Answer:
17 seconds
Explanation:
Given that:
The mass attached to the spring (m) = 0.30 kg
The spring constant (k) = 2.00 N/m
The damping constant (b) = 0.025 kg/s
The initial distance [tex]x_o[/tex] = 5.0 cm
The initial final amplitude [tex]A_f[/tex] = 2.5 cm and not 2.5 m, please note the mistake, if it is 2.5 m, our time taken will be -93.7 sec, and we do not want a negative time value.
To start with the angular frequency damping using the formula:
[tex]\omega_{\gamma}= \dfrac{b}{2m}[/tex]
[tex]\omega_{\gamma}= \dfrac{0.025 \ kg/s}{2(0.3 \ kg)}[/tex]
[tex]\omega_{\gamma}=4.167 \times 10^{-2} \ s^{-1}[/tex]
In the absence of damping, the angular frequency is:
[tex]\omega_o = \sqrt{\dfrac{k}{m}}[/tex]
[tex]\omega_o = \sqrt{\dfrac{2 \ N/m}{0.3 kg}} \\\\\omega_o = 2.581 \ s^{-1}[/tex]
The initial amplitude oscillation can be computed by using the formula:
[tex]A_i = e^{-\omega_{\gamma}t} x_o \sqrt{\dfrac{\omega_o^2}{\omega_o^2-\omega_f^2}}[/tex]
[tex]A_i = e^{-\omega_{\gamma}0} (5.0 \ cm) \sqrt{\dfrac{2.581^2}{2.581^2-(4.167*10^{-2})^2}}[/tex]
[tex]A_i = 5.0006 \ cm \\ \\ A_i = 5.001 \ cm[/tex]
The final amplitude, as well as the initial amplitude, can be illustrated by using the relation:
[tex]A_f = e^{-\omega_{\gamma}t}A_i\\ \\ e^{-\omega_{\gamma}t} = \dfrac{2. 5 \ cm}{5.001 cm}\\ \\ = 0.4999\\ \\ \implies -\omega_{\gamma}t_f = \mathsf{In (0.4999)} \\ \\ t_f = \dfrac{\mathsf{-In (0.4999)}}{4.167*10^{-2} \ s^{-1}} \\ \\[/tex]
[tex]t_f = 16.64 \ sec \\ \\ \mathbf{t_f \simeq 17 sec}[/tex]
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
An eagle is flying horizontally at a speed of 3.40 m/s when the fish in her talons wiggles loose and falls into the lake 5.50 m below. Calculate the velocity (in m/s) of the fish relative to the water when it hits the water. (Assume that the eagle is flying in the x-direction and that the y-direction is up.)
Answer:
The velocity of the fish when it hits the water is:
v = 10.93 m/s and 71.88 ° below the x-direction.
Explanation:
Let's find the velocity of the fish in the y-direction.
[tex]v_{fy}^{2}=v_{iy}^{2}-2gh[/tex]
Here, v(iy) of the fish is zero, and the heigh h = 5.50 m, then the velocity will be:
[tex]v_{fy}^{2}=0-2(9.81)(5.50)[/tex]
[tex]v_{fy}^{2}=-2(9.81)(5.50)[/tex]
[tex]v_{fy}=-10.39 \: m/s[/tex]
Now, we know that the velocity in the x-direction is constant, so we can calculate the velocity of the fish when it hits the water.
[tex]v=\sqrt{v_{x}^{2}+v_{y}^{2}}[/tex]
[tex]v=\sqrt{3.40^{2}+(-10.39)^{2}}[/tex]
[tex]v=10.93 \: m/s[/tex]
And the direction will be:
[tex]\theta=tan^{-1}(\frac{10.39}{3.40})[/tex]
[tex]\theta=71.88^{\circ}[/tex]
The angle is 71.88 ° belox the x-direction.
I hope it helps you!
An object that gives off electromagnetic waves based on its temperature
demonstrates which phenomenon?
A. Emission spectra
B. Blackbody radiation
C. Quantum mechanics
D. Photoelectric effect
Answer:
B) Blackbody radiation
MARK THIS AS BRAINLIEST PLEAWESSS :)
Answer:
Black body radiation
Explanation:
Which of the following would experience induced magnestism mostly easily
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!!
