Answer:
D. Entropy is a measure of disorder or randomness
Explanation:
The answer is entropy. It is the method to measure any disorder or randomness of any system. Basically it is concept of physics and chemistry but can also be used in other subject matters as well. In formula terms, it is denoted by S and is measurement value is "Joule per kelvin". The values could be both positive and negative as well.
All the other options are incorrect as heat, energy and temperature are the different concepts of science which do not relates with any measurement.
In a mass spectrometer, ions are directed into a magnetic field, where they curve and strike a detector. If a variety of singly ionized atoms travel at the same speed through the magnetic field, would you expect them all to be deflected by the same amount, or would different ions be bent to different amounts? Defend your answer.
Answer:
the trajectory depends on the velocity and mass of the ions, therefore, they should be in different quantities
Explanation:
A mass spectrometer works based on the electrical force to accelerate the ions to a certain speed and the magnetic force to give a circular trajectory, let's use Newton's second law
F = m a
q v B = m v2 / r
r = (q / m) B / v
Here we can see that the radius of curvature of the ions depends on the velocity, but also on the mass of each of them. Therefore, even when they enter with the same speed they deviate in different trajectories depending on their mass.
The heavier ions will have smaller radii than the lighter ions
Consequently, the trajectory depends on the velocity and mass of the ions, therefore, they should be in different quantities
A mass attached to the end of a spring is oscillating with a period of 2.25 s on a horizontal frictionless surface. The mass was released from rest at t = 0 from the position x = 0.0360 m. (a) Determine the location of the mass at t = 5.47 s? (b) Determine if the mass is moving in the positive or negative x direction at t = 5.47 s? positive x direction negative x direction
The location of the mass at t = 5.47s can be determined by the equation for the position associated with SHM. To ascertain the direction of motion at the same time, the velocity function (the derivative of the position function) can be used. If the velocity value is positive, the mass moves in the positive x direction; if negative, it moves in the negative x direction.
Explanation:The problem describes the motion of a mass attached to a spring - specifically, it's simple harmonic motion (SHM). This can be understood in terms of the mass moving back and forth around a central, or equilibrium, position.
a) To determine the location of the mass at a specific time, t = 5.47s, we need to use the equation for the position x(t) associated with SHM: x(t) = A * cos(2 * π * t / T), where A is the amplitude (0.0360m in this case), π is the mathematical constant Pi, t is the time, and T is the period (2.25s in this case). Substituting the given values into this equation will give the position of the mass at t = 5.47s.
b) To determine the direction of motion at that time, we use the derivative of the position function, which gives us the velocity function: v(t) = dx(t) / dt = -A * (2 * π / T) * sin(2 * π * t / T). If this value is positive, then the mass is moving in the positive x direction; if negative, then it's moving in the negative x direction.
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(a) What is the length of a simple pendulum that oscillates with a period of 3.2 s on Earth, where the acceleration due to gravity is 9.80 m/s2, and on Mars, where the acceleration due to gravity is 3.70 m/s2?
Length (Earth)=?
Length (Mars)=?
(b) What mass would you need to suspend from a spring with a force constant of 20 N/m in order for the mass-spring system to oscillate with a period of 3.2 s on Earth, where the acceleration due to gravity is 9.80 m/s2, and on Mars, where the acceleration due to gravity is 3.70 m/s2?
Mass (Earth)=?
Mass (Mars)=?
Answer:
Explanation:
The expression relating length and time period
T = 2π [tex]\sqrt{\frac{l}{g} }[/tex]
3.2 = [tex]2\pi \sqrt{\frac{l}{9.8} }[/tex]
l = 2.54 m
On Mars g = 3.7
[tex]3.2 = 2\pi \sqrt{\frac{L}{3.7} }[/tex]
L = .96 m
b )
Expression for elastic constant and time period is as follows
[tex]T = 2\pi \sqrt{\frac{m}{k} }[/tex]
[tex]3.2=2\pi \sqrt{\frac{m}{20} }[/tex]
m = 5.19 N/s
Time period of oscillation due to spring is not dependent on g , so same time period will be found on Mars as that on the earth.
One at your summer lunar space camp activities is to launch n 1170-kg rocket from the surface of the Moon. You are a serious space camper and you launch a serious rocket: II reaches an altitude of 211 km. What gain m gravitational potential memo does the launch accomplish? The mass and radios of the Moon are 7.36 x 10^22 kg and 1740 km, respectively.
To solve this problem it is necessary to apply the concepts related to gravitational potential energy.
The change in gravitational potential energy is given by,
[tex]\Delta PE = PE_f - PE_i[/tex]
Where,
[tex]PE = \frac{GMm}{R}[/tex]
Here,
G = Gravitational Universal Constant
M = Mass of Earth
m = Mass of Object
R = Radius
Replacing we have that
[tex]\Delta PE = \frac{GMm}{R+h} -\frac{GMm}{R}[/tex]
Note that h is the height for this object. Then replacing with our values we have,
[tex]\Delta PE = \frac{GMm}{R+h} -\frac{GMm}{R}[/tex]
[tex]\Delta PE = GMm(\frac{1}{R} -\frac{1}{R+h})[/tex]
[tex]\Delta PE = (6.65*10^{-11})(7.36*10^{22})(1170)(\frac{1}{1740*10^3} -\frac{1}{211*10^3+1740*10^3})[/tex]
[tex]\Delta PE = 57264.48*10^{11}(5.1255*10^{-7}-5.747*10^{-7})[/tex]
[tex]\Delta PE = 3.56*10^8J[/tex]
Therefore the gravitational potential is [tex]3.56*10^8J[/tex]
5.37 At steady state, a power cycle develops a power output of 10 kW while receiving energy by heat transfer at the rate of 10 kJ per cycle of operation from a source at temperature T. The cycle rejects energy by heat transfer to cooling water at a lower temperature of 300 K. If there are 100 cycles per minute, what is the minimum theoretical value for T, in K?
