Answer:
D.frequency.
Explanation:
The number of times a wave passes through a particular point in a time of 1 second is called the frequency.
Amplitude is the maximum height the wave reaches from the reference axis of the wave
Wavelength is the distance between the two upper (crest) or lower (troughs) points of a wave.
Crest is the top most part of a wave.
Hence, the question is referring to frequency.
Organ pipe A, with both ends open, has a fundamental frequency of 270 Hz. The third harmonic of organ pipe B, with one end open, has the same frequency as the second harmonic of pipe A. How long are___________.
(a) pipe A
(b) pipe B
Answer
Organ Pipe A has both end open.
Organ Pipe B has one end open.
speed of sound, v = 343 m/s
The fundamental frequency = 270 Hz
wavelength of the pipe when both end open
λ = 2 L
we know,
[tex]\lambda = \dfrac{v}{f}[/tex]
now,
[tex]L = \dfrac{v}{2f}[/tex]
inserting all the values
[tex]L = \dfrac{343}{2\times 270}[/tex]
L_A = 0.635 m
length of pipe A is equal to 0.635 m
b) since third harmonic of pipe B is equal to second harmonic of pipe A
[tex]f_B = f_A[/tex]
[tex]\dfrac{n_BV}{4L_B} = \dfrac{n_AV}{2L_A} [/tex]
[tex]L_B= \dfrac{2n_BL_A}{4n_A}[/tex]
[tex]L_B= \dfrac{2\times 3 \times 0.635}{4\times 2}[/tex]
L_B = 0.476 m
length of pipe B is equal to 0.476 m.
A +6.00 μC point charge is moving at a constant 9.00 ×10^6m/s in the + y-direction, relative to a reference frame.
At the instant when the point charge is at the origin of this reference frame, what is the magnetic field B it produces at the following points?
A) x=0.500m,y=0, z=0
B) x=0, y=-0.500m, z=0
C) x=0, y=0, z=+0.500m
D) x=0, y=-0.500m, z=+0.500m
Explanation:
We know magnetic field due to a charge q moving with a velocity v [tex]=\dfrac{\mu_oq}{4\pi r^3}(v \times r)[/tex].
Case A:
r=0.5i
Putting value of r, q and v .
We get, [tex]B=-2.34\times 10^{-5} \ k.[/tex] [tex](\ j\times i=-k)[/tex]
Case B:
r=0.5 j
B=0 [tex]( j\times j=0)[/tex]
Case C:
We get, [tex]B=2.34\times 10^{-5} \ i\ (Since\ , j \times k=i)[/tex]
Case D:
We get , [tex]B=8.28\times 10^{-6} \ T \ i[/tex].
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For tax and accounting purposes, corporations depreciate the value of equipment each year. One method used is called "linear depreciation," where the value decreases over time in a linear manner. Suppose that two years after purchase, an industrial milling machine is worth $820,000, and five years after purchase, the machine is worth $430,000. Find a formula for the machine value V.
Answer:
[tex]V=-130000x+1080000[/tex]
Explanation:
Linear Dependence
Some variables are known or assumed to have linear dependence which means the graph of the ordered pairs (x,V) is a straight line.
If we know two points of the line, we can come up with the exact equation and therefore make predictions for other values of x
The linear depreciation gives us these points (2,820000) and (5,430000)
The general equation of the line is
[tex]V=mx+b[/tex]
Where V is the machine value and x is the number of years after purchase. We need to find the values of m and b.
Replacing the first point
[tex]820000=m(2)+b[/tex]
[tex]2m+b=820000[/tex]
Replacing the second point
[tex]5m+b=430000[/tex]
Subtracting them
[tex]-3m=390000[/tex]
[tex]m=-130000[/tex]
Replacing in any of the equations, say, the first one
[tex]2(-130000)+b=820000[/tex]
Solving for b
[tex]b=820000+260000[/tex]
[tex]b=1080000[/tex]
The formula for the machine value V is
[tex]\boxed{V=-130000x+1080000}[/tex]
Formula for the machine value V in x year is v = -130000(x) + 1,080,000
Liner depreciation based problem:What information do we have?
Cost of machine after 2 year = $820,000
Cost of machine after 5 year = $430,000
Liner depreciation equation
v = mx + b
After 2 year
820,000 = m(2) + b
820,000 = 2m + b ......... Eq1
After 5 year
430,000 = m(5) + b
430,000 = 5m + b ......... Eq2
Eq2 - Eq1
3m = -390,000
m = -130,000
From Eq
430,000 = 5m + b
430,000 = 5(-130,000) + b
b = 1,080,000
Liner equation of tha cost.
Amount of machine = mx + b
Amount of machine = -130000(x) + 1,080,000
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A block with mass m=1.50 kg is initially at rest on a horizontal frictionless surface at x =0 , where x is the horizontal coordinate. A horizontally directed force is then applied to the block. The force is not constant: instead, its magnitude as a function of position is described by the relationship F(x)=(α-βx 2 )i , where x is given in units of meters, α = 2.50 N , β = 1.00 N/m2 , and i is the unit vector in the x direction.
a) What is the kinetic energy of the block as it passes through x=2.00 m?
b) What is the maximum speed of the block in the interval during which it moves from its initial position to x=2.00 m?
