We have that for the Question, it can be said that the amount by which the length of the stack decreases is
[tex]dl'=3.621*10^{-4}m[/tex]
From the question we are told
A copper (Young's modulus 1.1 x 1011 N/m2) cylinder and a brass (Young's modulus 9.0 x 1010 N/m2) cylinder are stacked end to end, as in the drawing. Each cylinder has a radius of 0.24 cm.
A compressive force of F = 7900 N is applied to the right end of the brass cylinder. Find the amount by which the length of the stack decreases.
Generally the equation for copper cylinder is mathematically given as
[tex]dl=\frac{Flo}{yA}[/tex]
[tex]dl=\frac{7900*3*10^-^2}{1.1*10^{11}*\pi(0.24*10^{-2})^2}[/tex]
[tex]dl=1.19064778*10^-^4[/tex]
Generally the equation for brass cylinder is mathematically given as
[tex]dl=\frac{7900*5*10^-^2}{9*10^{10}*\pi(0.24*10^{-2})^2}[/tex]
[tex]dl=2.43*10^{-4}[/tex]
Therefore Total change in length
[tex]dl'=1.191*10^-^4+(2.43*10^{-4})[/tex]
[tex]dl'=3.621*10^{-4}m[/tex]
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Which wave has the highest frequency?
Answer:its E
Explanation:
1. A race car accelerated from rest to 34.7 m/s by the time it crossed the finish line. The race car moved in a straight line and traveled from the starting line to the finish line in 7.81 seconds. What was the acceleration of the race car?
Answer: 4.44302176697 m/s^2
Explanation:
formula for acceleration is (final velocity - initial velocity)/time. So in your case:
(34.7m/s - 0m/s)/ 7.81s = 4.44302176697 m/s^2
Which of the following correctly describes the law of conservation of matter?
Group of answer choices
A.) Matter can only be created, but not destroyed in chemical equations
B.) Matter can neither be created nor destroyed during a chemical reaction
C.) Matter can be destroyed during a chemical reaction as long as it is created again in a different reaction
D.) Matter makes up 98% of stuff in the universe, so it has to be created and destroyed sometimes
Answer:
C
Explanation:
Tama Po yan don't worry
Inertia causes bodies to slow in their motion, unless they’re pushed by a force.
Inertia resists changes to the state of motion of a body.
Inertia decelerates a body.
Inertia is due to friction.
Answer:
ill go with B
Explanation:
inertia is the ability of a body to remain at rest or in a constant speed in a constant direction when no force is acting on it
why would a balloon attract your hair without touching it?
Answer:
Hold the balloon (negatively charged) just above your head so your hair (positively charged) will be attracted to it and stand up on end. ... The can will start to roll towards the balloon without touching it. The negatively charged balloon repels the electrons of the can so that a positive charge is near the balloon.
Explanation:
I gave my explanation along with the answer.
Alejandro Kirk is the catcher for the Blue Jays’ baseball team. He exerts a forward force on the 0.145-kg baseball to bring it to rest from a speed of 38.2 m/s. During the process, his hand recoils a distance of 0.135 m.
Determine is the acceleration of the ball.
Determine the force applied by Alejandro.
The kinematics and Newton's second law allow to find the results for the questions about the motion of the ball are:
The acceleration is: a = 5.4 10³ m / s² The force is: F = 7.84 10² N
Given parameters
Mass of the ball m = 0.145 kg Ball starts speed vo = 38.2 m / s Setback distance x = 0.135 mTo find
Acceleration The force
Kinematics studies the movement of bodies, looking for relationships between the position, speed and acceleration of bodies.
v² = v₀² - 2 ax
Wher vo es la initial veloicity, a the acceleration y x is the distance,
When the ball stops the velocity is zero.
0 = v₀² - 2ax
a = [tex]\frac{v_o^2}{2x}[/tex]
Let's calculate
a = [tex]\frac{38.2^2}{2 \ 0.135}[/tex]
a = 5.4 10³ m / s²
Newton's second law establishes a relationship between force, mass, and acceleration of the body.
F = ma
F = 0.145 5.4 10³
F = 7.84 10² N
In conclusion using kinematics and Newton's second law we can find the results for the questions about the motion of the ball are:
The acceleration is: a = 5.4 10³ m / s² The force is: F = 7.84 10² N
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if the magntidude of the sum and difference of two non zero vectors A and B is equal. what can we say about the two vectors
Answer:
A + B = B - A
The above equation must hold for the sum and difference to be zero.