Answer: minimum theoretical value of T = 750k
Explanation:
Assuming the the cycle is reversible and is ideal then
Wnet/Qh = Nmin .... equa 1
Equation 1 can be rewritten as
(Th -TL)/ Th ...equation 2
Th= temp of hot reservoir
TL= temp of low reservoir= 300
Wnet = power generated=10kw
He = energy transfer=10kj per cycle
Qhe = power transfer = (100/60)*10Kj = 16.67kw
Sub into equat 1
Nmin = 10/16.67 = 0.6
Sub Nmin into equation 2
Th = -TL/(Nmin -1) = -300k/(0.6 - 1)
Th =750k
The escape velocity on earth is 11.2 km/s. What fraction of the escape velocity is the rms speed of H2 at a temperature of 31.0 degrees Celsius on the earth? Note that virtually all the molecules will have escaped the earth's atmosphere if this fraction exceeds 0.15.
To solve this problem it is necessary to apply the concept related to root mean square velocity, which can be expressed as
[tex]v_{rms} = \sqrt{\frac{3RT}{n}}[/tex]
Where,
T = Temperature
R = Gas ideal constant
n = Number of moles in grams.
Our values are given as
[tex]v_e =11.2km/s = 11200m/s[/tex]
The temperature is
[tex]T = 30\°C = 30+273 = 303K[/tex]
Therefore the root mean square velocity would be
[tex]v_{rms} = \sqrt{\frac{3(8.314)(303)}{0.002}}[/tex]
[tex]v_{rms} = 1943.9m/s[/tex]
The fraction of velocity then can be calculated between the escape velocity and the root mean square velocity
[tex]\alpha = \frac{v_{rms}}{v_e}[/tex]
[tex]\alpha = \frac{1943.9}{11200}[/tex]
[tex]\alpha = 0.1736[/tex]
Therefore the fraction of the scape velocity on the earth for molecula hydrogen is 0.1736
Argon makes up 0.93% by volume of air. Calculate its solubility (in mol/L) in water at 20°C and 1.0 atm. The Henry's law constant for Ar under these conditions is 1.5 × 10−3 mol/L·atm. Enter your answer in scientific notation.
Answer:
S= 1.40x10⁻⁵mol/L
Explanation:
The Henry's Law is given by the next expression:
[tex] S = k_{H} \cdot p [/tex] (1)
where S: is the solubility or concentration of Ar in water, [tex]k_{H}[/tex]: is Henry's law constant and p: is the pressure of the Ar
Since the argon is 0.93%, we need to multiply the equation (1) by this percent:
[tex] S = 1.5 \cdot 10^{-3} \frac{mol}{L\cdot atm} \cdot 1.0atm \cdot \frac{0.93}{100} = 1.40 \cdot 10^{-5} \frac{mol}{L} [/tex]
Therefore, the argon solubility in water is 1.40x10⁻⁵mol/L.
Have a nice day!
A disk of radius R = 11 cm is pulled along a frictionless surface with a force of F = 16 N by a string wrapped around the edge.At the instant when d = 30 cm of string has unwound off the disk, what is the torque exerted about the center of the disk?
The torque exerted about the center of the disk is 1.76 Nm.
Explanation:To determine the torque exerted about the center of the disk, we need to calculate the perpendicular component of the force applied by the string. Since the force is applied tangentially, it can be resolved into two components: the perpendicular component and the parallel component. The perpendicular component creates the torque.
The perpendicular component of the force is equal to the tension in the string multiplied by the sine of the angle between the force and the radius. In this case, the angle is 90 degrees, so sin(90) = 1. Therefore, the torque is equal to the tension in the string multiplied by the radius of the disk.
Given that the force applied by the string is 16 N and the radius of the disk is 11 cm (0.11 m), the torque exerted about the center of the disk is 16 N * 0.11 m = 1.76 Nm.
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The torque exerted on the disk is calculated using the formula Torque = Force x Distance, where the distance is the radius of the disk. The torque in this situation is 1.76 N.m.
Explanation:In your question, we're being asked to find the torque exerted on a disk with radius R that's being pulled along a frictionless surface by a force F. The torque exerted on an object can be calculated using the formula Torque = Force x Distance, where distance refers to the perpendicular distance from the axis of rotation to the line of action of the force. In this case, the force is being applied at the edge of the disk, and the radius of the disk serves as the lever arm distance. Therefore, the torque exerted about the center of the disk is Torque = F * R = 16 N * 0.11 m = 1.76 N.m. Please note that the radius was converted from cm to m in order to keep the units consistent.
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You have a flashlight that uses 0.75 watts of power and requires 1.5 volt battery.How many joules of energy are used by the flashlight in 15 minutes? Hint: Joules are watts and seconds and you are computing the difference between the electric energy at two points A and B.What the total charge q0? Hint: the statement of the problem provides delta V.How many particles, each having a charge of 1.60 x 10^-6, are needed to produce the total charge at q0?
Answer:
(a) Energy will be 675 J
(B) charge will be 450 C
(C) Total number of particles will be [tex]281.25\times 10^6[/tex]
Explanation:
We have given that a flashlight uses 0.75 watts of power
So power P = 0.75 watt
Voltage is given as V = 1.5 volt
Time is given as t = 15 minutes
We know that 1 minute = 60 sec
So 15 minutes = [tex]15\times 60=900sec[/tex]
(A) We know that energy is given by [tex]E=P\times T=0.75\times 900=675j[/tex]
(b) We know that energy is also given by [tex]E=QV[/tex]
So [tex]675=Q\times 1.5[/tex]
[tex]Q=450C[/tex]
Now we have given charge on each particle [tex]=1.6\times 10^{-6}C[/tex]
So number of charge particle [tex]n=\frac{450}{1.6\times 10^{-6}}=281.25\times 10^6[/tex]
⦁ A baseball is struck by a bat 46 cm from the axis of rotation when the angular velocity of the bat is 70 rad/s. If the ball is hit at a height of 1.2 m at an angle of 45 degrees, will the ball clear a 1.2 m fence 110 m away (assume the initial velocity of the ball is the same as the linear velocity of the bat at the point at which it is struck?