Answer:
(a) 2.33 J.
(b) 1.87 m/s.
Explanation:
(a)
[tex]F(x) = (2.5 - x^2)\^i[/tex]
The force as a function of position is given above. Since the force is a function of position, we can assume that we will use work-energy theorem.
[tex]W = \Delta K = K_2 - K_1\\\int\limits^2_0 {F(x)} \, dx = \frac{1}{2}mv_2 - 0\\\int\limits^2_0 {(2.5-x^2)} \, dx = (2.5x - \frac{x^3}{3})\left \{ {{x=2} \atop {x=0}} \right. = (5 - 8/3) = 7/3[/tex]
Therefore, the kinetic energy of the block is 2.33 J.
(b) In order to find the maximum speed in this interval, we need to investigate the acceleration of the block. Since acceleration is the derivative of velocity, velocity is at its maximum when acceleration is zero.
From Newton's Second Law:
[tex]F = ma\\a(x) = F(x)/m = 1.66 - 0.66x^2[/tex]
In order this to be zero:
[tex]1.66 - 0.66x^2 = 0\\x = 1.58~m[/tex]
The velocity of the block at x = 1.58 m can be found by work-energy theorem.
[tex]\int\limits^{1.58}_0 {(2.5-x^2)} \, dx = \frac{1}{2}(1.5)v^2\\ (2.5x-\frac{x^3}{3})\left \{ {{x=1.58} \atop {x=0}} \right. = 2.63 J\\2.63 = \frac{1}{2}(1.5)v^2\\v = 1.87~m/s[/tex]
Why is a frequency distribution useful? It allows researchers to see the "shape" of the data. It tells researchers how often the mean occurs in a set of data. It can visually identify the mean. It ignores outliers.
Answer: Frequency distribution helps to understand data. some complex data needs to be converted into intervals to see its frequency of occurrence. it helps to find the mean median and mode of the data. It helps in the analysis of data and its practical implementation.
Explanation:
Similar to what you see in your textbook, you can generally omit the multiplication symbol as you answer questions online, except when the symbol is needed to make your meaning clear. For example, 1⋅105 is not the same as 1105. When you need to be explicit, type * (Shift + 8) to insert the multiplication operator. You will see a multiplication dot (⋅) appear in the answer box. Do not use the symbol ×. For example, for the expression ma, typing m ⋅ a would be correct, but mxa would be incorrect.Enter the expression ma .
Final answer:
Type m · a to correctly enter the multiplication of m and a, ensuring clarity in expressions, especially when dealing with scientific notation which simplifies multiplication and division by manipulating coefficients and powers of ten separately.
Explanation:
When entering the expression m times a, you should use proper notation to differentiate between multiplication and adjacent variables or numbers. Instead of using 'x' as the multiplication symbol, as you might on paper, use the multiplication dot, represented by typing m · a with your keyboard. This avoids confusion, especially in contexts like scientific notation, where clarity is crucial. For example, 1·105 is not the same as 1 followed by 105. So, it is important to insert the multiplication dot to clarify that you are multiplying 1 by 10 raised to the 5th power.
In scientific notation, multiplication and division are simplified by separately manipulating the coefficients and the powers of ten. For instance, to multiply two numbers in scientific notation, such as (3 × 105) × (2 × 10°), you would multiply 3 by 2 to get 6, and then add the exponents (5 + 0) to get 5, resulting in an answer of 6 · 105. This simplifies calculations incredibly, especially with large numbers.
On a summer day in Narragansett, Rhode Island, the air temperature is 74 F and the barometric pressure is 14.5 lbf/in . Estimate the air density in kg/m.
There are some mistakes in the question as units pressure and air density are not written correctly.The correct question is here
On a summer day in Narragansett, Rhode Island, the air temperature is 74°F and the barometric pressure is 14.5 lbf/in². Estimate the air density in kg/m³.
Answer:
p=1.175 kg/m³
Explanation:
Given data
Temperature =74 F
Barometric pressure=14.5 lbf/in²
To find
Air density
Solution
From ideal gas law
[tex]pV=mRT\\p=(m/V)RT\\[/tex]
As mass/volume is pressure So
[tex]p=\frac{P}{RT}[/tex]
First we need to convert barometric pressure lbf/in² to N/m²
So
[tex]Pressure=(14.5lbf/in^{2} )*(\frac{6894.76N/m^{2} }{lbf/in^{2} } )\\Pressure=99974.02N/m^{2}[/tex]
Now Substitute the given value and pressure to find air density
[tex]p=\frac{P}{RT}\\ p=\frac{99974.02N/m^{2} }{(287J/kg.K)(\frac{74^{o}F-32}{1.8}+273.15 )}\\ p=\frac{99974.02N/m^{2} }{(287J/kg.K)*296.48K}\\ p=1.175kg/m^{3}[/tex]
given that a car's distance from a stop sign t seconds since it started moving is represented by d=t2-2t determine the car's average speed over each interval of elapsed time
Answer:
Average speed =t-2t
Explanation:
Equation given for Average speed = Total distance/Elapsed time
Given: Total distance=d=t^2-2t
Elapsed time= t
Substituting into the equation
Average speed = t^2-2t/t
Dividing both numerator an denomination with t
Average speed=t-2t
A kicker punts a football from the very center of the field to the sideline 39 yards downfield.