15. A car is stationary at the top of a hill with the engine
switched off. The brakes are released and the car rolls down
the hill. At which labelled point does the car have the greatest
kinetic energy? *
A
Answer:
kinetic energy
Explanation:
hop it is helpful
mark me brainlist
A 3.25-gram bullet traveling at 345 ms-1 strikes and enters a 2.50-kg crate. The crate slides 0.75 m along a wood floor until it comes to rest.
Required:
a. What is the coefficient of dynamic friction between crate and the floor?
b. What is the average force applied by the crate on the bullet during collision if the bullet penetrates the 1.10cm into the crate?
Answer:
a) μ = 0.0136, b) F = 22.8 N
Explanation:
This exercise must be solved in parts. Let's start by using conservation of moment.
a) We define a system formed by the downward and the box, therefore the forces during the collision are internal and the momentum is conserved
initial instant. Before the crash
p₀ = m v₀
final instant. After inelastic shock
p_f = (m + M) v
the moment is preserved
p₀ = p_f
m v₀ = (m + M) v
v = [tex]\frac{m}{m + M} \ v_o[/tex]
We look for the speed of the block with the bullet inside
v = [tex]\frac{0.00325}{0.00325 + 2.50 } \ 345[/tex]
v = 0.448 m / s
Now we use the relationship between work and kinetic energy for the block with the bullet
in this journey the force that acts is the friction
W = ΔK
W = ½ (m + M) [tex]v_f^2[/tex] - ½ (m + M) v₀²
the final speed of the block is zero
the work between the friction force and the displacement is negative, because the friction always opposes the displacement
W = - fr x
we substitute
- fr x = 0 - ½ (m + M) vo²
fr = ½ (m + M) v₀² / x
the friction force is
fr = μ N
μ = fr / N
equilibrium condition
N - W = 0
N = W
N = (m + M) g
we substitute
μ = ½ v₀² / x g
we calculate
μ = ½ 0.448 ^ 2 / 0.75 9.8
μ = 0.0136
b) Let's use the relationship between work and the variation of the kinetic energy of the block
W = ΔK
initial block velocity is zero vo = 0
F x₁ = ½ M v² - 0
F = [tex]\frac{1}{2} M \frac{x}{y} \frac{v^2}{x1}[/tex]
F = ½ 2.50 0.448² / 0.0110
F = 22.8 N
A rod of 3.0-m length and a square (2.0 mm X 2.0 mm) cross section is made of a material with a resistivitYof 6.0 X 10-8 Ω m. If a potential difference of 0.60 V is placed across the ends of the rod, at what rate is heat generated in the rod in watt.
Select one:
a. 24
b. 12
c. 4
d. 8
Q}Write any two conversation which are followed while writing the units and symbols?Please tell me the answer of this question
Answer:
Units named after scientists are written in lowercase but their symbols are written as capital
Unit of power is watt, since it is named after a scientist, then symbol will be written as W
Farad, symbol = F
Henry, symbol = H
Units whose names aren't culled from that of scientists are written in lower case and their symbols are also in lower case.
Units such as meter, kilogram should be represented with symbols in small letters as: m and kg respectively.
Explanation:
Kindly check answer
A stone of mass 0.2 kg falls with an acceleration of 10.0 m/s. How big is the force that causes this acceleration?
Answer:
[tex]\boxed {\boxed {\sf 2 \ Newtons}}[/tex]
Explanation:
According to Newton's Second Law of Motion, force is the product of mass and acceleration.
[tex]F= m \times a[/tex]
The mass of the stone is 0.2 kilograms and the acceleration is 10.0 meters per square second.
m= 0.2 kg a= 10.0 m/s²Substitute the values into the formula.
[tex]F= 0.2 \ kg * 10.0 \ m/s^2[/tex]
Multiply.
[tex]F=2 \ kg*m/s^2[/tex]
Convert the units.
1 kilogram meter per square second (kg*m/s²) is equal to 1 Newton (N)Our answer of 2 kg*m/s² is equal to 2 N[tex]F= 2 \ N[/tex]
The force is 2 Newtons.
