Answer:
8.953 s
Explanation:
Here a time of 8953 ms is given.
Some of the prefixes of the SI units are
mili = 10⁻³
micro = 10⁻⁶
nano = 10⁻⁹
kilo = 10³
The number is 8953.0
Here, the only solution where the number of significant figures is mili
1 milisecond = 1000 second
[tex]1\ second=\frac{1}{1000}\ milisecond[/tex]
[tex]\\\Rightarrow 8953\ milisecond=\frac{8953}{1000}\ second\\ =8.953\ second[/tex]
So 8953 ms = 8.953 s
Give a critical discussion on why hardness test needs to perform in engineering practice.
b) Discuss 2 possible source of errors (other than parallax error) and how it's can effect the result.
Answer:
Explanation:
a) Hardness is a measure of the resistance of a material to permanent deformation (plastic) on its surface,
Hardness tests play an important role in material testing, quality control and component acceptance.
Hardness test are needed to be perform as a quality assurance procedure, to validate materials are according to the specific hardness required,
We depend on the data to verify the quality of the components to determine if a material has the necessary properties for its intended use.
Through the years, the establishment of increasingly productive and effective means of testing, has given way to new cutting-edge methods that perform and interpret hardness tests more effectively than ever. The result is a greater capacity and dependence on "letting the instrument do the work", contributing to substantial increases in performance and consistency and continuing to make hardness tests very useful in industrial and R&D applications.
b)
Instrumental errors: Instrument calibration is extremely important. An instrument with expired calibration may be generating erroneous data systematically.Enviromental error: An example is when surface preparation of the sample to be tested is poor, then the error can be presented when measuring the indentation on the sample to determine the hardness value."The hardness test is a critical aspect of engineering practice for several reasons:
1. Material Performance: Hardness is a measure of a material's resistance to deformation, which is directly related to its wear resistance, strength, and durability. By performing hardness tests, engineers can predict how a material will perform under service conditions.
2. Quality Control: Hardness testing is used to ensure that materials meet the required specifications. It is a non-destructive (in some cases) or semi-destructive method to verify the quality of the material without compromising its integrity.
3. Material Selection: The results of hardness tests help engineers to select appropriate materials for different applications. A material that is too soft or too hard for a particular application can lead to premature failure or unnecessary costs.
4. Heat Treatment Verification: Heat treatment processes such as quenching, annealing, and tempering are used to alter the mechanical properties of materials. Hardness testing is used to confirm that the desired properties have been achieved after such treatments.
5. Failure Analysis: In the event of a material failure, hardness testing can provide valuable information about the cause of the failure. It can indicate if the material was too brittle or too soft for the intended application.
6. Comparative Analysis: Hardness tests provide a simple way to compare the properties of different materials or different batches of the same material.
7. Research and Development: In the development of new materials, hardness testing is an essential tool for characterizing materials and understanding their properties.
B) Two possible sources of error in hardness testing, other than parallax error, and their effects on the results are:
1. Indenter Geometry Errors: The geometry of the indenter (e.g., ball, cone, or pyramid) must be precise according to the test standard being used. Wear, damage, or incorrect shape of the indenter can lead to inaccurate measurements. For instance, a worn or blunt indenter will give lower hardness values than the actual hardness of the material.
2. Loading Rate and Duration Errors: The rate at which the load is applied and the duration for which it is held can significantly affect the hardness measurement. Some materials exhibit time-dependent plasticity (creep), which can affect the size of the indentation. If the loading rate or duration is not consistent with the standards or previous tests, the results may not be comparable or may not accurately represent the material's hardness. For example, applying the load too quickly may result in an underestimation of hardness due to the material's inability to fully resist the indenter.
To minimize these errors, it is essential to follow standardized testing procedures, maintain and calibrate equipment regularly, and ensure that the test conditions are consistent. Additionally, operators should be well-trained and aware of the potential sources of error to ensure the accuracy and reliability of the hardness test results."
An aircraft is in a steady level turn at a flight speed of 200 ft/s and a turn rate about the local vertical of 5 deg/s. Thrust frm the engine is along the flight direction. Given the weight of the aircraft as 50,000 lb and L/D of 10, determine the lift, drag and thrust required for the given equilibrium flight. Assume g = 32.2 ft/s
Answer:
L =4166.66 N and D = 46296.29 N and T = 96047.34 N
Explanation:
given,
flight speed = 200 ft/s
turn rate = 5 deg/s
weight of aircraft = 50000 lb
L/D = 10
for equilibrium in the horizontal position
w - L cos ∅ - D sin ∅ = 0 .............(1)
D cos ∅ - L sin ∅ = 0................(2)
L/D = tan ∅ = 10
∅ = 84.29°
50000 - L cos 84.29°- D sin 84.29°= 0
D cos 84.29° - L sin 84.29° = 0
on solving the above equation we get
L =4166.66 N and D = 46296.29 N
thrust force calculation:
T = W sin ∅ + D
= 50000×sin 84.29 + 46296.29
T = 96047.34 N
The expression "1in. = 2.54cm" is properly called: A. an equation B. a conversion factor C. a rule of thumb D. an equivalence statement
Answer:
B. a conversion factor
Explanation:
The expression "1 in. = 2.54 cm" is called a conversion factor.
With this expression inches can be converted to centimeters.
Inversely the expression can also be used to convert centimeters to inches
By rearranging the equation we get
[tex]\frac{1}{2.54}\ inches=1\ cm\\\Rightarrow 1\ cm=0.394\ inches[/tex]
The statement which shows the equal nature of two different expressions is called an equation
Rule of thumb is a general approximation of a result of a test or experiment.