To determine if the ball will clear the fence, calculate the range and maximum height of the ball's trajectory using given initial velocity and angle. Compare the calculated range with the distance to the fence.
Explanation:To determine whether the ball will clear the fence, we need to calculate the range and maximum height of the ball's trajectory. Using the given initial velocity of the ball (which is equal to the linear velocity of the bat), we can decompose it into horizontal and vertical components. The horizontal component remains constant throughout the flight, while the vertical component is affected by gravity. We can use the equations of motion to find the range and maximum height.
First, let's calculate the horizontal distance (range) the ball will travel. We can use the formula: range = (initial horizontal velocity) * (time of flight). Since the initial velocity and angle are given, we can calculate the initial horizontal velocity using trigonometry:
initial horizontal velocity = initial velocity * cos(angle)Next, we need to find the time of flight. Since the vertical displacement is zero at the peak of the trajectory, we can use the equation: time of flight = 2 * (initial vertical velocity) / (acceleration due to gravity). From the given height and angle, we can calculate the initial vertical velocity using trigonometry:
initial vertical velocity = initial velocity * sin(angle)Now, we can substitute the calculated values into the formula for range and solve for the time of flight:
range = (initial velocity * cos(angle)) * (2 * (initial velocity * sin(angle)) / (acceleration due to gravity))Finally, we can compare the calculated range with the distance to the fence. If the range is greater than the distance to the fence, the ball will clear the fence.
A vertical spring with k=490N/m is standing on the ground. You are holding a 5.0kg block just above the spring, not quite touching it. Part A How far does the spring compress if you let go of the block suddenly? Express your answer to two significant figures and include the appropriate units.
Answer:
Compression in the spring, x = 0.20 m
Explanation:
Given that,
Spring constant of the spring, k = 490 N/m
Mass of the block, m = 5 kg
To find,
Compression in the spring.
Solution,
Since the block is suddenly dropped on the spring gravitational potential energy of block converts into elastic potential energy of spring. Its expression is given by :
[tex]mgx=\dfrac{1}{2}kx^2[/tex]
Where
x is the compression in the spring
[tex]x=\dfrac{2mg}{k}[/tex]
[tex]x=\dfrac{2\times 5\times 9.8}{490}[/tex]
x = 0.20 m
So, the compression in the spring due to block is 0.20 meters.
The vertical spring with spring constant of 490N/m will compress at a distance of 0.2m.
How to calculate distance of spring?According to this question, the following information were given:
Spring constant of the spring, k = 490 N/mMass of the block, m = 5 kgTo calculate the compression in the spring, we deduce that the gravitational potential energy of the spring converts into elastic potential energy. Its expression is given by:
mgx = ½kx²
Where;
x = compression in the springx = 2mg/k
x = 2 × 5 × 9.8/490
x = 98/490
x = 0.2m
Therefore, the spring will compress at a distance of 0.2m.
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A brass rod with a length of 1.22 m and a cross-sectional area of 2.19 cm2 is fastened end to end to a nickel rod with length L and cross-sectional area 0.520 cm2 . The compound rod is subjected to equal and opposite pulls of magnitude 5.00×104 N at its ends. Find the length L of the nickel rod if the elongations of the two rods are equal. What is the stress in the brass rod? What is the stress in the nickel rod? What is the strain in the brass rod? What is the strain in the nickel rod?
Answer:
a) L₂ = 0.676 m
b) σ₁ = 2.28*10⁸ N/m²
σ₂ = 9.62*10⁸ N/m²
c) ε₁ = 0.00253678
ε₂ = 0.00457875
Explanation:
Given info
L₁ = 1.22 m
A₁ = 2.19 cm² = 2.19*10⁻⁴ m²
L₂ = ?
A₂ = 0.52 cm² = 0.52*10⁻⁴ m²
P = 5.00*10⁴ N
E₁ = 9*10¹⁰ N/m²
E₂ = 2.1*10¹¹ N/m²
In order to get the length L of the nickel rod if the elongations of the two rods are equal, we can say that
ΔL₁ = ΔL₂ ⇒ P*L₁/(A₁*E₁) = P*L₂/(A₂*E₂)
⇒ L₂ = A₂*E₂*L₁ / (A₁*E₁)
⇒ L₂ = (0.52*10⁻⁴ m²)*(2.1*10¹¹ N/m²)*(1.22 m) / (2.19*10⁻⁴ m²*9*10¹⁰ N/m²)
⇒ L₂ = 0.676 m
The stress in the brass rod is obtained as follows
σ₁ = P/A₁ ⇒ σ = 5.00*10⁴ N / 2.19*10⁻⁴ m² = 2.28*10⁸ N/m²
The stress in the niquel rod is obtained as follows
σ₂ = P/A₂ ⇒ σ = 5.00*10⁴ N / 0.52*10⁻⁴ m² = 9.62*10⁸ N/m²
The strain in the brass rod is obtained as follows
σ₁ = E₁*ε₁ ⇒ ε₁ = σ₁ / E₁
⇒ ε₁ = 2.28*10⁸ N/m² / 9*10¹⁰ N/m² = 0.00253678
The strain in the niquel rod is obtained as follows
σ₂ = E₂*ε₂ ⇒ ε₂ = σ₂ / E₂
⇒ ε₂ = 9.62*10⁸ N/m² / 2.1*10¹¹ N/m² = 0.00457875
A rod of mass M = 4 kg, length L = 1.8 meters, and moment of inertia ML2/12 is free to move on a frictionless surface. The rod is at rest when a puck of mass m = 0.4 kg approaches with a speed vi = 20 m/s perpendicular to the rod's length and strikes the rod at a point d = 0.3 meters from its lower end. After the collision, the puck moves backward with a speed vf = 10 m/s. The rod's center of mass moves forward with speed v and rotates with angular speed ω. (a) What is the speed v of the rod's center of mass after the collision?