What is the magnitude of the net displacement of the ball? (a,football field is 53 yards wide)
What is the angle between the direction of the net displacement of the ball and the central line of the feild.
Answer:
L = 47.15 yards
∅ = 55.80°
Explanation:
Given
y = 39 yards
x = 53/2 yards
L = ?
We apply the formula
L = √(x²+y²)
L = √((53/2 yards)²+(39 yards)²)
L = 47.15 yards
The angle between the direction of the net displacement of the ball and the central line of the feild is obtained as follows
∅ = tg⁻¹(y/x)
∅ = tg⁻¹(39/(53/2)) = 55.80°
A toy train rolls around a horizontal 1.0-m-diameter track.The coefficient of rolling friction is 0.10.a) What is the magnitude of the trains angular acceleration afterit is released?b)How long does it take the train to stop if it's released with anangular speed of 30 rpm?In step one of the question, the solution supplied states thattangential acceleration is the coefficient of rolling frictionmultiplied by gravity. Why is this so?
Answer:
1.962 rad / s², 1.6 s
Explanation:
Radius of the part = 1.0m / 2 = 0.5 m
angular speed = 30 rpm = 30 rpm × (2πrad / rev) × 1 minutes / 60 seconds = 3.142 rads⁻¹
μk = frictional force / normal ( mg )
normal is the force acting upward against the force of gravity
frictional force = - μk mg
since the body came to rest then
Fnet + Ff = 0
Fnet = - Ff
Fnet = ma
ma = - μk mg
a = - μkg where g = 9.81
a = - 0.1 × 9.81 = 0.981 m/s²
magnitude of angular acceleration = tangential acceleration / radius = 0.981 / 0.5 = 1.962 rad / s²
b) time for the train to come to rest = angular velocity / angular angular acceleration = 3.142/ 1.962 = 1.6 s
The equation earlier derived answer this question
Fnet + Ff = 0 since the body came to a rest
Fnet = - Ff and Ff = - μk mg, Fnet = ma
ma = - μk mg
m cancel m on both side
a = - μkg since it magnitude
a = μkg
The coefficient of rolling friction multiplied by gravity gives the tangential acceleration in rolling motion due to the relationship between frictional force and the object's acceleration.
The reason why the coefficient of rolling friction multiplied by gravity gives the tangential acceleration is due to the relationship between the frictional force and the force responsible for the acceleration of the object. In the case of rolling motion, the frictional force opposes the rotation and contributes to the overall acceleration.
A farmer lifts his hay bales into the top loft of his barn by walking his horse forward with a constant velocity of 1 ft/s. Determine the velocity and acceleration of the hay bale when the horse is 10 ft away from the__________.
When an object moves with constant velocity, its acceleration is zero. So, the hay bale's velocity is 1 ft/s, and its acceleration is 0 ft/s2. These values persist irrespective of the horse's distance from the barn.
Explanation:In the case of the farmer's horse and the hay bale, the horse is moving with a constant velocity of 1 ft/s. When an object moves with constant velocity, its acceleration is zero. This principle stems from the fundamental definition of acceleration as the rate of change of velocity over time. Since the horse's velocity is not changing, acceleration is zero.
Moving on to the velocity of the hay bale: if we suppose that the horse's movement directly influences the lifting of the hay bale, the bale's velocity would also be 1 ft/s. Provided that the system of lifting the bale is suitable for the task, it does not matter how far the horse is away from the barn; the velocity and acceleration values persist. Therefore, regardless of whether the horse is 10 ft away or 100 ft away, the velocity of the hay bale remains 1 ft/s and acceleration 0 ft/s2.
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The hay bale being lifted by the farmer's horse, which is walking at a constant velocity of 1 ft/s, would have the sameconstant velocity with an acceleration of 0 m/s², because there is no change in velocity. Velocity vectors for the horse's movement would remain consistent, indicating the absence of acceleration.
The question involves determining the velocity and acceleration of a hay bale being lifted into a barn loft by a horse walking at a constant velocity. When dealing with such problems in physics, we typically look for changes in velocity over time to find acceleration. In this scenario, since the horse is moving with a constant velocity of 1 ft/s and no information is provided about forces acting on the hay bale or if there is any change in the speed of the horse, we can infer that the hay bale being lifted is moving at the same constant velocity of 1 ft/s, with an acceleration of 0 m/s² (or ft/s²) since there is no change in speed. This is because acceleration is defined as the rate of change of velocity over time, and a constant velocity means no change in velocity.
To conceptualize this, imagine sketching out a series of velocity vectors for the horse's path from one point to the next, indicating consistent motion with no alterations above or below the horizontal axis, representing that there is no acceleration occurring. An example in the given information refers to a racehorse accelerating from rest to 15.0 m/s, indicating that its acceleration can be calculated by dividing the change in velocity by the change in time, giving an average acceleration. But for the farmer's horse walking forward at a constant pace, the situation is quite different since there is no increase in speed.