Hi your answer is B on edge
A football is kicked from ground level with an initial velocity of 22.0 m/s at angle of 58.5° above the horizontal. How long, in seconds, is the football in the air before it hits the ground? Ignore air resistance. _______ seconds
Final answer:
To find the time the football is in the air before hitting the ground, we can analyze the vertical motion using the given initial velocity and launch angle.
Explanation:
To find the time it takes for the football to hit the ground, we need to analyze the vertical motion of the football. We can use the formula:
t = (2 * vy) / g
where t is the time, vy is the vertical component of the initial velocity, and g is the acceleration due to gravity.
Given that the initial velocity is 22.0 m/s and the launch angle is 58.5°, we can find the vertical component of the velocity using the formula:
vy = v * sin(θ)
where v is the initial velocity and θ is the launch angle.
Using this information, we can calculate the time it takes for the football to hit the ground.
Part A) A small turbo prop commuter airplane, starting from rest on a New York airport runway, accelerated for 19.0s before taking off. It's speed at takeoff is 50.0 m/s. Calculate the acceleration of the plane , assuming it remains constant. Part B) In this problem, how far did the plane move while accelerating for 19.0s?
Answer:
a)a=2.63m/s^2
b)x=475.25m
Explanation:
To solve the first part of this problem we use the following equation
Vf=final speed=50m/s
Vo=initial speed=0
t=time
a=aceleration
a=(Vf-Vo)/t
a=(50-0)/19=2.63m/s^2
B )
For the second part of this problem we use the following equation
X=(Vf^2-Vo^2)/2a
X=(50^2-0^2)/(2*2.63)=475.25m
What is the difference between Reynolds equation set and Navier Stokes equation?
Answer:
Navier Stokes equation
( 1 ) it is a partial differential equation that is describe the flow of incompressible fluids
Reynolds equation
(1) it is partial differential equation that governs the pressure distribution of thin viscous fluid in lubrication
Explanation:
Navier Stokes equation
( 1 ) it is a partial differential equation that is describe the flow of incompressible fluids
(2) Navier Stokes equation is used to model weather and ocean current and water flow in the pipe and air flow around wing
( 3) equation is
[tex]\nabla .\overrightarrow{v} = 0[/tex] momentum equation
[tex]\rho \frac{d\overrightarrow{v}}{dt} = \nabla p + \rho \overrightarrow{g} + \mu \nabla ^2 v^2[/tex]
here [tex]\nabla p[/tex] is pressure gradient and [tex]\rho \overrightarrow{g}[/tex] is body force and [tex]\mu \nabla ^2 v^2[/tex] is diffusion term
and
Reynolds equation
(1) it is partial differential equation that governs the pressure distribution of thin viscous fluid in lubrication
(2) it is drive in 1886 from Navier Stokes law
(3) equation is attach
here
An earthquake emits both S-waves and P-waves which travel at different speeds through the Earth. A P-wave travels at 9000 m/s and an S-wave travels at 5000 m/s. If P-waves are received at a seismic station 2.00 minute before an S wave arrives, how far away is the earthquake center?
To determine the distance to the epicenter of an earthquake, we can use the time difference between the arrival of P-waves and S-waves. In this case, the P-wave arrives 2.00 minutes before the S-wave. The distance of the earthquake center can be calculated by multiplying the time difference by the speed difference between the two waves.
Explanation:To determine the distance to the epicenter of an earthquake, we can use the time difference between the arrival of P-waves and S-waves. In this case, the P-wave arrives 2.00 minutes before the S-wave. We know that the speed of the P-wave is 9000 m/s and the speed of the S-wave is 5000 m/s.
We can calculate the distance using the formula: distance = speed × time.
So, the distance of the earthquake center can be calculated as follows:
Time difference = 2.00 minutes = 120 secondsDistance = (9000 m/s - 5000 m/s) × 120 secondsDistance = 4000 m/s × 120 secondsDistance = 480000 meters or 480 kilometersLearn more about Calculating distance to earthquake center here:https://brainly.com/question/31108695
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A small metal bead, labeled A, has a charge of 28 nC .It is touched to metal bead B, initially neutral, so that the two beads share the 28 nC charge, but not necessarily equally. When the two beads are then placed 5.0 cmapart, the force between them is 4.8×10^−4 N . Assume that A has a greater charge. What is the charge qA and qB on the beads?
Answer:
Explanation:
Let the charge on bead A be q nC and the charge on bead B be 28nC - qnC
Force F between them
4.8\times10^{-4} = [tex]\frac{9\times10^9\times q\times(28-q)\times10^{-18}}{(5\times10^{-2})^2}[/tex]
=120 x 10⁻⁸ = 9 x q(28 - q ) x 10⁻⁹
133.33 = 28q - q²
q²- 28q +133.33 = 0
It is a quadratic equation , which has two solution
q_A = 21.91 x 10⁻⁹C or q_B = 6.09 x 10⁻⁹ C
The charges [tex]q_A[/tex] = 21.907nC and [tex]q_B[/tex] = 6.093nC are the charges on the beads.
We have to use Coulomb's Law and the principle of charge conservation.