Answer:
The speed of the rod's center of mass after the collision is 6 m/s.
Explanation:
Given that,
Mass of rod = 4 kg
Length l = 1.8 m
Moment of inertia [tex]I=\dfrac{ML^2}{12}[/tex]
Mass of puck = 0.4 kg
Initial speed= 20 m/s
Distance = 0.3 m
Final speed = 10 m/s
(a). We need to calculate the speed v of the rod's center of mass after the collision
As there is no external force acting on the system so, linear and angular momentum of the system will be conserved.
Using conservation of momentum
[tex]m_{i}v_{i}=m_{f}v_{f}+Mv[/tex]
Put the value into the formula
[tex]0.4\times20=-0.4\times10+2v[/tex]
[tex]v=\dfrac{8+4}{2}[/tex]
[tex]v=6\ m/s[/tex]
Hence, The speed of the rod's center of mass after the collision is 6 m/s.
Interactive LearningWare 10.1 reviews the concepts involved in this problem. A spring stretches by 0.0161 m when a 3.74-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 3.84 Hz?
Answer:
m = 3.91 kg
Explanation:
Given that,
Mass of the object, m = 3.74 kg
Stretching in the spring, x = 0.0161 m
The frequency of vibration, f = 3.84 Hz
When the object is suspended, the gravitational force is balanced by the spring force as :
[tex]mg=kx[/tex]
[tex]k=\dfrac{mg}{x}[/tex]
[tex]k=\dfrac{3.74\times 9.8}{0.0161}[/tex]
k = 2276.52 N/m
The frequency of vibration is given by :
[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}[/tex]
[tex]m=\dfrac{k}{4\pi^2f^2}[/tex]
[tex]m=\dfrac{2276.52}{4\pi^2\times (3.84)^2}[/tex]
m = 3.91 kg
So, the mass of the object is 3.91 kg. Hence, this is the required solution.
Energy is to be stored in a flywheel in the shape of a uniform solid disk with a radius of R = 1.22 m and a mass of 67.0 kg . To prevent structural failure of the flywheel, the maximum allowed radial acceleration of a point on its rim is 3510 m/s2 .What is the maximum kinetic energy that can be stored in the fly-wheel?
The maximum kinetic energy that can be stored in the flywheel is 27,909.78 J.
Explanation:To calculate the maximum kinetic energy that can be stored in the flywheel, we need to find the moment of inertia. For a uniform solid disk, the moment of inertia is given by the formula I = 0.5 * m * r^2, where m is the mass of the disk and r is the radius. Plugging in the values, we get I = 0.5 * 67.0 kg * (1.22 m)^2 = 49.518 kg·m².
The maximum kinetic energy is equal to 0.5 * I * ω^2, where ω is the angular velocity. Since the radial acceleration of a point on the rim is given as 3510 m/s², we can calculate ω using the formula a = r * ω^2. Rearranging the formula, we get ω = √(a / r) = √(3510 m/s² / 1.22 m) = 33.24 rad/s.
Plugging in the values, the maximum kinetic energy is 0.5 * 49.518 kg·m² * (33.24 rad/s)^2 = 27,909.78 J.
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The maximum kinetic energy that can be stored in the flywheel is approximately 71,791 Joules.
To determine the maximum kinetic energy that can be stored in the flywheel, we need to use the concepts of rotational motion. The relevant formulas are:
Kinetic Energy of a rotating object: KE = 1/2 I ω², where I is the moment of inertia and ω is the angular velocity.
Moment of Inertia for a solid disk: I = 1/2 m R², where m is the mass and R is the radius.
Relation between radial acceleration and angular velocity: a = ω² R.
We are given:
Radius (R): 1.22 m
Mass (m): 67.0 kg
Maximum radial acceleration (a): 3510 m/s²
First, solve for angular velocity (ω) using the radial acceleration formula:
ω² = a / R
⇒ ω² = 3510 m/s² / 1.22 m
⇒ ω² ≈ 2877.05 s⁻²
⇒ ω ≈ 53.61 rad/s
Next, calculate the moment of inertia (I) for the flywheel:
I = 1/2 m R² => I = 0.5 × 67.0 kg × (1.22 m)²
⇒ I ≈ 49.917 kgm²
Finally, calculate the maximum kinetic energy:
KE = 1/2 I ω² => KE = 1/2 × 49.917 kgm² × (53.61 rad/s)²
⇒ KE ≈ 71,791 Joules
Thus, the maximum kinetic energy that can be stored in the flywheel is approximately 71,791 Joules.
An electric fan is turned off, and its angular velocity decreases uniformly from 550 rev/min to 180 rev/min in a time interval of length 4.30 s.
A.) Find the angular acceleration in revolutions per second per second.
= -1.43 rev/s^2
B.) Find the number of revolutions made by the fan blades during the time that they are slowing down in Part A.
= _________? rev
C.) How many more seconds are required for the fan to come to rest if the angular acceleration remains constant at the value calculated in Part A?
= ____________? s
The angular acceleration is -1.43 rev/s². The number of revolutions made by the fan blades is -1870.5 rev. It will take an additional 125.87 seconds for the fan to come to rest.
Explanation:To find the angular acceleration, we can use the formula:
Angular acceleration = (final angular velocity - initial angular velocity) / time
Substituting the given values:
Angular acceleration = (180 rev/min - 550 rev/min) / 4.30 s = -1.43 rev/s²
For part B, the number of revolutions made by the fan blades can be found using the formula:
Number of revolutions = (final angular velocity - initial angular velocity) * time
Substituting the given values:
Number of revolutions = (180 rev/min - 550 rev/min) * 4.30 s = -1870.5 rev
For part C, the time required for the fan to come to rest can be found using the formula:
Time = (final angular velocity - initial angular velocity) / angular acceleration
Substituting the given values:
Time = (0 rev/min - 180 rev/min) / -1.43 rev/s² = 125.87 s
The Young�s modulus for steel is 20.7*1010 N/m2. How much will a 2.5 mm diameter wire 12 cm long, be strained when it supports a load of 450 N?