How can one of the vectors not be zero?????
how much force is required to accelerate a 12 kg mass at 5 m/s 2
Answer:
60 N
Explanation:
This is just Newton's Second Law
F = m*a
F = ?
m = 12 kg
a = 5 m/^2
F = 5*12 = 60 Newtons
Taking Earth to be a perfect sphere, find the linear speed of a point located on the 32nd parallel, as a result of Earth’s rotation. The 32nd parallel north is a circle of latitude that is 32 degrees north of the Earth's equatorial plane. Take the Earth's radius to be 6,371,000 m
The linear speed of a point located on the 32nd parallel, as a result of Earth’s rotation is 392.91 m/s
We find the linear speed of a point on the 32nd parallel from v = rω where r = radius of 32nd parallel north = Rcos32° (since it is the radius of the small circle at the 32nd parallel) where R = radius of earth = 6,371,000 m and ω = angular speed of the earth = 2π/T where T = period of earth = 24h = 24 × 60 × 60 s = 86400 s.
So, v = rω
v = Rcos32° × 2π/T
v = 2πRcos32°/T
substituting the values of the variables into the equation, we have
v = 2πRcos32°/T
v = 2π × 6,371,000 m × cos32°/86400 s.
v = 2π × 6,371,000 m × 0.8480/86400 s.
v = 33947512.5035 m/86400 s
v = 392.91 m/s
So, the linear speed of a point located on the 32nd parallel, as a result of Earth’s rotation is 392.91 m/s
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Help help fast please first answer gets brainiest also needs to be clearly stated
The launch speed of a projectile is three times the speed it has at its maximum height.what is the elevation angle at launch?
Answer:
Given: a projectile of initial launch velocity(V) and launch angle ∅ and no air resistance. At the maximum height, the projectile would have a zero contribution of speed from the vertical component(Vy) Therefore, if we say Vx=Vcos∅ is the only speed the projectile has at the instant of maximum height then we can replace Vx with 1/5V and write 1/5V=Vcos∅. Solving for the the launch angle ∅, gives Inverse Cos(1/5)=78.5 degrees.
Does the dependenceof the orbital energy on nmake sense in terms of the total number of nodes in each type of orbital?.
Answer:The 2s orbital different than the 1s orbital because the 2s orbital extends farther from the nucleus than the 1s.
As we move away from the nucleus, the values of the principal quantum number (n) continues to increase.
As the principal quantum number (n) increases, the orbital becomes farther away from the nucleus.
The higher energy orbital are always larger than the orbitals closer to the nucleus.
Hence, the 2s orbital different than the 1s orbital because the 2s orbital extends farther from the nucleus than the 1s.
Explanation:
How can magnets attract and repeal without touching each other?
Answer:
The hyperactivity of these electrons give magnets the ability to attract and repel
What are the advantages and/or disadvantages of the radiation that technology has had for the human economy?
why do we believe 90 percent of the mass of the milky way is in the form of dark matter?
Answer:
The orbital speeds of stars far from the galactic center are surprisingly high, suggesting that these stars are feeling gravitational effects from unseen matter in the halo.
Column A
Column B
1.
Frequency
a. Ohm
2.
Amplitude for sound
b. Decibel
3
Force
C. Joule
4
Distance
d. Newton
5.
Current
e. Amperdiet
6.
Resistance
f. Meter
7.
Voltage
g. Hertz
8.
Energy
h. Volt
Answer:its 3 and 5
Explanation:
j
A two-dimensional crystal has a rectangular unit cell, with spacings between the centers of the atoms of 0.5 nm in one direction (e.g., the x direction), and 0.4 nm in the other direction (e.g., the y direction). Presuming that the crystal has 1000 unit cells in each direction, sketch a representative portion of the reciprocal lattice on a scale drawing, showing the dimensions, units, and directions (i.e., kx and ky)
So directly the cell density value in a rectangle is:
[tex]5*10^{17} cells/cm^{2}[/tex]
In this exercise, we must first calculate the area of the rectangle, remembering that you need to convert from nm to cm is:
[tex]A=B*H\\A=(0.4*10^{-7})*(0.5*10^{-7})= 2*10^{-15}[/tex]
Next, let's calculate the density from the calculated area:
[tex]\frac{N}{A} = \frac{1000}{2*10^{-15} } = 5*10^{17}[/tex]
In this way, the calculation represents that there is:
[tex]5*10^{17} cells/cm^{2}[/tex]
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Which type of solid is dry ice (solid carbon dioxide)? ionic atomic molecular none of the above.
Answer: molecular solid
Explanation:
The Earth system is an example of a(n)
How to find direction of magnetic field.
unlock your third eye and youll be able to see it
A man is sitting on a chair with wheels. He grabs a 2.1 kg book from the desk and throws
the book at a speed of 7.2 m/s. His chair moves backward. The man has a mass of 70 kg and
the chair has a mass of 9.2 kg. What is the speed of the man and the chair after the book is
thrown?
The speed of the man and the chair after the book is thrown is 0.2 m/s.
The given parameters:
mass of the book, m₁ = 2.1 kgspeed of the book, u₁ = 7.2 m/smass of the man, M = 70 kgmass of the book, m = 9.2 kgThe total mass of the man and the book is calculated as follows;
m₂ = 70 kg + 9.2 kg
m₂ = 79.2 kg
The speed of the man and the chair after the book is thrown is determined by applying the principle of conservation of linear momentum;
[tex]m_1 u_1 = m_2u_2\\\\u_2 = \frac{m_1u_1}{m_2} \\\\u_2 = \frac{2.1 \times 7.2}{79.2} \\\\u_2 = 0.2 \ m/s[/tex]
Thus, the speed of the man and the chair after the book is thrown is 0.2 m/s.