1. Conservation of Charge:
The total charge before and after they are touched must be the same. Initially, bead A has a charge of 28 nC, and bead B is neutral. After touching, let:
[tex]\[ q_A + q_B = 28 \, \text{nC} \][/tex]
2. Coulomb's Law:
The force between two charges is given by Coulomb's Law:
[tex]\[ F = k \frac{q_A q_B}{r^2} \][/tex]
Let's set up the equation:
[tex]\[4.8 \times 10^{-4} = 8.99 \times 10^9 \frac{q_A q_B}{(0.05)^2}\][/tex]
Rearrange to solve for [tex]\( q_A q_B \)[/tex]:
[tex]\[q_A q_B = \frac{4.8 \times 10^{-4} \times (0.05)^2}{8.99 \times 10^9}\][/tex]
[tex]\[q_A q_B = \frac{4.8 \times 10^{-4} \times 0.0025}{8.99 \times 10^9}\][/tex]
[tex]\[q_A q_B = \frac{1.2 \times 10^{-6}}{8.99 \times 10^9}\][/tex]
[tex]\[q_A q_B = 1.3348 \times 10^{-16} \, \text{C}^2\][/tex]
Now, we have two equations:
1. [tex]\( q_A + q_B = 28 \times 10^{-9} \, \text{C} \)[/tex]
2. [tex]\( q_A q_B = 1.3348 \times 10^{-16} \, \text{C}^2 \)[/tex]
To solve these, we can substitute [tex]\( q_B = 28 \times 10^{-9} \, \text{C} - q_A \)[/tex] into the second equation:
[tex]\[q_A \left( 28 \times 10^{-9} - q_A \right) = 1.3348 \times 10^{-16}\][/tex]
[tex]\[28 \times 10^{-9} q_A - q_A^2 = 1.3348 \times 10^{-16}\][/tex]
[tex]\[q_A^2 - 28 \times 10^{-9} q_A + 1.3348 \times 10^{-16} = 0\][/tex]
This is a quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex], where:
[tex]\[a = 1, \quad b = -28 \times 10^{-9}, \quad c = 1.3348 \times 10^{-16}\][/tex]
Solve using the quadratic formula [tex]\( q_A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[q_A = \frac{28 \times 10^{-9} \pm \sqrt{(28 \times 10^{-9})^2 - 4 \times 1 \times 1.3348 \times 10^{-16}}}{2}\][/tex]
[tex]\[q_A = \frac{28 \times 10^{-9} \pm \sqrt{7.84 \times 10^{-16} - 5.3392 \times 10^{-16}}}{2}\][/tex]
[tex]\[q_A = \frac{28 \times 10^{-9} \pm \sqrt{2.5008 \times 10^{-16}}}{2}\][/tex]
[tex]\[q_A = \frac{28 \times 10^{-9} \pm 1.5814 \times 10^{-8}}{2}\][/tex]
This gives us two solutions for [tex]\( q_A \)[/tex]:
1. [tex]\( q_A = \frac{28 \times 10^{-9} + 1.5814 \times 10^{-8}}{2} \\q_A = \frac{4.3814 \times 10^{-8}}{2} \\q_A = 2.1907 \times 10^{-8} \, \text{C} \\q_A = 21.907 \, \text{nC} \)[/tex]
2. [tex]\( q_A = \frac{28 \times 10^{-9} - 1.5814 \times 10^{-8}}{2} \\q_A = \frac{1.2186 \times 10^{-8}}{2} \\q_A = 6.093 \times 10^{-9} \, \text{C} \\q_A = 6.093 \, \text{nC} \)[/tex]
Since bead A has the greater charge, we take:
[tex]\[q_A = 21.907 \, \text{nC}\][/tex]
And the charge on bead B:
[tex]\[q_B = 28 \, \text{nC} - 21.907 \, \text{nC} = 6.093 \, \text{nC}\][/tex]
So, the charges on the beads are:
[tex]\[q_A = 21.907 \, \text{nC}, \quad q_B = 6.093 \, \text{nC}\][/tex]
Two train whistles have identical frequencies of 175 Hz. When one train is at rest in the station and the other is moving nearby, a commuter standing on the station platform hears beats with a frequency of 4.05 beats/s when the whistles operate together. What are the two possible speeds and directions the moving train can have? slower speed m/s Correct: Your answer is correct. faster speed m/s Changed: Your submitted answer was incorrect. Your current answer has not been submitted.
Answer:
The speed of the train is 7.75 m/s towards station.
The speed of the train is 8.12 m/s away from the station.
Explanation:
Given that,
Frequency of the whistles f= 175 Hz
Beat frequency [tex]\Delta f= 4.05 Hz[/tex]
Speed of observer = 0
We need to calculate the frequency
Using formula of beat frequency
[tex]\Delta f=f'-f[/tex]
[tex]f'=\Delta f+f[/tex]
[tex]f'=4.05+175[/tex]
[tex]f'=179.05\ Hz[/tex]
When the train moving towards station, then the frequency heard is more than the actual
Using Doppler effect
[tex]f'=f(\dfrac{v-v_{o}}{v-v_{s}})[/tex]
[tex]v=v-\dfrac{vf}{f'}[/tex]
Put the value into the formula
[tex]v=343-\dfrac{343\times175}{179.05}[/tex]
[tex]v=7.75\ m/s[/tex]
The speed of the train is 7.75 m/s towards station.
When the train moving away form the station
Again beat frequency
[tex]\Delta f=f-f'[/tex]
[tex]f'=f-\Delta [/tex]
[tex]f'=175-4.05[/tex]
[tex]f'=170.95\ Hz[/tex]
We need to calculate the speed
Using Doppler effect
[tex]f'=f(\dfrac{v-v_{o}}{v+v_{s}})[/tex]
[tex]v=\dfrac{vf}{f'}-v[/tex]
Put the value into the formula
[tex]v=\dfrac{343\times175}{170.95}-343[/tex]
[tex]v=8.12\ m/s[/tex]
The speed of the train is 8.12 m/s away from the station.
Hence, This is the required solution.
Consider two concentric spheres whose radii differ by t = 1 nm, the "thickness" of the interface. At what radius (in nm) of sphere does the volume of a shell equal that of the interior sphere? Assume the shell thickness to be t = 1 nm.
Consider two concentric spheres whose radii differ by t = 1 nm, the "thickness" of the interface. On log-log axes, plot the ratio of the shell volume to the interior sphere volume for an interior sphere radius of 10 microns to 1 nm. Assume the shell thickness to be t = 1 nm.