Answer:
ΔL = 53.14*10⁻⁶ m
Explanation:
Given
E = 20.7*10¹⁰ N/m²
D = 2.5 mm = 2.5*10⁻³ m
L = 12 cm = 0.12 m
P = 450 N
ΔL = ?
We can use the formula
ΔL = P*L / (A*E)
where A = π*D² / 4 = π*(2.5*10⁻³ m)² / 4
⇒ A = 4.908*10⁻⁶ m²
then
ΔL = (450 N)*(0.12 m) / (4.908*10⁻⁶ m²*20.7*10¹⁰ N/m²) = 53.14*10⁻⁶ m
A ball of mass 0.250 kg and a velocity of + 5.00 m/s collides head-on with a ball of mass 0.800 kg that is initially at rest. No external forces act on the balls. After the collision, the velocity of the ball which was originally at rest is + 2.38 m/s. What is the velocity of the 0.250 kg ball?
Answer:
[tex]v_1=-2.616\ m/s[/tex]
Explanation:
Given that,
Mass of ball 1, [tex]m_1=0.25\ kg[/tex]
Initial speed of ball 1, [tex]u_1=5\ m/s[/tex]
Mass of ball 2, [tex]m_2=0.8\ kg[/tex]
Initial speed of ball 2, [tex]u_2=0[/tex] (at rest)
After the collision,
Final speed of ball 2, [tex]v_2=2.38\ m/s[/tex]
Let [tex]v_2[/tex] is the final speed of ball 1.
Initial momentum of the system is :
[tex]p_i=m_1u_1+m_2u_2[/tex]
[tex]p_i=0.25\times 5+0[/tex]
[tex]p_i=1.25\ m/s[/tex]
Final momentum of the system is :
[tex]p_f=m_1v_1+m_2v_2[/tex]
[tex]p_f=0.25\times v_1+0.8\times 2.38[/tex]
[tex]p_f=0.25 v_1+1.904[/tex]
According the law of conservation of linear momentum :
initial momentum = final momentum
[tex]1.25=0.25 v_1+1.904[/tex]
[tex]v_1=-2.616\ m/s[/tex]
So, the final velocity of ball 1 is (-2.616)m/s.
Answer:
mass of ball 1= 0.25 kg
initial speed of ball 1= 5. m/s
mass ball 2= 0.8 kg
initial speed ball 2= 0 ( it was at rest)
final speed of ball 2= 2.38m/s
the formula is:
m1 . v 1 + m2 . v2 = m1 . v1f+ m2 . v2f
0.25 . 5 + 0.8 . 0 = o.25 . x + 0.8 . 2.38
-2.6288 m/s
Explanation:
i got it right on the flvs
Electricity is distributed from electrical substations to neighborhoods at 1.6×104 V . This is a 60 Hz oscillating (AC) voltage. Neighborhood transformers, seen on utility poles, step this voltage down to the 120 V that is delivered to your house. Part A How many turns does the primary coil on the transformer have if the secondary coil has 140 turns? Express your answer using two significant figures. n prim n p r i m = nothing turns SubmitRequest Answer Part B No energy is lost in an ideal transformer, so the output power P out from the secondary coil equals the input power P in to the primary coil. Suppose a neighborhood transformer delivers 230 A at 120 V . What is the current in the 1.6×104 V line from the substation? Express your answer using two significant figures. I i I i = nothing A SubmitRequest Answer Provide Feedback Next
Answer:
(A) 18667 turns
(B) 1.7 A
Solution:
As per the question:
Voltage at which the electricity is distributed, [tex]V_{p} = 1.6\times 10^{4}\ Hz[/tex]
Frequency of the oscillating voltage, f = 60 Hz
Step down voltage, [tex]V_{s} = 120\ V[/tex]
No. of turns in the secondary coil, [tex]N_{s} = 140\ turns[/tex]
Current in the secondary coil, [tex]I_{s} = 230\ A[/tex]
Now,
(A) To calculate the primary no. of turns, we use the relation:
[tex]\frac{V_{s}}{V_{p}} = \frac{N_{s}}{N_{p}}[/tex]
[tex]N_{p} = \frac{V_{p}}{V_{s}}\times N_{s}[/tex]
[tex]N_{p} = \frac{1.6\times 10^{4}}{120}\times 140 = 18,667\ turns[/tex]
(B) To calculate the current in the primary coil, [tex]I_{p}[/tex], we use the relation:
[tex]\frac{V_{p}}{V_{s}} = \frac{I_{s}}{I_{p}}[/tex]
[tex]I_{p} = \frac{V_{s}}{V_{p}} \times {I_{s}}[/tex]
[tex]I_{p} = \frac{120}{1.6\times 10^{4}} \times 230 = 1.7\ A[/tex]
A scene in a movie has a stuntman falling through a floor onto a bed in the room below. The plan is to have the actor fall on his back, but you have been hired to investigate the safety of this stunt. When you examine the mattress, you see that it effectively has a spring constant of 65144 N/m for the area likely to be impacted by the stuntman, but cannot depress more than 12.89 cm without injuring him. To approach this problem, consider a simplified version of the situation. A mass falls through a height of 3.32 m before landing on a spring of force constant 65144 N/m. Calculate the maximum mass that can fall on the mattress without exceeding the maximum compression distance.
=______________________ kg
Answer:
The maximum mass that can fall on the mattress without exceeding the maximum compression distance is 16.6 kg
Explanation:
Hi there!