A glass of milk has what kind of energy?
A. Chemical Potential Energy
B. Kinetic Energy
C. Elastic Potential Energy
D. Radiant Energy
The laboratory exercise for this chapter addresses kinematic motion. You have experienced these motions in everyday life. Instead of a discussion board requiring posts in a forum, this assignment has been modified to accept your response to the following questions in this assignment. Be sure to clearly address each of the points below and show all of your work • What is the difference between a vector and a scalar quantity? • List two examples each of vector and scalar quantities • Write the due date of this assignment: Month and Day (For example, July 15 would be Month - 7. Day = 15) • For a building having a height equal to the quantity you have recorded for Day in meters in our example 15 meters), compute the time required for the ball to fall to the ground while experiencing acceleration due to gravity (g=9.8m/s) • How fast was the ball traveling when it hit the ground? Submit the kinetics Assignment by 11:59 p.m. (ET) on Monday,
Answer:
A) vectors: veloicty, force
scalar: speed, work
B) t = 1.75 s, C) v = - 17 2 m / s
Explanation:
We answer each part separately
A) A vector magnitude has magnitude and direction instead a scalar magnitude has only magnitude
vector quantities: the speed of a car number is the magnitude and direction is where it goes
Force, the number is the magnitude and above that applies gives direction
Scalar magnitude: how quickly the number of the speedometer of the car
Temperature, work
B) I = 15 m height to the soil and get to calculate time = 0
y = y₀ + v₀ t - ½ g t²
as the ball is loose its initial velocity is zero
0 = 0 +0 - ½ g t²
t = [tex]\sqrt{2y_o/g}[/tex]
t = [tex]\sqrt{2 \ 15/ 9.8}[/tex]
t = 1.75 s
C) the velocity to the reach the floor
v = vo - g t
v = 0 - g t
v = - 9.8 1.74
v = - 17 2 m / s
The negative signt iindicates that the speed goes down
An air-track glider attached to a spring oscillates between the 14.0 cm mark and the 71.0 cm mark on the track. The glider completes 12.0 oscillations in 34.0 s . You may want to review (Pages 391 - 393) . Part A What is the period of the oscillations
Answer:
A = 2,8333 s
Explanation:
El periodo es definido como el tiene que toma de dar una oscilación.
En este caso realiza varias osicilacion por lo cual debemos encontrar el promedio del perdono.
T = t/n
calculemos
A = 34,0/ 12,0
A = 2,8333 s
6. (6P) Formula Based Problem: A car is speeding down a highway at 30 m/s. A stray dog
runs into the road ahead of the driver. The driver hits the brake decelerating at a rate of
10m/s. The driver managed to stop the car right as it was about to hit the car. How far
away was the dog when the driver first hit the brake?
Answer:
45 m
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 30 m/s
Deceleration (a) = –10 m/s²
Final velocity (v) = 0 m/s
Distance (s) =?
The distance of the dog from the car before the driver hits the brake can be obtained as follow:
v² = u² + 2as
0² = 30² + (2 × –10 × s)
0 = 900 + (–20s)
0 = 900 – 20s
Collect like terms
0 – 900 = –20s
–900 = –20s
Divide both side by –20
s = –900 / –20
s = 45 m
Therefore, the dog was 50 m away when the driver first hit the brake
From 2 King 6:1-6, one of the disciples of Elisha was cutting a tree and the ax head fell into the water. While we do not know how high the ax head was when it fell into the water, we will work through a physics example of the ax head's vertical motion as if it were dropped into the water. ( Due date 09/07)
Write your name and date. The due date of this assignment is the height the ax head falls from in meters into the water. For example, if the due date is July 15, then the ax head fell 15 meters to the water.
Write Newton’s 2nd Law in Equation Form.
Write the quantity and units of average gravitational acceleration on the surface of Earth.
Given the ax head mentioned in the opening portion with the height being equal in numerical value of the due day of this assignment. How long does it take for the ax to fall to the river surface?
Compute the final speed of the ax when it hits the water.
Answer:
time of fall is 1.75 s and the velocity with which it strikes the water is 17.15 m/s.