Answer:
3.85 nm
Explanation:
The volume of a sphere is:
V = 4/3 * π * r^3
The volume of a shell is the volume of the big sphere minus the volume of the small sphere
Vs = 4/3 * π * r2^3 - 4/3 * π * r1^3
Vs = 4/3 * π * (r2^3 - r1^3)
If the difference between the radii is 1 nm
r2 = r1 + 1 nm
Vs = 4/3 * π * ((r1 + 1)^3 - r1^3)
Vs = 4/3 * π * ((r1 + 1)^3 - r1^3)
Vs = 4/3 * π * (r1^3 + 3*r1^2 + 3*r1 + 1 - r1^3)
Vs = 4/3 * π * (3*r1^2 + 3*r1 + 1)
The volme of the shell is equall to the volume of the inner shell:
4/3 * π * (r1^3) = 4/3 * π * (3*r1^2 + 3*r1 + 1)
r1^3 = 3*r1^2 + 3*r1 + 1
0 = -r1^3 + 3*r1^2 + 3*r1 + 1
Solving this equation electronically:
r1 = 3.85 nm
A 0.5 kg block is attached to a spring (k = 12.5 N/m). The damped frequency is 0.2% lower than the natural frequency, (a) What is the damping constant? (b) How does the amplitude vary with time? (c) Determine the critical damping constant?
For a mass-spring system with specified parameters, the damping constant is found to be 4.999 Ns/m. The amplitude reduces exponentially over time. The critical damping constant, defining the transition from underdamped to overdamped behavior, is 5 Ns/m.
Explanation:This problem pertains to a system of mass springs undergoing damped oscillations. The damping constant (b) can be calculated using the formula b = 2*m*ωd. Here ωd signifies the damped frequency and is given by ωd = ω * (1 - γ/2), where ω is the natural frequency and γ is the damping ratio of 0.2% = 0.002. The natural frequency ω = sqrt(k/m) = sqrt(12.5/0.5) = 5 rad/s. Substituting these values in the equation gives b = 2*0.5*4.999 = 4.999 Ns/m.
The amplitude (A) of a damped oscillation decreases exponentially with time due to energy loss, described by A(t) = A0 * e^(-γωt/2), wherein A0 is the initial amplitude.
The critical damping constant (bc), describing the boundary between underdamped and overdamped systems, is equal to 2*sqrt(m*k) = 2*sqrt(0.5*12.5) = 5 Ns/m.
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An electric field of intensity 3.0kN is applied along the x axis. Calculate the electric flux through a rectangular plane 0.350m wide and 0.700 m long if the following conditions are true: a) the plane is parallel to the yz plane b) The plane is parallel to thte xy plane c) The plane contains the y axis and its normal makes an angle of 30 degrees with the x axis.
Answer:
Explanation:
Area of plane = .35 x .7 = 0.245 m²
a) When plane is perpendicular to field ( plane is parallel to yz plane. )
Flux = field x area
3000 x .245 = 735 weber
b ) When plane is parallel to xy plane , the plane also becomes parallel to electric field so no flux will pass though the given plane.
Flux through the plane = 0
c ) Since normal to the plane makes 30 degree with x axis, it will also make 30 degree with direction of the field.
Flux through the plane
= electric field x area x cos 30
3000 x .245 x .866
636.51 weber.
What is the forward velocity of a 250 lb stone projectile that leaves a medieval catapult 55 degrees from the ground at 50 m/s?
Answer:
28.5 m/s
Explanation:
The speed has two orthogonal components, horizontal and vertical. To decompose the speed into these components we can use these trigonometric equations:
Vh = V * cos(a)
Vv = V * sin(a)
The forwards velocity is the horizontal component, so we use
Vh = V * cos(a)
Vh = 50 * cos(55)
Vh = 50 * 0.57
Vh = 28.5 m/s
A student gives a 5.0 kg box a brief push causing the box to move with an initial speed of 8.0 m/s along a rough surface. The box experiences a friction force of 30 N as it slows to a stop. How long does it take the box to stop?
Answer:
The time taken to stop the box equals 1.33 seconds.
Explanation:
Since frictional force always acts opposite to the motion of the box we can find the acceleration that the force produces using newton's second law of motion as shown below:
[tex]F=mass\times acceleration\\\\\therefore acceleration=\frac{Force}{mass}[/tex]
Given mass of box = 5.0 kg
Frictional force = 30 N
thus
[tex]acceleration=\frac{30}{5}=6m/s^{2}[/tex]
Now to find the time that the box requires to stop can be calculated by first equation of kinematics
The box will stop when it's final velocity becomes zero
[tex]v=u+at\\\\0=8-6\times t\\\\\therefore t=\frac{8}{6}=4/3seconds[/tex]
Here acceleration is taken as negative since it opposes the motion of the box since frictional force always opposes motion.
The acceleration of a particle is given by a = 3t – 4, where a is in meters per second squared and t is in seconds. Determine the velocity and displacement for the time t = 3.6 sec. The initial displacement at t = 0 is s0 = – 8 m, and the initial velocity is v0 = – 5 m/sec.
Answer:
[tex]v=0.04m/s\\[/tex]
[tex]s=-28.592m\\[/tex]
Explanation:
[tex]a = 3t-4[/tex]
[tex]v(t)=\int\limits^t_0 {a(t)} \, dt =3/2*t^{2}-4t+v_0\\[/tex]
if t=3.6s and initial velocity, v0, is -5m/s
[tex]v=0.04m/s\\[/tex]
[tex]s(t)=\int\limits^t_0 {v(t)} \, dt =1/2*t^{3}-2t^{2}+v_0*t+s_0\\[/tex]
if t=3.6s and the initial displacement, s0, is -8m:
[tex]s=-28.592m\\[/tex]
A small particle has charge -3.90 μC and mass 1.80×10^−4 kg . It moves from point A, where the electric potential is VA= 130 V , to point B, where the electric potential VB = 500 V. The electric force is the only force acting on the particle. The particle has a speed of 4.90 m/s at point A. What is its speed at point B? Express your answer in meters per second to three significant figures.