Due to conservation of energy, the potential energy (PE) of the mass at a height of 3.32 m will be transformed into elastic potential energy (EPE) when it falls on the mattress:
PE = EPE
m · g · h = 1/2 k · x²
Where:
m = mass.
g = acceleration due to gravity.
h = height.
k = spring constant.
x = compression distance
The maximum compression distance is 0.1289 m, then, the maximum elastic potential energy will be the following:
EPE =1/2 k · x²
EPE = 1/2 · 65144 N/m · (0.1289 m)² = 541.2 J
Then, using the equation of gravitational potential energy:
PE = m · g · h = 541.2 J
m = 541.2 J/ g · h
m = 541.2 kg · m²/s² / (9.8 m/s² · 3.32 m)
m = 16.6 kg
The maximum mass that can fall on the mattress without exceeding the maximum compression distance is 16.6 kg.
he drag characteristics of a torpedo are to be studied in a water tunnel using a 1 : 7 scale model. The tunnel operates with freshwater at 20 ˚C, whereas the prototype torpedo is to be used in seawater at 15.6 ˚C. To correctly simulate the behavior of the prototype moving with a velocity of 53 m/s, what velocity is required in the water tunnel?
Answer:20.03 m/s
Explanation:
Given
[tex]L_r=1:7[/tex]
velocity of Prototype [tex]v_p=53 m/s[/tex]
Taking Froude number same for both flow as it is a dimensionless number for different flow regimes in open Flow
[tex](\frac{v_m}{\sqrt{L_mg}})=(\frac{v_p}{\sqrt{L_pg}})[/tex]
[tex]v_m=v_p\times \sqrt{\frac{L_m}{L_p}}[/tex]
[tex]v_m=53\times \frac{1}{\sqrt{7}}[/tex]
[tex]v_m=20.03 m/s[/tex]
A transverse harmonic wave travels on a rope according to the following expression:
y(x,t) = 0.14sin(2.1x + 17.7t)
The mass density of the rope is μ = 0.104 kg/m. x and y are measured in meters and t in seconds.
1)
What is the amplitude of the wave?
m
Your submissions:
2)
What is the frequency of oscillation of the wave?
Hz
Your submissions:
3)
What is the wavelength of the wave?
m
Your submissions:
4)
What is the speed of the wave?
m/s
Your submissions:
5)
What is the tension in the rope?
N
Your submissions:
6)
At x = 3.4 m and t = 0.48 s, what is the velocity of the rope? (watch your sign)
m/s
Your submissions:
7)
At x = 3.4 m and t = 0.48 s, what is the acceleration of the rope? (watch your sign)
m/s2
Your submissions:
8)
What is the average speed of the rope during one complete oscillation of the rope?
m/s
Your submissions:
9)
In what direction is the wave traveling?
+x direction
-x direction
+y direction
-y direction
+z direction
-z direction
Your submissions:
10)
On the same rope, how would increasing the wavelength of the wave change the period of oscillation?
the period would increase
the period would decrease
the period would not change
This answer clarifies various aspects related to a transverse harmonic wave, such as amplitude, frequency, and tension in the rope. The wave is traveling in the -x direction. And, the period of oscillation will increase if the wavelength increases.
Explanation:To answer such questions, we need to understand the equation of the wave given by y(x,t) = A sin (kx + ωt). Here, 'A' represents the amplitude of the wave, 'k' is the wave number, and 'ω' is the angular frequency. We get these values by comparing the given equation with the standard wave equation.
The amplitude of the wave is given by A and here A=0.14 m.The frequency of the wave is given by f=ω/2π=17.7/2π ≈ 2.82 Hz.The wavelength of the wave λ = 2π/k = 2π/2.1 ≈ 2.99 m.The speed of the wave is given by the relation v=λf ≈ 2.99 m * 2.82 Hz ≈ 8.43 m/s.The tension in the rope is given by T = μv^2 = 0.104 kg/m * (8.43 m/s)^2 ≈ 7.4 N.The velocity and acceleration at given values of x and t can be obtained from the time derivative of your wave equation. They require more advanced calculations and depend on the given values of x and t.The average speed of the rope during one complete oscillation is 0, as every point on the rope returns to its initial position during one oscillation.The wave is travelling in the -x direction since the sign before 't' in the wave equation is positive.If the wavelength of the wave is increased, the period of oscillation will increase as the period T=1/f and frequency f=v/λ and v is constant for a given medium.Learn more about Transverse harmonic wave here:https://brainly.com/question/14014748
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A four-wheel-drive vehicle is transporting an injured hiker to the hospital from a point that is 30 km from the nearest point on a straight road. The hospital is 70 km down that road from that nearest point. If the vehicle can drive at 30 kph over the terrain and at 130 kph on the road, how far down the road should the vehicle aim to reach the road to minimize the time it takes to reach the hospital? (Round your answer to two decimal places.)
Final answer:
To minimize the time it takes to reach the hospital, the vehicle should aim to reach the road 30 km from the starting point.
Explanation:
To minimize the time it takes to reach the hospital, the four-wheel-drive vehicle should aim to reach the road at a distance that allows it to travel at its maximum speed on the road. The vehicle can drive at 30 kph over the terrain and at 130 kph on the road.
So, if the vehicle aims to reach the road after traveling 30 km over the terrain at 30 kph, it will take 1 hour. Then, on the road, it will travel at 130 kph for the remaining 40 km, which will take approximately 18.46 minutes. Therefore, the vehicle should aim to reach the road at a distance of 30 km from the starting point to minimize the time it takes to reach the hospital.
A 0.200-kg mass is attached to the end of a spring with a spring constant of 11 N/m. The mass is first examined (t = 0) when the mass is 17.0 cm from equilibrium in the positive x-direction, and is traveling at 2.0 m/s in the positive x-direction.
a) Write an equation x(t) that describes the position of this mass as a function of time. Express this function in terms of numerical values, trigonometric functions and the time variable "t".b) Repeat for v(t), the speed of the mass as a function of time.c) Repeat for a(t), the acceleration of the mass as a function of time.