To reiterate, if the horse maintains a constant velocity, the hay bale will have the same velocity (1 ft/s) and zero acceleration, provided that the system remains unaltered. Calculating acceleration in such cases is straightforward when there's a change in velocity; for constant velocity, the acceleration is simply zero.
A rocket has a mass 280(103) slugs on earth. Specify its mass in SI units, and its weight in SI units.
To solve this problem we will apply the concepts related to the conversion of units for which we will have that 1 slug is equal to 14.59kg. At the same time we will use Newton's second law for which weight is defined as the product between mass and acceleration (Due to gravity). This is then
A: Using the conversion ratio of slug to kilogram we have to,
[tex]1 slug = 14.59kg[/tex]
Then
[tex]m = 280*10^3 slugs (\frac{14.59kg}{1slug})[/tex]
[tex]m = 4.09*10^6kg[/tex]
B: Using Newton's second law we have to,
[tex]W = mg[/tex]
[tex]W = (4.09*10^6)(9.8)[/tex]
[tex]W = 40034960N\approx 4*10^7N[/tex]
Two point charges are released at rest in otherwise empty space. As time goes on, the magnitude of the force exerted on each of the particles decreases. Which of the following must be true?a. Both particles are negatively chargedb. Not enough infoc. The particles may be either both positively charged or both negatively chargedd. One particle is positively charged and the other is negatively chargede. Both particles are positively charged
Answer:
c.)
Explanation:
Assuming no other forces acting on the charges, if as time goes on, the magnitude of the force exerted on each of the particles decreases, as the electric force is inversely proportional to the square of the distance between charges, the only explanation is that this distance is increasing, so the force on both charges is repulsive.
When the force is repulsive, all we can say, is that both charges are of the same type, so they can be both positively charged or both negatively charged, as stated by the choice c).
Suppose that a particular artillery piece has a range R = 5580 yards . Find its range in miles. Use the facts that 1mile=5280ft and 3ft=1yard. Express your answer in miles to three significant figures..
Answer:
3.170 miles
Explanation:
Conversion from yards to feet:
5580 yards = 5580 yard * 3 ft/yard = 16740 feet
Conversion from feet to miles:
16740 feet = 16740*(1/5280) mile/ft = 3.170 miles.
Therefore, that particular artillery piece has a range R = 3.170 miles
When the balls are brought close together, without touching, the following observations are made: Balls B, C, and D are attracted to ball A. Balls B and D have no effect on each other. Balls B is attracted to ball C. What are the charge states (positive, negative, or neutral) of balls A, B, C, and D
Answer:
B and D is neutral. A and C have opposite charges.
Explanation:
Opposite charges attract each other and similar charges repel each other. Furthermore, a neutral object is attracted to a charged object whether negative or positive.
- If the balls B, C, and D is attracted to A, then their charge is either opposite to that of A or neutral.
- If the balls B and D have no effect on each other, then both balls are neutral.
- If B is neutral and attracted to C, then ball C have opposite charges than A.
Finally, B and D is neutral. A and C have opposite charges. Whether they are positive or negative cannot be determined by the information given in the question.
Total lung capacity of a typical adult is approximately 5.0L. Approximately 20% of the air is oxygen, as air is 20% oxygen. At sea level and at anaverage body temperature of 37°C. How many moles ofoxygen do the lungs contain at the end of an inflation?
Answer:
n=0.03928 moles
Moles of oxygen do the lungs contain at the end of an inflation are 0.03928 moles
Explanation:
The amount of oxygen which lung can have is 20% of 5 L which is the capacity of lungs
Volume of oxygen in lungs =V=5*20%= 1 L=[tex]1*10^{-3} m^3[/tex]
Temperature=T=[tex]37^oC=273+37=310K[/tex]
Pressure at sea level = P= 1 atm=[tex]1.0125*10^5 Pa[/tex]
R is universal Gas Constant =8.314 J/mol.K
Formula:
[tex]n=\frac{PV}{RT}\\n=\frac{(1.0125*10^5) *(1*10^{-3})}{(8.314)*310} \\n=0.03928 mol[/tex]
Moles of oxygen do the lungs contain at the end of an inflation are 0.03928 moles
To seven significant figures, the mass of a proton is 1.672623 x10^{-27} kg. Which of the following choices demonstrates correct rounding?
Answer:
[tex]1.67\times 10^{-27}kg[/tex]
Explanation:
We are given that mass of proton
[tex]1.672623\times 10^{-27}kg[/tex]
There are seven significant figures.
We have to round off.
If we round off to three significant figures
The thousandth place of given mass of proton is less than five therefore, digits on left side of thousandth place remains same and digits on right side of thousandth place and thousandth replace by zero
Therefore, the mass of proton can be written as
[tex]1.67\times 10^{-27}kg[/tex]
Hence, the mass of proton is [tex]1.67\times 10^{-27}kg[/tex]
To round the mass of a proton to seven significant figures, the correct way is to round up the last significant figure if it is 5 or greater, and if it is less than 5, simply drop the remaining digits.
Explanation:To round the mass of a proton, we look at the digit right after the desired number of significant figures, which in this case is the seventh significant figure. If this digit is 5 or greater, we round up the last significant figure. If it is less than 5, we simply drop the remaining digits.