Final answer:
By applying the conservation of energy principle, accounting for the work done by the electric field as change in potential energy, we calculate the change in kinetic energy to find the speed of the charged particle at a different electric potential.
Explanation:
To calculate the speed of the particle at point B, we need to consider the change in electric potential energy and convert it to kinetic energy. The work done on the particle by the electric field is equal to the change in potential energy, which is given by W = q(VB - VA), where q is the charge of the particle, VA and VB are the electric potentials at points A and B respectively. Since no other forces act on the particle, the work done by the electric field causes a change in kinetic energy, which is defined by ½mv22 - ½mv12 = q(VB - VA), where m is the mass of the particle and v1 and v2 are the speeds at points A and B.
Rearranging the equation to solve for v2 gives us v2 = √(2q(VB - VA)/m + v12). Substituting in the given values, we can find the speed at point B.
What is the diameter of a 12lb shot if the specific gravity is of the shot iron in the shot is 6.8, the density of fresh water 62.4lb/ft
Answer:
The diameter is 0.378 ft.
Explanation:
Given that,
Mass of shot = 12 lb
Density of fresh water = 62.4 lb/ft
Specific gravity = 6.8
We need to calculate the volume of shot
[tex]V = \dfrac{4}{3}\pi r^3\ ft^3[/tex]
The density of shot is
Using formula of density
[tex]\rho = \dfrac{m}{V}[/tex]
Put the value into the formula
[tex]\rho =\dfrac{12}{ \dfrac{4}{3}\pi r^3}[/tex]
We need to calculate the radius
Using formula of specific gravity
[tex]specific\ gravity =\dfrac{density\ of\ shot}{dnsity\ of\ water}[/tex]
Put the value into the formula
[tex]6.8=\dfrac{\dfrac{12}{\dfrac{4}{3}\pi r^3}}{62.4}[/tex]
[tex]r^3=\dfrac{12}{\dfrac{4}{3}\pi\times6.8\times62.4}[/tex]
[tex]r^3=0.0067514[/tex]
[tex]r =(0.0067514)^{\frac{1}{3}}[/tex]
[tex]r=0.1890\ ft[/tex]
The diameter will be
[tex]d = 2\times r[/tex]
[tex]d =2\times0.1890[/tex]
[tex]d =0.378\ ft[/tex]
Hence, The diameter is 0.378 ft.
Four springs with the following spring constants, 113.0 N/m, 65.0 N/m, 102.0 N/m, and 101.0 N/m are connected in series. What is their effective spring constant? 004393 Submit Answer Incorrect. Tries 4/99 Previous Tres If a mass of 0.31 kg is attached what will be the frequency of oscillation?
Answer:
[tex]K_e_q=22.75878093\frac{N}{m}[/tex]
[tex]f=1.363684118Hz[/tex]
Explanation:
In order to calculate the equivalent spring constant we need to use the next formula:
[tex]\frac{1}{K_e_q} =\frac{1}{K_1} +\frac{1}{K_2} +\frac{1}{K_3} +\frac{1}{K_4}[/tex]
Replacing the data provided:
[tex]\frac{1}{K_e_q} =\frac{1}{113} +\frac{1}{65} +\frac{1}{102} +\frac{1}{101}[/tex]
[tex]K_e_q=22.75878093\frac{N}{m}[/tex]
Finally, to calculate the frequency of oscillation we use this:
[tex]f=\frac{1}{2(pi)} \sqrt{\frac{k}{m} }[/tex]
Replacing m and k:
[tex]f=\frac{1}{2(pi)} \sqrt{\frac{22.75878093}{0.31} } =1.363684118Hz[/tex]
For two different inertial frames of reference would acceleration and change in velocity be different? How would energy be affected?
Answer and Explanation:
The inertial reference frame is one with constant velocity or non-accelerated frame of reference.
The value of acceleration and velocity change will vary in the two frames and will not be same.
As in case, we observe the acceleration and velocity of a moving train from the platform and the one observed in the train itself will be different.
In case of energy, it is dependent on the frame of reference but the energy change is independent of the frame of reference.
Aaron is flying a helicopter with an air-relative velocity of 12.5 m/s to the West. A constant wind is also blowing the helicopter towards the South at 4.55 m/s relative to the ground. What is the velocity vector of the helicopter relative to the ground? What is the speed and direction relative to the East? How long does it take to travel 1 km?
Answer:
Explanation:
Given
Velocity of helicopter relative to air 12.5 m/s to west
In vector form
[tex]V_{ha}=-12.5\hat{i}[/tex]
where [tex]V_h[/tex]= velocity of helicopter relative to ground
Also the velocity of air
[tex]V_a=4.55\hat{j}[/tex]
[tex]V_{ha}=V_h-V_a[/tex]
[tex]V_h=V_{ha}+V_a[/tex]
[tex]V_h=-12.5\hat{i}+4.55\hat{j}[/tex]
Speed relative to east[tex]=\sqrt{12.5^2+4.55^2}[/tex]
=13.302
For Direction
[tex]tan\theta =\frac{4.55}{-12.5}[/tex]
[tex]180-\theta =20[/tex]
[tex]\theta =160^{\circ}[/tex] relative to east
For 1 km travel it takes
[tex]t=\frac{1000}{13.302}=75.17 s[/tex]
Consider a river flowing toward a lake at an average velocity of 3 m/s at a rate of 500m3/sat a location 90 m above the lake surface. Determine the total mechanical energy of the river water per unit mass and the power generation potential of the entire river at that location.