Answer:
a) x (t) = 0.3187 cos (7.416 t + 1.008) , b) v = -2,363 sin (7,416 t + 1,008)
c) a = - 17.52 cos (7.416t + 1.008)
Explanation:
The spring mass system creates a harmonic oscillator that is described by the equation
x = Acos (wt + φ)
Where is the amplitude, w the angular velocity and fi the phase
a) Let's reduce the SI system
x = 17.0 cm (1 m / 100 cm) = 0.170 m
The angular velocity is given by
w = √ (k / m)
w = √ 11 / 0.200
w = 7.416 rad / s
Let's look for the terms of the equation with the data for time zero (t = 0 s)
0.170 = A cos φ
Body speed can be obtained by derivatives
v = dx / dt
v = -A w sin (wt + φ)
2.0 = -A 7.416 sin φ
Let's write the two equations
0.170 = A cos φ
2.0 / 7.416 = -A sin φ
Let's divide those equations
tan φ= 2.0 / (7.416 0.170)
φ= tan⁻¹ (1,586)
φ= 1.008 rad
We calculate A
A = 0.170 / cos φ
A = 0.170 / cos 1.008
A = 0.3187 m
With these values we write the equation of motion
x (t) = 0.3187 cos (7.416 t + 1.008)
b) the speed can be found by derivatives
v = dx / dt
v = - 0.3187 7.416 sin (7.416 t +1.008)
v = -2,363 sin (7,416 t + 1,008)
c) the acceleration we look for conserved
a = dv / dt
a = -2,363 7,416 cos (7,416 t + 1,008)
a = - 17.52 cos (7.416t + 1.008)
Sheila (m=56.8 kg) is in her saucer sled moving at 12.6 m/s at the bottom of the sledding hill near Bluebird Lake. She approaches a long embankment inclined upward at 16° above the horizontal. As she slides up the embankment, she encounters a coefficient of friction of 0.128. Determine the height to which she will travel before coming to rest.
Final answer:
To find the height Sheila will ascend before stopping, we use energy conservation, calculating her initial kinetic energy and the work done against friction to solve for the final potential energy, which includes the height she reaches on the incline.
Explanation:
To determine the height Sheila will travel up the embankment before coming to rest, we need to apply the principles of energy conservation and include the work done against friction. Sheila's initial kinetic energy (KE) at the bottom of the hill is converted into gravitational potential energy (PE) and the work done against friction as she moves up the incline.
We can calculate the initial kinetic energy using the formula KE = 1/2 * m * v2, where m is her mass and v is her velocity. Then, we find the work done against friction, which is equal to the force of friction times the distance traveled (Wfriction = Ffriction * d). The force of friction is calculated as the coefficient of friction multiplied by the normal force, which, on an incline, is the component of the gravitational force perpendicular to the slope.
Using these concepts, we can set up the equation: KEinitial = PEfinal + Wfriction. Solving for the final potential energy gives us PEfinal = m * g * h, where h is the height she will reach. The height can be determined by rearranging this equation after calculating the work done against friction and knowing the initial kinetic energy.
It is important to note that since no values are given for distances or the length of the incline and we are assuming a constant coefficient of friction, we find the height as a function of distance she travels up the slope until she comes to rest.
A potter's wheel having a radius 0.47 m and a moment of inertia of 14.2 kg · m2 is rotating freely at 53 rev/min. The potter can stop the wheel in 8.0 s by pressing a wet rag against the rim and exerting a radially inward force of 68 N. Find the effective coefficient of kinetic friction between the wheel and the wet rag.
Answer:
the effective coefficient of kinetic friction between the wheel and the wet rag is 0.31
Explanation:
given information:
radius, r = 0.47 m
moment of inertia, I = 14.2 kg[tex]m^{2}[/tex]
angular velocity, ω0 = 53 rev/min = 53 x 2π/60 = 5.5 rad/s
time, t = 8.0 s
inward force, N = 68 N
τ = I α
F r = I α, F is friction force, F = μ N
μ N r = I α
μ = I α / N r
We have to find α
ωt = ω0 + αt. ωt = 0 because the wheel stop after 8 s
0 = 5.5 + α 8
α = -5.5/8 = 0.69 [tex]rad/s^{2}[/tex]
Now we can calcultae the coeffcient of kinetik friction
μ = I α / N r
= (14.2) (0.69) / (68) (0.47)
= 0.31
When water freezes, its volume increases by 9.05% (that is, ΔV / V0 = 9.05 × 10-2). What force per unit area is water capable of exerting on a container when it freezes? (It is acceptable to use the bulk modulus of water, B = 2.2 × 109 N/m2, for this problem.) Give your answer in N/cm2.
Answer:
P = 1.99 10⁸ Pa
Explanation:
The definition of the bulk module is
B = - P / (ΔV / V)
The negative sign is included for which balk module is positive, P is the pressure and V that volume
They tell us that the variation in volume is 9.05%, that is
ΔV / V = 9.0Δ5 / 100 = 0.0905
P = - B DV / V
P = 2.2 10⁹ (0.0905)
P = 1.99 10⁸ Pa
Final answer:
The force per unit area water exerts on a container when it freezes is calculated using the bulk modulus of water, resulting in a pressure of 1.99 × 10⁴ N/cm². This explains why freezing water can cause significant damage to rigid structures.
Explanation:
When water freezes, its volume increases by 9.05% (ΔV/V = 9.05 × 10⁻²). The force per unit area that water can exert on a container when it freezes can be calculated using the bulk modulus of water, which is given as B = 2.2 × 10⁹ N/m². The change in volume fraction (ΔV/V) is directly related to the change in pressure (ΔP) and the bulk modulus (B) by the formula ΔP = -B(ΔV/V), where the negative sign indicates a reduction in volume leads to an increase in pressure, but in this case, we're dealing with an expansion, so ΔV/V is positive.