Explanation:
Height, h = 15 m
Newton's second law
Force = mass x acceleration
The unit of gravitational force is Newton and the value is m x g.
where, m is the mas and g is the acceleration due to gravity.
Let the time of fall is t.
Use second equation of motion
[tex]s= u t +0.5 at^2\\\\15 = 0 +0.5\times 9.8\times t^{2}\\\\t = 1.75 s[/tex]
Let the final speed is v.
Use third equation of motion
[tex]v^2 = u^2 + 2 a s\\\\v^2 = 0 + 2 \times 9.8\times 15\\\\v =17.15 m/s[/tex]
The decibel level of a jackhammer is 125 dB relative to the threshold of hearing. Determine the decibel level if three jackhammers operate side by side.
Answer:
130 dB
Explanation:
The equation for decibel level is given by:
[tex]D=10log(\frac{I}{I_n} )\\\\Where\ D\ is\ the \ decibel\ level\ in\ dB, I\ is\ the\ intensity\ in \ W/m^2, \\I_n\ is\ threshold\ intensity\ to\ the\ human\ ear=1*10^{-12}W/m^2\\\\Given\ that\ D=125dB, hence:\\\\125=10log(\frac{I}{1*10^{-12}} )\\\\12.5=log(\frac{I}{1*10^{-12}} )\\\\I=3.2\ W/m^2[/tex]
The intensity for 1 jack hammer is 3.2 W/m², therefore for 3 jack hammers, the intensity = 3 * 3.2 = 9.6 W/m²
[tex]D=10\ log(\frac{I}{I_n} )\\\\D=10*log(\frac{9.6}{1*10^{-12}} )\\\\D=130\ dB[/tex]
Our solar system is made up of the Sun, 8 planets, and other bodies such as
asteroids orbiting the Sun. The solar system is very large compared to anything we see on
Earth. The distance between planets is measured in astronomical units (AU). One AU is
equal to 149.6 million kilometers, the average distance between the Sun and Earth. Scale
models are useful for helping us understand the size of the solar system.
Mr. Wilson’s science class made a scale model of the solar system. They went out to the
school’s football field, and they used the chart shown below to mark out the scale distance
from the Sun to each planet
The average orbital radius for a planet is its distance from the Sun. Which statement BEST
describes the relationship between the planets and their average orbital radii?
A. The planets are evenly spaced in the solar system.
B. Closer to the Sun, the planets are regularly spaced apart.
C. The planets closest to each other are the ones farthest from the sun.
D. Farther from the Sun, the planets are spaced farther apart from each other.
the answer to the question is D
What is the minimum value of the friction coefficient between the boxes that will keep them from slipping when the 100 N force is applied
Answer:
The friction coefficient's minimum value will be "0.173".
Explanation:
The given query seems to be incomplete. Below is the attached file of the complete question.
According to the question,
(a)
The net friction force's magnitude will be:
⇒ [tex]F_{net}=ma[/tex]
[tex]=5\times 1.7[/tex]
[tex]=8.5 \ N[/tex]
(b)
For m₃,
⇒ [tex]ma=\mu m_3 g[/tex]
Or,
⇒ [tex]\mu=\frac{a}{g}[/tex]
[tex]=\frac{1.7}{9.8}[/tex]
[tex]=0.173[/tex]
Coherent light with wavelength 603 nm passes through two very narrow slits, and the interference pattern is observed on a screen a distance of 3.00 m from the slits. The first-order bright fringe is a distance of 4.84 mm from the center of the central bright fringe.
Required:
For what wavelength of light will the first-order dark fringe (the first dark fringe next to a central maximum) be observed at this same point on the screen?