Answer:
mechanical energy per unit mass is 887.4 J/kg
power generated is 443.7 MW
Explanation:
given data
average velocity = 3 m/s
rate = 500 m³/s
height h = 90 m
to find out
total mechanical energy and power generation potential
solution
we know that mechanical energy is sum of potential energy and kinetic energy
so
E = [tex]\frac{1}{2}[/tex]×m×v² + m×g×h .............1
and energy per mass unit is
E/m = [tex]\frac{1}{2}[/tex]×v² + g×h
put here value
E/m = [tex]\frac{1}{2}[/tex]×3² + 9.81×90
E/m = 887.4 J/kg
so mechanical energy per unit mass is 887.4 J/kg
and
power generated is express as
power generated = energy per unit mass ×rate×density
power generated = 887.4× 500× 1000
power generated = 443700000
so power generated is 443.7 MW
The total mechanical energy per unit mass of the river water is the sum of the potential and kinetic energy. The potential energy is calculated using the height and the gravitational constant, while kinetic energy is calculated using velocity. The power generation potential of the river is the total mechanical energy multiplied by the volume flow rate.
Explanation:The total mechanical energy per unit mass of the river water at this location can be determined using the principle of mechanical energy conservation and the knowledge of potential and kinetic energy. The total mechanical energy (E) equals the potential energy (PE) plus the kinetic energy (KE) per unit mass.
The potential energy is given by: PE = mgh, where m is the mass, g is the acceleration due to gravity (~9.81 m/s^2), and h is the height (90 m above lake surface). But since we are looking for energy per unit mass, m cancels out and we get PE = gh.
The kinetic energy is given by: KE = 0.5mv^2, where v is the velocity (3 m/s). In per unit mass terms, this simplifies to KE = 0.5v^2.
Therefore, E = PE + KE = gh + 0.5v^2.
The power generation potential can be calculated using the equation: P = E * volume flow rate (in our case, 500 m^3/s).
So, P = (gh + 0.5v^2) * 500.
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A janitor standing on the top floor of a building wishes to determine the depth of the elevator shaft. They drop a rock from rest and hear it hit bottom after 2.42 s. (a) How far (in m) is it from where they drop the rock to the bottom of the shaft? The speed of sound at the temperature of air in the shaft is 336 m/s. (Round your answer to at least three significant figures. Use g = 9.80 m/s2 as needed.) m (b) If the travel time for the sound is ignored, what percent error is introduced in the determination of depth of the shaft? %
Answer:
Part a)
H = 26.8 m
Part b)
error = 7.18 %
Explanation:
Part a)
As the stone is dropped from height H then time taken by it to hit the floor is given as
[tex]t_1 = \sqrt{\frac{2H}{g}}[/tex]
now the sound will come back to the observer in the time
[tex]t_2 = \frac{H}{v}[/tex]
so we will have
[tex]t_1 + t_2 = 2.42[/tex]
[tex]\sqrt{\frac{2H}{g}} + \frac{H}{v} = 2.42[/tex]
so we have
[tex]\sqrt{\frac{2H}{9.81}} + \frac{H}{336} = 2.42[/tex]
solve above equation for H
[tex]H = 26.8 m[/tex]
Part b)
If sound reflection part is ignored then in that case
[tex]H = \frac{1}{2}gt^2[/tex]
[tex]H = \frac{1}{2}(9.81)(2.42^2)[/tex]
[tex]H = 28.7 m[/tex]
so here percentage error in height calculation is given as
[tex]percentage = \frac{28.7 - 26.8}{26.8} \times 100[/tex]
[tex]percentage = 7.18 [/tex]
The janitor can determine the depth of the elevator shaft by timing the drop of the rock and using the displacement for free-falling objects formula, considering the time for the sound to travel up the shaft. The percent error if the sound travel time is ignored can be calculated by comparing the depth calculated with and without the sound travel time.
Explanation:The janitor wants to determine how far it is from the point they dropped the rock to the bottom of the elevator shaft by measuring the time it takes for the rock to drop and hit the bottom. Here we can use the formula for the displacement of items in free-fall: d = 0.5gt², where d is the displacement (depth of the shaft), g is the acceleration due to gravity (9.8 m/s²), and t is the time it takes for the rock to fall (2.42s).
So, the depth of the shaft is d = 0.5 x 9.80 m/s² x (2.42 s)² = 28.63 m. However, we need to account for the time it takes for the sound to travel back up the shaft. The speed of sound is given as 336 m/s, and the time it takes to travel a certain distance is the distance divided by the speed, so in this case, it’s 28.63 m / 336 m/s = 0.085 s. Subtracting this from the total time gives us the true fall time of the rock, which we can plug back into the displacement formula to get the corrected depth of the shaft.
The percent error introduced if we ignore the travel time for the sound is then the depth obtained with sound travel time subtracted from the depth obtained without it, divided by the depth obtained with sound travel time, times 100.
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What is the net charge of a metal ball if there are 21,749 extra electrons in it?
Answer:
[tex]Q=3.47\times 10^{-15}\ C[/tex]
Explanation:
Given that,
Number of extra electrons, n = 21749
We need to find the net charge on the metal ball. Let Q is the net charge.
We know that the charge on an electron is [tex]q=1.6\times 10^{-19}\ C[/tex]
To find the net charge if there are n number of extra electrons is :
Q = n × q
[tex]Q=21749\times 1.6\times 10^{-19}\ C[/tex]
[tex]Q=3.47\times 10^{-15}\ C[/tex]
So, the net charge on the metal ball is [tex]3.47\times 10^{-15}\ C[/tex]. Hence, this is the required solution.