To find the pressure in N/cm², we first calculate the pressure in Pascals (Pa) using the given values:
ΔP = B(ΔV/V)
= (2.2 × 10⁹ N/m²)(9.05 × 10⁻²)
= 1.99 × 10⁸ Pa.
Since 1 Pa equals 1 N/m² and 1 N/m² equals 0.0001 N/cm²,
the pressure in N/cm² is 1.99 × 10⁴ N/cm².
78. A film of oil on water will appear dark when it is very thin, because the path length difference becomes small compared with the wavelength of light and there is a phase shift at the top surface. If it becomes dark when the path length difference is less than one-fourth the wavelength, what is the thickest the oil can be and appear dark at all visible wavelengths? Oil has an index of refraction of 1.40.
To solve this problem it is necessary to apply the concepts related to the condition of path difference for destructive interference between the two reflected waves from the top and bottom of a surface.
Mathematically this expression can be described under the equation
[tex]\delta = 2nt[/tex]
Where
n = Refractive index
t = Thickness
In terms of the wavelength the path difference of the reflected waves can be described as
[tex]\delta = \frac{\lambda}{4}[/tex]
Where
\lambda = Wavelenght
Equation the two equations we have that
[tex]2nt = \frac{\lambda}{4}[/tex]
[tex]t = \frac{\lambda}{8n}[/tex]
Our values are given as
[tex]\lambda = 380nm \rightarrow[/tex] Wavelength of light
[tex]n = 1.4[/tex]
[tex]t = \frac{380nm}{8*1.4}[/tex]
[tex]t = 33.93nm[/tex]
Therefore the minimum thickness of the oil for destructive interference to occur is approximately 34.0 nm
The maximum thickness of the oil film that will appear dark at all visible wavelengths is approximately 33.9 nm, ensuring destructive interference for all visible light.
Explanation:The oil film appears dark when its thickness is such that the path length difference is less than one-fourth the wavelength of the light. This is due to the destructive interference of light waves. Given that the thickest the oil can be and still appear dark when the path length difference is one-fourth the shortest visible wavelength in oil, we need to find this thickness value.
The shortest visible wavelength of light in air is typically about 380 nm, which is violet. However, the speed and consequently the wavelength of light will decrease in oil due to its refractive index being higher. Therefore, the wavelength of light in oil is shorter than in air and can be found by dividing the wavelength in the air by the index of refraction, which gives about 271 nm for violet light in oil.
Since the path length difference is twice the film thickness, the maximum thickness of the oil should be one-eighth the shortest visible wavelength in oil to appear dark, which is approximately 33.9 nm. This calculation ensures destructive interference for all visible wavelengths and hence, the film will appear black or dark at all visible wavelengths.
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A truck is traveling horizontally to the right. When the truck starts to slow down, the crate on the (frictionless) truck bed starts to slide. In what direction could the net force be on the crate?
a. Straight up (the normal force).
b. Horizontal and to the left.
c. Straight down (because of gravity).
d. Horizontal and to the right.
e. No direction. The net force is zero.
To solve this problem it is necessary to apply Newton's first law which warns that every body remains in continuous motion or at rest until an external force acts on it.
From the statement it is said that there is no friction on the crate, then the sum of forces in the horizontal direction will be zero. Here as the truck is slowing down, there is not net horizontal force on the crate, it means that the crate is at rest.
The correct answer is e. No direction. The net force is zero.
When the truck slows down, the crate experiences a 'force' that is horizontal and to the left due to relative motion and the principle of inertia.
Explanation:The net force acting on the crate would be horizontal and to the left (option b). This is due to the principle of inertia. When the truck slows down, the crate tries to maintain its original speed and direction, which is horizontal and to the right. This will make the crate slide towards the back of the truck, giving a semblance of a force acting to the left on the crate. Note that this is a result of the relative motion between the crate and the truck and not an actual force acting on the crate.
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Briefly answer the following questions. a) A clock is mounted on the wall. As you look at it, what is the direction of the angular velocity vector of the second hand? What is the value of the angular acceleration of the second hand? b) If the angular acceleration of a rigid body is zero, what is the functional form of the angular velocity? c) What if another planet the same size as Earth were put into orbit around the Sun along with Earth. Would the moment of inertia of the system increase, decrease, or stay the same? d) Does increasing the number of blades on a propeller increase or decrease its moment of inertia, and why? e) Can you think of a body that has the same moment of inertia for all possible axes? If so, give an example, and if not, explain why this is not possible. Can you think of a body that has the same moment of inertia for all axes passing through a certain point? If so, give an example and indicate where the point is located. f) To maximize the moment of inertia of a flywheel while minimizing its weight, what shape and distribution of mass should it have? Explain. g) Is it possible to change the translational) kinetic energy of an object without changing its rotational kinetic energy? What about the reverse of it?
Answer:
Explanation:
a ) The direction of angular velocity vector of second hand will be along the line going into the plane of dial perpendicular to it.
b ) If the angular acceleration of a rigid body is zero, the angular velocity will remain constant.
c ) If another planet the same size as Earth were put into orbit around the Sun along with Earth the moment of inertia of the system will increase because the mass of the system increases. Moment of inertia depends upon mass and its distribution around the axis.
d ) Increasing the number of blades on a propeller increases the moment of inertia , because both mass and mass distribution around axis of rotation increases.
e ) It is not possible that a body has the same moment of inertia for all possible axes because a body can not remain symmetrical about all axes possible. Sphere has same moment of inertia about all axes passing through its centre.
f ) To maximize the moment of inertia of a flywheel while minimizing its weight, the shape and distribution of mass should be such that maximum mass of the body may be situated at far end of the body from axis of rotation . So flywheel must have thick outer boundaries and this should be
attached with axis with the help of thin rods .
g ) When the body is rotating at the same place , its translational kinetic energy is zero but its rotational energy can be increased
at the same place.