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The specific volume of 5 kg of water vapor at 1.5 MPa, 440°C is 0.2160 m3 /kg. Determine (a) the volume, in m3 , occupied by the water vapor, (b) the amount of water vapor present, in gram moles, and (c) the number of molecules.
(a) The volume occupied by the water vapor is V = 1.08 m³.
(b) The amount of water vapor present is 277.78 gram-moles.
(c) The number of molecules is [tex]\rm\(1.67305 \times 10^{26}\)[/tex] molecules.
Given:
Specific volume (v) = 0.2160 m³/kg
Mass of water vapor (m) = 5 kg
One gram-mole of water vapor = 18 g
Avogadro's number [tex]\rm (\(N_A\))[/tex] = [tex]\rm \(6.023 \times 10^{23}\)[/tex]
(a) The volume occupied by the water vapor (V) can be calculated using the formula: [tex]\rm \[ V = m \times v \][/tex]
Substitute the values:
[tex]\rm \[ V = 5 \, \text{kg} \times 0.2160 \, \text{m³/kg} \]\\\rm V = 1.08 \, \text m\³[/tex]
(b) The amount of water vapour present in gram moles can be calculated using the formula:
[tex]\rm \[ \text{Amount of water vapor} = \frac{m}{\text{One gram-mole of water vapor}} \][/tex]
Substitute the value of one gram-mole of water vapor [tex](\(18 \, \text{g}\))[/tex]:
[tex]\[ \text{Amount of water vapor} = \frac{5 \, \text{kg} \times 1000}{18 \, \text{g}} \]\\\\\ \text{Amount of water vapor} = 277.78 \, \text{gram-moles} \][/tex]
(c) The number of molecules can be calculated using Avogadro's number:
[tex]\rm \[ \text{Number of molecules in 1 gram-mole} = N_A \\= 6.023 \times 10^{23} \, \text{molecules} \][/tex]
So, the number of molecules in 1 gram of water vapor is [tex]\rm \(\frac{N_A}{18}\)[/tex] molecules.
The number of molecules in [tex]\rm \(5 \, \text{kg}\)[/tex] of water vapor is:
[tex]\rm \[ \text{Number of molecules} = 5 \times 1000 \times \frac{N_A}{18} \, \text{molecules} \][/tex]
Now, calculate the number of molecules:
[tex]\rm \[ \text{Number of molecules} = 1.67305 \times 10^{26} \, \text{molecules} \][/tex]
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If the x-component of a force vector is 2.16 newtons and its y-component is 4.11 newtons, then what is its magnitude? F = newtons
Answer:
Magnitude of resultant vector will be 4.643 N
Explanation:
We have given x-component of the force vector is given [tex]F_{X}=2.16N[/tex]
And y component of the force vector is given [tex]F_{y}=4.11N[/tex]
We have to find the magnitude of resultant force F
Resultant force will be equal to vector sum of magnitude of x competent and y component of force vector
So F = [tex]\sqrt{F_X^2+F_Y^2}=\sqrt{2.16^2+4.11^2}=4.643N[/tex]
So magnitude of resultant vector will be 4.643 N
Assume the equation x 5 At3 1 Bt describes the motion of a particular object, with x having the dimension of length and t having the dimension of time. Determine the dimen- sions of the constants A and B. (b) Determine the dimen- sions of the derivative dx/dt 5 3At2 1 B.
Answer:
(a) A = m/s^3, B = m/s.
(b) dx/dt = m/s.
Explanation:
(a)
[tex]x = At^3 + Bt\\m = As^3 + Bs\\m = (\frac{m}{s^3})s^3 + (\frac{m}{s})s[/tex]
Therefore, the dimension of A is m/s^3, and of B is m/s in order to satisfy the above equation.
(b) [tex]\frac{dx}{dt} = 3At^2 + B = 3(\frac{m}{s^3})s^2 + \frac{m}{s} = m/s[/tex]
This makes sense, because the position function has a unit of 'm'. The derivative of the position function is velocity, and its unit is m/s.
1. Suppose you have a pipe producing standing sound waves. Two adjacent harmonics of standing waves (i.e., no standing waves in between these) have wavelengths 2.000 meters and 1.500 meters.
Which wavelength corresponds to a higher mode?
Answer:
1.5 m
Explanation:
given,
wavelength of of the standing waves
λ₁ = 2 m
λ₂ = 1.5 m
to find wavelength corresponding to higher mode.
we know,
[tex]wavelength\ \alpha\ \dfrac{1}{n}[/tex]
where n is the mode number.
From the above expression we can say that wavelength is inversely proportional to mode.
Hence , the wavelength corresponding to higher mode is equal to 1.5 m
A 0.23 kg mass at the end of a spring oscillates 2.0 times per second with an amplitude of 0.15 m
Part A
Determine the speed when it passes the equilibrium point.
Part B
Determine the speed when it is 0.10 m from equilibrium.
Part C
Determine the total energy of the system.
Part D
Determine the equation describing the motion of the mass, assuming that at t=0, x was a maximum.