Answer:
λ = 4.023 10⁻⁷ m
Explanation:
The double-slit interference phenomenon is described by
d sin θ = (m + ½) λ destructive interference
d sin θ = m λ constructive interference
we can use trigonometry
tan θ = y / L
how these experiments occur for small angles
tan θ = sin θ/cos θ = sin θ
sin θ = y / L
we substitute
d y / L = (m + ½) λ destructive interference
d y / L = m λ constructive interference
with the expression for constructive interference we look for the separation of the slits
d = m λ L / y
d = 1 603 10⁻⁹ 3 /4.84 10⁻³
d = 3.738 10⁻⁴ m
Now let's analyze the case where the distance for constructive and destructive interference occurs at the same point y = 4.84 mm = 4.84 10⁻³m
d y / L = (m + ½) λ
λ = [tex]\frac{ d \ y}{L\ (m+ 1/2) }[/tex]
the first strip is for m = 1
let's calculate
λ = [tex]\frac{3.738 \ 10^{-4} 4.84 \ 10^{-3} }{ 3 \ ( 1 + 0.5) }[/tex]
λ = 4.023 10⁻⁷ m
Six artificial satellites complete one circular orbit around a space station in the same amount of time. Each satellite has mass m and radius of orbit L. The satellites fire rockets that provide the force needed to maintain a circular orbit around the space station. The gravitational force is negligible.
Rank the net force acting on each satellite from their rockets. Rank from largest to smallest.
a. m=200 kg and L= 5000m
b. m=400 kg and L=2500m
c. m=100kg and L=2500m
d. m=100kg and L=10000m
e. m=800kg and L=5000m
f. m=300kg and L=7500m
Answer:
b = e > a = c = f > d
Explanation:
Since the satellites complete one circular orbit in the same amount of time, their speed is the same.
The force needed to maintain the orbit is the centripetal force given by F = mv²/L where m = mass of artificial satellite and L = radius of orbit.
So, for artificial satellite a
a. m=200 kg and L= 5000m
F = mv²/L
F = 200 kgv²/5000 m
F = 0.04v² N
So, for artificial satellite b
b. m=400 kg and L= 2500m
F = mv²/L
F = 400 kgv²/2500 m
F = 0.16v² N
So, for artificial satellite c
c. m=100 kg and L= 2500m
F = mv²/L
F = 100 kgv²/2500 m
F = 0.04v² N
So, for artificial satellite d
d. m=100 kg and L= 10000m
F = mv²/L
F = 100 kgv²/10000 m
F = 0.01v² N
So, for artificial satellite e
e. m=800 kg and L= 5000m
F = mv²/L
F = 800 kgv²/5000 m
F = 0.16v² N
So, for artificial satellite f
f. m=300 kg and L= 7500m
F = mv²/L
F = 300 kgv²/7500 m
F = 0.04v² N
So, the net force are in the order b = e > a = c = f > d
A typical ceiling fan running at high speed has an airflow of about 2.00 ✕ 103 ft3/min, meaning that about 2.00 ✕ 103 cubic feet of air move over the fan blades each minute.
Determine the fan's airflow in m3/s.
Answer:
0.94 m³/s
Explanation:
From the question given above, the following data were obtained:
Air flow (in ft³/min) = 2×10³ ft³/min
Air flow (in m³/s) =.?
Next, we shall convert 2×10³ ft³/min to m³/min. This can be obtained as follow:
35.315 ft³/min = 1 m³/min
Therefore,
2×10³ ft³/min = 2×10³ ft³/min × 1 m³/min / 35.315 ft³/min
2×10³ ft³/min = 56.63 m³/min
Finally, we shall convert 56.63 m³/min to m³/s. This can be obtained as follow:
1 m³/min = 1/60 m³/s
Therefore,
56.63 m³/min = 56.63 m³/min × 1/60 m³/s ÷ 1 m³/min
56.63 m³/min = 0.94 m³/s
Thus, 2×10³ ft³/minis equivalent to 0.94 m³/s.
Two water-slide riders, A and B, start from rest at the same time and same height, h but on differently shaped slides.
Required:
a. Which rider is traveling faster at the bottom?
b. Which rider makes it to the bottom first? Ignore friction and assume both slides have the same path length.
Answer:
a. None
b. Both
Explanation:
a. Which rider is traveling faster at the bottom?
Since both riders fall from the same height, h, their potential energy, U at the top equals their kinetic energy, K at the bottom.
U = mgh and K = 1/2mv²
Since U is he same for both water-slide riders, then K will be the same and thus their speed at the bottom will be the same. This is shown below.
K = U
1/2mv² = mgh
v² = 2gh
v =√(2gh) where v = speed of rider at the bottom, g = acceleration due to gravity and h = height of slide.
