - v ^2 = 2 (1.50 m/s^2) (49.3 m)
==> v ^2 = (13.5 m/s)^2 - 2 (1.50 m/s^2) (49.3 m)
==> v = √((13.5 m/s)^2 - 2 (1.50 m/s^2) (49.3 m))
==> v ≈ 5.86 m/s
An electric device, which heats water by immersing a resistance wire in the water, generates 20 cal of heat
per second when an electric potential difference of 6 V is placed across its leads. What is the resistance in Ω
of the heater wire? (Note: 1 cal = 4.186 J)
Select one:
a. 0.86
b. 0.17
c. 0.29
d. 0.43
Answer:
1 cal/s =4.184w
p=50 cal/s =2093w
v=12v
P = V*I
I =P/V
I = 17.43 A
P =1²*R
R = P/I²
R = 0.68A 5 kg box drops a distance of 10 m to the ground. If 70% of the initial potential energy goes into increasing the internal energy of the box, determine the magnitude of the increase.
Answer:
Explanation:
From the given information:
The initial PE [tex](PE)_i[/tex] = m×g×h
= 5 kg × 9.81 m/s² × 10 m
= 490.5 J
The change in Potential energy P.E of the box is:
ΔP.E = [tex]P.E_f -P.E_i[/tex]
ΔP.E = 0 - [tex](PE)_i[/tex]
ΔP.E = [tex]-P.E_i[/tex]
If we take a look at conservation of total energy for determining the change in the internal energy of the box;
[tex]\Delta P.E + \Delta K.E + \Delta U = 0[/tex]
[tex]\Delta U = -\Delta P.E - \Delta K.E[/tex]
this can be re-written as:
[tex]\Delta U =- (-\Delta P.E_i) - \Delta K.E[/tex]
Here, K.E = 0
Also, 70% goes into raising the internal energy for the box;
Thus,
[tex]\Delta U =(70\%) \Delta P.E_i-0[/tex]
[tex]\Delta U =(0.70) (490.5)[/tex]
ΔU = 343.35 J
Thus, the magnitude of the increase is = 343.35 J
If I am going to explore Mars what will I need?
Answer:
uhm oxygen tank, a suit (like an astronaut suit) uh food, and a space ship
Explanation:
FROM THE _____ WHOLE WATER CYCLE STARTS ALL OVER AGAIN
From the water whole water cycle starts again.
Most possibly water should be the answer.Consider a swimmer that swims a complete round-trip lap of a 50 m long pool in 100 seconds. What is the swimmers average speed and average velocity?
Answer:
The average speed is 1 m/s
The average velocity is 0
Explanation:
Given;
length of the pool, L = 50 m
time taken for the motion, t = 100 s
The total distance = 50 m + 50 m
The total distance = 100 m
The average speed = total distance / total time
= 100 / 100
= 1 m/s
The average velocity = change in displacement / change in time
change in displacement = 50 m - 50 m = 0
The average velocity = 0 / 100
The average velocity = 0
g Is a nucleus that absorbs at 4.13 δ more shielded or less shielded than a nucleus that absorbs at 11.45 δ? _________ Does the nucleus that absorbs at 4.13 δ require a stronger applied field or a weaker applied field to come into resonance than the nucleus that absorbs at 11.45 δ?
Answer: A nucleus that absorbs at [tex]11.45\delta[/tex] is less shielded and a nucleus that absorbs at [tex]4.13\delta[/tex] will require a stronger applied field
Explanation:
While interpreting the data in NMR, the positions of signals are studied.
The nucleus/ protons having a higher value of [tex]\delta[/tex] are said to be less shielded. They are said to be upfield.
The nucleus/protons having a lower value of [tex]\delta[/tex] are said to be more shielded. They are said to be downfield.
So, a nucleus that absorbs at [tex]11.45\delta[/tex] is less shielded by the nucleus that absorbs at [tex]4.13\delta[/tex]
Also, the less shielded nucleus/protons will require a weak applied field to come into resonance than the more shielded nucleus/protons
So, a nucleus that absorbs at [tex]4.13\delta[/tex] will require a stronger applied field to come into resonance than the nucleus that absorbs at [tex]11.45\delta[/tex]