Determine the equation describing the motion of the mass, assuming that at , was a maximum.
x(t)=(0.075m)cos[2π(2.0Hz)t]
x(t)=(0.15m)sin[2π(2.0Hz)t]
x(t)=(0.15m)cos[2π(2.0Hz)t]
x(t)=(0.15m)cos[(2.0Hz)t]
Answer:
A) v = 1.885 m/s
B) v = 0.39 m/s
C) E = 0.03 J
D) [tex]x(t) = (0.15m)\cos(2\pi (2.0Hz)t)[/tex]
Explanation:
Part A
We will use the conservation of energy to find the speed at equilibrium.
[tex]K_{eq} + U_{eq} = K_A + U_A\\
\frac{1}{2}mv^2 + 0 = 0 + \frac{1}{2}kA^2\\
v = \sqrt{\frac{k}{m}}A[/tex]
where [tex]\omega = \sqrt{k/m}[/tex] and [tex]\omega = 2\pi f[/tex]
Therefore,
[tex] v = 2\pi f A = 2(3.14)(2)(0.15) = 1.885~m/s[/tex]
Part B
The conservation of energy will be used again.
[tex]K_1 + U_1 = K_2 + U_2\\
\frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \frac{1}{2}kA^2\\
mv^2 + kx^2 = kA^2\\
(0.23)v^2 + k(0.10)^2 = k(0.15)^2\\
v^2 = \frac{k(0.15)^2-(0.10)^2}{0.23}\\
v = \sqrt{0.054k}[/tex]
where [tex]k = \omega^2 m = (2\pi f)^2 m = 2(3.14)(2)(0.23) = 2.89[/tex]
Therefore, v = 0.39 m/s.
Part C
Total energy of the system is equal to the potential energy at amplitude.
[tex]E = \frac{1}{2}kA^2 = \frac{1}{2}(2.89)(0.15)^2 = 0.03~J[/tex]
Part D
The general equation of motion in simple harmonic motion is
[tex]x(t) = A\cos(\omega t + \phi)\\
x(t) = (0.15m)\cos(2\pi (2.0Hz)t + \phi)[/tex]
where [tex]\phi[/tex] is the phase angle to be determined by the initial conditions. In this case, the initial condition is that at t = 0, x is maximum. Therefore,
[tex]x(t) = (0.15m)\cos(2\pi (2.0Hz)t)[/tex]
The speed of the mass when passing the equilibrium point is approximately 1.88 m/s, and when it is 0.10 m from equilibrium, the speed is approximately 1.26 m/s. The equation describing the motion of the mass when x was a maximum at t=0 is x(t) = (0.15 m)cos[2π(2.0 Hz)t].
Explanation:Part A: To determine the speed when the mass passes the equilibrium point, we can use the formula for the velocity of an object in simple harmonic motion, which is v = ωA, where ω is the angular frequency and A is the amplitude. In this case, the angular frequency is given by ω = 2πf, where f is the frequency. So, ω = 2π(2.0 Hz) = 4π rad/s. The amplitude is 0.15 m. Substituting these values into the formula, we get v = (4π rad/s)(0.15 m) ≈ 1.88 m/s.
Part B: To determine the speed when the mass is 0.10 m from equilibrium, we can again use the formula for velocity. Using the same angular frequency ω, we find that v = (4π rad/s)(0.10 m) ≈ 1.26 m/s.
Part C: The total energy of the system can be found using the formula for the total energy of an object in simple harmonic motion, which is E = (1/2)kA², where k is the spring constant and A is the amplitude. In this case, k is not given, so we cannot determine the total energy.
Part D: The equation describing the motion of the mass, assuming that at t=0, x was a maximum, is given by x(t) = (0.15 m)cos[2π(2.0 Hz)t].
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You are pulling a box so it moves at increasing speed. Compare the work (W1) you need to do to accelerate it from 0 m/s to speed v to the work (W2) needed to accelerate it from speed v to the speed of 2v in the presence of friction. Compare the work () you need to do to accelerate it from 0 to speed to the work () needed to accelerate it from speed to the speed of 2 in the presence of friction. W2 > W1 W1 > W2 W1
Answer:
Explanation:
In absence of friction
Work done by force is given by change in kinetic energy of box
According work Energy theorem change in kinetic Energy of object is equal to work done by all the forces
[tex]W_1=\frac{1}{2}mv^2-0[/tex]
[tex]W_1=\frac{1}{2}mv^2[/tex]
where m=mass of object
[tex]W_2=\frac{1}{2}m(2v)^2-\frac{1}{2}mv^2[/tex]
[tex]W_2=\frac{3}{2}mv^2[/tex]
[tex]\frac{W_2}{W_1}=\frac{\frac{3}{2}mv^2}{\frac{1}{2}mv^2}[/tex]
[tex]W_2>W_1[/tex]
so work done in second case is three times of first case
(b)In Presence of friction
[tex]W_f+W_1=\frac{1}{2}mv^2[/tex]
suppose [tex]f_r[/tex] is the friction force and d is the displacement
so [tex]W_f=f_r\cdot d\cos 180[/tex]
[tex]W_f=-f_r\cdot d[/tex]
[tex]W_1=\frac{1}{2}mv^2+f_r\cdot d[/tex]
for second case when speed increases v to 2v
[tex]W_2=\frac{3}{2}mv^2+f_r\cdot d[/tex]
thus [tex]W_2>W_1[/tex] when friction is present
The work done to accelerate the box from zero is lesser than the work done to accelerate the box from velocity [tex]v[/tex].