At what distance from a long straight wire carrying a
currentof 5.0A is the magnitude of the magnetic field due to the
wireequal to the strength of the Earth's magnetic field of about
5.0 x10^-5 T?
Answer:
The distance is 2 cm
Solution:
According to the question:
Magnetic field of Earth, B_{E} = [tex]5.0\times 10^{- 5} T[/tex]
Current, I = 5.0 A
We know that the formula of magnetic field is given by:
[tex]B = \farc{\mu_{o}I}{2\pi d}[/tex]
where
d = distance from current carrying wire
Now,
[tex]d = \frac{\mu_{o}I}{2\pi B}[/tex]
[tex]d = \frac{4\pi\times 10^{- 7}\times 5.0}{2\pi\times 5.0\times 10^{- 5}}[/tex]
d = 0.02 m 2 cm
The distance from the wire where the magnitude of the magnetic field equals the Earth's magnetic field is approximately 0.4 meters.
Explanation:To find the distance from the wire where the magnitude of the magnetic field is equal to the strength of the Earth's magnetic field, we can use the equation:
B = μ0 * I / (2π * r)
Where B is the magnetic field, μ0 is the permeability of free space (4π x [tex]10^-7[/tex]A), I is the current, and r is the distance from the wire.
Plugging in the given values, we have:
B_wire = μ0 * 5.0A / (2π * r) and B_earth = 5.0 x [tex]10^-5 T[/tex]
Setting B_wire equal to B_earth and solving for r:
5.0 x[tex]10^-5[/tex]* 5.0A / (2π * r)
Solving for r, we find that the distance from the wire where the magnitude of the magnetic field is equal to the Earth's is approximately 0.4 meters.
What is the amount of heat, in Sl units, necessary to melt 1 lb of ice?
Answer:
Heat required to melt 1 lb of ice is 151.469 KJ
Explanation:
We have given mass of ice = 1 lb
We know that 1 lb = 0.4535 kg
Latent heat of fusion for ice =334 KJ/kg
Amount if heat for fusion of ice is given by
[tex]Q=mL[/tex], here m is mass of ice and L is latent heat of fusion
So heat [tex]Q=mL=0.4535\times 334=151.469kj[/tex]
So heat required to melt 1 lb of ice is equal to 151.469 KJ
An engine extracts 441.3kJ of heat from the burning of fuel each cycle, but rejects 259.8 kJ of heat (exhaust, friction,etc) during each cycle. What is the thermal efficiency of the engine?
Answer:
[tex]\eta = 0.411[/tex]
Explanation:
As we know that efficiency is defined as the ratio of output useful work and the input energy to the engine
So here we know that the
input energy = 441.3 kJ
energy rejected = 259.8 kJ
so we have
[tex]Q_1 - Q_2 = W[/tex]
[tex]W = 441.3 kJ - 259.8 kJ = 181.5 kJ[/tex]
now efficiency is defined as
[tex]\eta = \frac{W}{Q_1}[/tex]
[tex]\eta = \frac{181.5}{441.3}[/tex]
[tex]\eta = 0.411[/tex]
An object, initially at rest, moves with a constant acceleration of 10 m/s2. How far will it travel in (a) 2.0 s and (b) 4.0 s? If this object had an initial velocity of 4 m/s, how far will it travel in (C) 2.0 s and (d) 4.0 s?
Answer:
(a) d = 20 m
(b) d' = 80 m
(c) x = 28 m
(d) x' = 96 m
Solution:
As per the question:
Initial velocity of the object, v = 0
Constant acceleration of the object, [tex]a_{c} = 10 m/s^{2}[/tex]
(a) Distance traveled, d in t = 2.0 s is given by the second eqn of motion:
[tex]d = vt + \frac{1}{2}a_{c}t^{2}[/tex]
[tex]d = 0.t + \frac{1}{2}\times 10\times 2^{2} = 20 m[/tex]
(b) Distance traveled, d' in t = 4.0 s is given by the second eqn of motion:
[tex]d' = vt + \frac{1}{2}a_{c}t^{2}[/tex]
[tex]d' = 0.t + \frac{1}{2}\times 10\times 4^{2} = 80 m[/tex]
Now, when initial velocity, v = 4 m/s, then
(c) Distance traveled, x in t = 2.0 s is given by the second eqn of motion:
[tex]x = vt + \frac{1}{2}a_{c}t^{2}[/tex]
[tex]x = 4\times 2.0 + \frac{1}{2}\times 10\times 2^{2} = 28 m[/tex]
(d) Distance traveled, x' in t = 4.0 s is given by the second eqn of motion:
[tex]x = vt + \frac{1}{2}a_{c}t^{2}[/tex]
[tex]x = 4\times 4.0 + \frac{1}{2}\times 10\times 4^{2} = 96 m[/tex]
A soft tennis ball is dropped onto a hard floor from a height of 1.95 m and rebounds to a height of 1.55 m. (Assume that the positive direction is upward.) (a) Calculate its velocity just before it strikes the floor. (b) Calculate its velocity just after it leaves the floor on its way back up. (c) Calculate its acceleration during contact with the floor if that contact lasts 3.50 ms. (d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?