Since the height is the same, so their speed at the bottom is the same. So, none of the riders travels faster than the other since they have the same speed at the bottom.
b. Which rider makes it to the bottom first? Ignore friction and assume both slides have the same path length.
Since the path length of the water slides are the same and friction is neglected, both water-slide rider get to the bottom at the same time since the distance moved is the same and they both start from rest.
So, both riders make it to the bottom at the same time.
The acceleration due to gravity acts vertically downwards, and the component of gravity acceleration is larger when the slope is steeper.
a. Rider Bb. Rider B
Reasons:
The acceleration of the riders are due to gravity
The component acceleration due to gravity acting on a slope is a = g·sin(θ)
As the steepness of the slope increase, the angle, θ, and sin(θ) increases, therefore, the acceleration increases.
Rider A is on a slide with gentle slope, such that if the slide is flat, rider A will be stationary.
The shape of the water slide rider B is on is steeper, and therefore, rider B is accelerating more than rider A. The higher acceleration of rider B, gives rider B a higher speed than rider, such that rider B, is riding faster than rider ATherefore;
a. The rider that is travelling faster at the bottom is rider B
b. Given that friction is ignored, and the path have the same length to the
bottom, the rider that makes to the bottom first is the rider that is moving
faster, which is rider B
Learn more here:
https://brainly.com/question/13218675
If the open body is postively charged and another body is negatively charged ,free electrons tend to
Answer:
The free electrons tend to Negatively Charged body to Positively Charged body
Explanation:
As they have different charges, Electrons are more attracted to the Positive or it's opposite charge.
Tiny organisms were collected from thermal vents deep in the Pacific Ocean. The characteristics below describe these primitive organisms.
1) They are all unicellular.
2) They live in harsh environments.
3) Most have a cell wall.
In which domain would these organisms be found? (SC.6.L.15.1)
A.Archaea
B.Bacteria
C.Eukarya
D.Protista
Answer:
A. Archaea
Explanation:
Living organisms have been taxonomically classified into a highest ranking taxa called DOMAIN. The domains are as follows: Eukarya, Prokarya (bacteria) and Archaea. Although both Domains Bacteria and Archaea are unicellular, the Archaea is specifically characterized by their ability to survive in harsh weather conditions like the thermal vents tiny organisms were collected from.
Also, members of the domain Archaea have a unique cell wall different from other organisms like bacteria. Hence, based on the matching characteristics of the collected tiny organisms, they would be found in the Domain ARCHAEA.
The application of solid-state physics that is the study of the arrangement of atoms in a solid is
O metallurgy.
O quantum mechanics.
O crystallography
electromagnetism.
Answer:
crystallography.......
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
make ansentance rkdloebebjekeoejbe
Answer:
the man has returned from his trip
Answer:
just did by typing this lol
galileo was a contemporary of
On a particle level, what happens when thermal conduction occurs within a
solid?
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]
An insulated tank contains 50 kg of water, which is stirred by a paddle wheel at 300 rpm while transmitting a torque of 0.1 kN-m. At the same time, an electric resistance heater inside the tank operates at 110 V, drawing a current of 2 A. Determine the rate of heat transfer after the system achieves steady state.
Answer:
the rate of heat transfer after the system achieves steady state is -3.36 kW
Explanation:
Given the data in the question;
mass of water m = 50 kg
N = 300 rpm
Torque T = 0.1 kNm
V = 110 V
I = 2 A
Electric work supplied W₁ = PV = 2 × 110 = 220 W = 0.22 kW
Now, work supplied by paddle wheel W₂ is;
W₂ = 2πNT/60
W₂ = (2π × 0.1 × 300) / 60
W₂ = 188.495559 / 60
W₂ = 3.14 kW
So the total work will be;
W = 0.22 + 3.14
W = 3.36 kW
Hence total work done on the system is 3.36 kW.
At steady state, the properties of the system does not change so the heat transfer will be 3.36 KW.
The heat will be rejected by the system so the sign of heat will be negative.
i.e Q = -3.36 kW
Therefore, the rate of heat transfer after the system achieves steady state is -3.36 kW