From Work- energy theorem,Work can be defined as the change in the kinetic energy of the object.
So,
In the first Scenario,
[tex]W_1 = \dfrac 12 mv^2- 0\\\\W_1 = \dfrac 12 mv^2[/tex]
In the second scenario,
[tex]W_2 = \dfrac 12 m(2v^2) -\dfrac 12 mv^2\\\\W_2 = \dfrac 32 mv^2[/tex]
Now compare both scenarios,
[tex]\dfrac {W_1}{W_2}=\dfrac{ \dfrac 32 mv^2}{\dfrac12 mv^2}[/tex]
So, [tex]W_1 <W_2[/tex]
Therefore, the work done to accelerate the box from zero is lesser than the work done to accelerate the box from velocity [tex]v[/tex].
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A circular-shaped object of mass 10 kg has an inner radius of 13 cm and an outer radius of 26 cm. Three forces (acting perpendicular to the axis of rotation) of magnitudes 12 N, 26 N, and 16 N act on the object, as shown. The force of magnitude 26 N acts 30❦ below the horizontal. 12 N 16 N 26 N 30❦ ϝ Find the magnitude of the net torque on the wheel about the axle through the center of the object. Answer in units of N · m
The net torque on the circular object, considering the three forces acting on it, is calculated to be 1.46 N × m in an anti-clockwise direction.
To find the net torque on the circular object about its axle, we will first consider each of the three forces acting on it separately and determine the torque produced by each force. Torque (Τ) is defined by the equation Τ = r × F × sin(θ), where r is the radius at which the force is applied, F is the magnitude of the force, and θ is the angle between the force and the direction of the radius.
The 12 N force acts on the outer radius, so r = 0.26 m (26 cm converted to meters). Since the force is perpendicular to the radius, θ = 90 degrees, and the sin(90) = 1. Therefore, the torque from this force is Τ = 0.26 m × 12 N × 1 = 3.12 N × m.
The 16 N force also acts on the outer radius at 90 degrees to the radius, so its torque is Τ = 0.26 m × 16 N × 1 = 4.16 N × m.
The 26 N force acts at an angle 30 degrees below the horizontal, so it makes an angle of 60 degrees with the radius. Hence, the torque from this force is Τ = 0.26 m × 26 N × sin(60) = 5.82 N × m (rounded to two decimal places).
To find the net torque, we need to consider the direction of each torque. Both the 12 N and 16 N forces produce torque in the same direction (let's say clockwise), whereas the 30 degree component of the 26 N force produces torque in the opposite direction (counter-clockwise).
Net torque = Torque from 12 N + Torque from 16 N - Torque from 26 N = 3.12 N × m + 4.16 N × m - 5.82 N × m = 1.46 N × m (anti-clockwise).
The sled is pulled up a steeperhill of the sameheightas the hill described above.
How will the velocity of the sled at the bottom of the hill (after it has slid down) compare to that of the sled at the bottom of the original hill?
Choose the best answer below.
A. The speed at the bottom is greater for the steeper hill.
B. The speed at the bottom is the same for both hills.
C. The speed at the bottom is greater for the original hill because the sled travels further.
D. There is not enough information given to say which speed at the bottom is faster.
E. None of these descriptions is correct.
Answer: B. The speed at the bottom is the same for both hills.
Explanation A sled which can also be called a sledge is a vehicle built with a smooth body on the side touching the ground underside which makes it easy for it to be able to slide towards the ground when on a slant hill. The velocity of the sled after it has slid down the Hill and the velocity at the bottom of the Hill will be the same because both point have the same level of steepness. Velocity of a material is directly proportional to the mass of a body and inversely proportional to the time traveled.
The speed of the sled at the bottom of both hills will be the same because the total energy, determined by the height of the hill and not its steepness, remains constant.
Explanation:The best answer to this question is B. The speed at the bottom is the same for both hills. This is because the total energy of the sled (kinetic plus gravitational potential energy) must remain constant if we assume no friction or air resistance. The height of the hill, not its steepness, determines the total energy. The sled will convert all its potential energy (which depends on the height) into kinetic energy (which affects the speed) as it travels downhill. Hence, if the two hills are of the same height, the sled will have the same speed at the bottom of both hills.
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If Star A is magnitude 1.0 and Star B is magnitude 9.6 , which is brighter and by what factor?
Answer:
Star A is brighter than Star B by a factor of 2754.22
Explanation:
Lets assume,
the magnitude of star A = m₁ = 1
the magnitude of star B = m₂ = 9.6
the apparent brightness of star A and star B are b₁ and b₂ respectively
Then, relation between the difference of magnitudes and apparent brightness of two stars are related as give below: [tex](m_{2} - m_{1}) = 2.5\log_{10}(b_{1}/b_{2})[/tex]
The current magnitude scale followed was formalized by Sir Norman Pogson in 1856. On this scale a magnitude 1 star is 2.512 times brighter than magnitude 2 star. A magnitude 2 star is 2.512 time brighter than a magnitude 3 star. That means a magnitude 1 star is (2.512x2.512) brighter than magnitude 3 bright star.