Answer:
a)[tex]v=6.19m/s[/tex]
b)[tex]v=5.51m/s[/tex]
c)[tex]a=3.3*10^{3}m/s^{2}[/tex]
d)[tex]x=5.78*10^{-3}m[/tex]
Explanation:
h1=195m
h2=1.55
a) Velocity just before the ball strikes the floor:
Conservation of the energy law
[tex]E_{o}=E_{f}[/tex]
[tex]E_{o}=mgh_{1}[/tex]
[tex]E_{f}=1/2*mv^{2}[/tex]
so:
[tex]v=\sqrt{2gh_{1}}=\sqrt{2*9.81*1.95}=6.19m/s[/tex]
b) Velocity just after the ball leaves the floor:
[tex]E_{o}=E_{f}[/tex]
[tex]E_{o}=1/2*mv^{2}[/tex]
[tex]E_{f}=mgh_{2}[/tex]
so:
[tex]v=\sqrt{2gh_{2}}=\sqrt{2*9.81*1.55}=5.51m/s[/tex]
c) Relation between Impulse, I, and momentum, p:
[tex]I=\Delta p\\ F*t=m(v_{f}-v{o})\\ (ma)*t=m(v_{f}-v{o})\\\\ a=\frac{ v_{f}-v{o}}{t}=\frac{ 5.51- (-6.19)}{3.5*10^{-3}}=3.3*10^{3}m/s^{2}[/tex]
d) The compression of the ball:
The time elapsed between the ball touching the ground and it is fully compressed, is half the time the ball is in contact with the ground.
[tex]t_{2}=t/2=3.5/2=1.75ms[/tex]
Kinematics equation:
[tex]x(t)=v_{o}t+1/2*a*t_{2}^{2}[/tex]
Vo is the velocity when the ball strike the floor, we found it at a) 6.19m/s.
a, is the acceleration found at c) but we should to use it with a negative sense, because its direction is negative a Vo, a=-3.3*10^3
So:
[tex]x=6.19*1.75*10^{-3}-1/2*3.3*10^{3}*(1.75*10^{-3})^2=5.78*10^{-3}m[/tex]
If the atmospheric pressure in a tank is 23 atmospheres at an altitude of 1,000 feet, the air temperature in the tank is 700F, and the volume of the tank is 800 f3, determine the weight of the air in the tank.
Answer:
W = 289.70 kg
Explanation:
Given data:
Pressure in tank = 23 atm
Altitude 1000 ft
Air temperature in tank T = 700 F
Volume of tank = 800 ft^3 = 22.654 m^3
from ideal gas equation we have
PV =n RT
Therefore number of mole inside the tank is
[tex]\frac{1}{n} = \frac{RT}{PV}[/tex]
[tex] = \frac{8.206\times 10^{-5}} 644.261}{23\times 22.654}[/tex]
[tex]= 1.02\times 10^{-4}[/tex]
[tex]n = 10^4 mole[/tex]
we know that 1 mole of air weight is 28.97 g
therefore, tank air weight is [tex]W = 10^4\times 28.91 g = 289700 g[/tex]
W = 289.70 kg
A charge of 5 µC is on the y axis at y = 7 cm, and a second charge of -5 µC is on the y axis at y = -7 cm. Find the force on a charge of 2 µC on the x axis at x = 3 cm.
Answer:
The force on X Fx=0 N
The force on Y Fy=-2.18 N
Explanation:
We have an array of charges, we will use the coulomb's formula to solve this:
[tex]F=k*\frac{Q*Q'}{r^2}\\where:\\k=coulomb constant\\r=distance\\Q=charge[/tex]
but we first have to find the distance and the angule of the charge respect the charges on the Y axis:
[tex]r=\sqrt{(7*10^{-2}m)^2+ (3*10^{-2}m)^2} \\r=7.62cm=0.0762m[/tex]
we can notice that it is the same distance from both charges on Y axis.
we can find the angle with:
[tex]\alpha = arctg(\frac{7cm}{3cm})=66.80^o[/tex]
for the charge of 5µC [tex]\alpha =-66.80^o[/tex]
for the charge of -5µC [tex]\alpha =66.80^o[/tex]
the net force on the X axis will be:
[tex]F_{x5u}=9*10^9*\frac{5*10^{-6}*2*10^{-6}}{0.0762^2}*cos(-66.80)\\F_{x5u}=0.465N[/tex]
and
[tex]F_{x(-5u)}=9*10^9*\frac{-5*10^{-6}*2*10^{-6}}{0.0762^2}*cos(66.80)\\F_{x(-5u)}=-0.465N[/tex]
So the net force on X will be Zero.
for the force on Y we have:
[tex]F_{y5u}=9*10^9*\frac{5*10^{-6}*2*10^{-6}}{0.0762^2}*sin(-66.80)\\F_{y5u}=-1.09N[/tex]
and
[tex]F_{y(-5u)}=9*10^9*\frac{-5*10^{-6}*2*10^{-6}}{0.0762^2}*sin(66.80)\\F_{y(-5u)}=-1.09N[/tex]
Fy=[tex]F_{y5u}+F_{y(-5u)}[/tex]
So the net force on Y is Fy=-2.18N
To find the force on a charge of 2 µC on the x axis at x = 3 cm, we can use Coulomb's Law. The force is calculated to be 5.12 N.
Explanation:To find the force on a charge of 2 µC on the x axis at x = 3 cm, we can use Coulomb's Law. Coulomb's Law states that the force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them.
The formula for Coulomb's Law is:
F = k * |q1 * q2| / r^2
Where F is the force, k is the electrostatic constant (k ≈ 8.99 * 10^9 N * m² / C²), q1 and q2 are the magnitudes of the charges, and r is the distance between them.
In this case, one charge is 5 µC and the other charge is -5 µC, with a distance of 14 cm between them (7 cm on the positive y-axis and 7 cm on the negative y-axis). The charge we're interested in is 2 µC at x = 3 cm. Plugging these values into the formula:
F = (8.99 * 10^9 N * m² / C²) * |(5 µC) * (2 µC)| / (0.14 m)²
F = 5.12 N
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