We need to find the factor by which star A is brighter than star B. Using the equation given above,
[tex](9.6 - 1) = 2.5\log_{10}(b_{1}/b_{2})[/tex]
[tex]\frac{8.6}{2.5} = \log_{10}(b_{1}/b_{2})[/tex]
[tex]\log_{10}(b_{1}/b_{2}) = 3.44[/tex]
Thus,
[tex](b_{1}/b_{2}) = 2754.22[/tex]
It means star A is 2754.22 time brighter than Star B.
Star A is brighter than Star B by approximately 2512 times.
Explanation:The brightness of stars is measured using the magnitude scale. The smaller the magnitude, the brighter the star. In this case, Star A has a magnitude of 1.0 and Star B has a magnitude of 9.6. Since Star A has a smaller magnitude, it is brighter than Star B. The difference in magnitude between the two stars can be calculated using the formula:
difference = 2.512^(m₁ - m₂)
where m₁ is the magnitude of Star A and m₂ is the magnitude of Star B. Plugging in the values, we get:
difference = 2.512^(1.0 - 9.6) ≈ 2512
Therefore, Star A is approximately 2512 times brighter than Star B.
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The electric force between two identical positively charged ions is 33 × 10 − 9 N when they are 0.50 nm apart. How many electrons are missing from each ion?
Answer:
There are 5.98 electrons are missing from each ions.
Explanation:
Given that,
The electric force between two identical charges is, [tex]F=33\times 10^{-9}\ N[/tex]
The distance between the charges, [tex]d=0.5\ nm=0.5\times 10^{-9}\ m[/tex]
The electric force between charges is given by :
[tex]F=\dfrac{kq^2}{d^2}[/tex]
[tex]q=\sqrt{\dfrac{Fd^2}{k}}[/tex]
[tex]q=\sqrt{\dfrac{33\times 10^{-9}\times (0.5\times 10^{-9})^2}{9\times 10^9}}[/tex]
[tex]q=9.57\times 10^{-19}\ C[/tex]
Let there are n number of electrons are missing from each ions. It is given by :
[tex]n=\dfrac{q}{e}[/tex]
[tex]n=\dfrac{9.57\times 10^{-19}}{1.6\times 10^{-19}}[/tex]
n = 5.98 electrons
So, there are 5.98 electrons are missing from each ions. Hence, this is the required solution.
Water is pumped from a lake to a storage tank 15 m above at a rate of 90 L/s while consuming 15.4 kW of electric power. Disregarding any frictional losses in the pipes and any changes in kinetic energy, answer the questions that follow.
a) the overall efficiency of the pump motor unit.
b) the pressure difference between the inlet and the exit of the pump.
Answer:
B
Explanation:
How do the properties of an electromagnetic wave change as a result of increasing the period of the wave?
As the period of an electromagnetic wave increases, the wavelength of the wave also increases. However, the frequency of the wave remains the same. The energy carried by the wave is inversely proportional to its wavelength, so as the wavelength increases, the energy of the wave decreases.
Explanation:Electromagnetic waves have properties such as wavelength and frequency. As the period of an electromagnetic wave increases, the wavelength of the wave also increases. This means that the distance between successive crests or troughs of the wave becomes larger. However, the frequency of the wave remains the same. The energy carried by the wave is inversely proportional to its wavelength, so as the wavelength increases, the energy of the wave decreases.
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determine the required values of the tension in ac and ad so that the resultant of the three forces applied at A is verticalFigure:A wire is coupled from three sides
Answer:
Fac = 21 KN
Fad = 64.29 KN
Explanation:
Step 1: Find the unit vectors in BA , CA, and DA directions
BA = -16 i + 48 j - 12 k
CA = -16 i + 48 j + 24 k
DA = 14 i + 48 j
The following magnitudes
mag (BA) = sqrt ( 16 ^2 + 48^2 + 12^2 ) = 52
mag (CA) = sqrt ( 16 ^2 + 48^2 + 24^2 ) = 56
mag (DA) = sqrt ( 14 ^2 + 48^2 ) = 50
Now compute unit vectors
unit (BA) = (-4 / 13) i + (12 / 13) j - (3 / 13) k
unit (CA) = (-2 / 7) i + (6 / 7) j - (3 / 7) k
unit (DA) = (7 / 25) i + (24 / 25) j
Step 2: Compute Force Vectors F(ml) dot unit( ML )
Fab = (-4*Fab / 13) i + (12*Fab / 13) j - (3*Fab / 13) k
Fac = (-2*Fac / 7) i + (6*Fac / 7) j - (3*Fac / 7) k
Fad = (7*Fad / 25) i + (24*Fad / 25) j
Step 3: Equilibrium force equations in x and z direction
Fr,x = 0 = 39*( -4 / 13 ) - Fac*( 2 / 7 ) + Fad*( 7 / 25 )
Fr,z = 0 = 39*( -3 / 13 ) + Fad*( 3 / 7 )
Step 4: Solve the above equations for Fca and Fda
Fca = 21 KN
Fda = 64.29 KN