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1. When a 50,000 kg rocket blasts off there is a large force that acts on the rocket at the moment
of take off. In 0.137 seconds, the rocket suddenly has traveled 4.2 meters. What force from
the rocket engines caused the acceleration? (Hint: first find acceleration.)
Hi there!
We can begin by calculating the acceleration using the kinematic equation:
d = vit + 1/2at²
Since the rocket starts from rest, we can rewrite this as:
d = 1/2at²
Plug in given values:
4.2 = 1/2a(0.137²) ⇒ a = 447.55 m/s²
Now, we can use Newton's Second Law to find the appropriate force:
∑F = m · a
∑F = 50,000 · 447.55 = 22,377,500 N
What type of energy transfer causes a sunburn?
Radiation, because particles of hot air move in space
Radiation, because electromagnetic waves transfer heat
Conduction, because hot air comes in contact with the skin
Conduction, because invisible light carries heat from the sun
Pacificaly what is the right answer A? B? C? or D? which one
Answer:
radiation
Explanation:
Answer:
B is most likely forgive me if im wrong
Explanation:
What is the difference between a low tide and a high tide
Answer:
High water level during a tide is called High tide.
Low water level during a tide is called Low tide.
A 800N mountain climber scales a 140m cliff. How much work is done by the mountain climber?
Answer:
112 [kJ]
Explanation:
1) related formula is: A=F*S, where A - required work [J], F - force [N], S - distance [m].
2) according to the formula above: A=140*800=112000[J]=112 [kJ]
Answer:
112000 J
Explanation:
800N x 140m
A transformer is being built to triple the voltage produced. Which data are collected to investigate whether the transformer can achieve the goal? (1 point)
A) the ratio of the number of turns in the primary and secondary coils
B) only the number of turns in the secondary coil and the magnetic field of the core
C) only the number of turns in the primary coil and the magnetic field of the core
D) the ratio of the number of turns in the primary and secondary coils and the magnetic field of the core
Answer:
the ratio of the number of turns in the primary and secondary coils.
Explanation:
hope i'm not too late!!
How much work is needed to stop a 1,110 kg car that is moving straight down the
highway at 59.0 km/h?
Answer:
382.74 kJ.
Explanation:
The work that must be done to stop an 1100 kg car travelling at 59 km/h is - 382.74 kJ.
The work needed to stop an 1100 kilograms car moving with a velocity of 59.0 km/hr would be 147.72 kilojoules.
What is work done?
The total amount of energy transferred when a force is applied to move an object through some distance
The work done is the multiplication of applied force with displacement.
Work Done = Force * Displacement
As given in the problem we have to find out how much work is needed to stop a 1,110-kilogram car that is moving straight down the highway at 59.0 kilometers/hour
the mass of the car = 1100 kilograms
the velocity of the car = 59 kilometers/hour
KE = 0.5mv²
=0.5×1100×(59×1000/3600)²
= 147.72 kilojoules
Thus, the work needed to stop an 1100 kilograms car would be 147.72 kilojoules
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A particle is constrained to move round a circle radius 382400km and makes a single revolution in 27.3 days. (i). Find the velocity in ms-1. (ii). Find its acceleration
The velocity and acceleration of the particle moving round the circle is;
Velocity = 162.12 m/s
Velocity = 162.12 m/sAcceleration = 6.873 × 10^(-5) m/s²
We are given;
Radius of circle; 382400 km = 382400000 m
Time; t = 27.3 days = 27.3 × 86400 s = 2358720 s
Now, formula for velocity is;
Velocity = distance/time
Thus;
I) velocity = 382400000/2358720
Velocity = 162.12 m/s
II) Acceleration is centripetal acceleration and is given by the formula;
a = v²/r
a = 162.12²/382400000
a = 6.873 × 10^(-5) m/s²
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how many seconds does it take an object to fall 98m if it start from the rest .(using gravity 10m/s²?
Answer:
Neither.
Explanation:
After 10 seconds the object will have travelled 490.5m, not 98.1. The vertical distance covered in freefall by an object in time t is given by Distance= 1/2 x 9.81 x t^2. Simply multiplying 9.81 by the time will NOT give the distance covered.
The acceleration of the object will not change, it will always be 9.81 m/s^2. Its speed will keep increasing by 9.81 m/s every second, thats what the ‘per second squared*’ means. The speed after 10 seconds will be 98.1 m/s
To summarise, after 10 seconds, the object will have covered 490.5 metres, and it will have a final speed of 98.1 m/s. It’s acceleration will remain 9.81 m/s^2 at all times
A 0.14-MIN baseball is dropped from rest. It has a momentum of 0.90 kg⋅m/skg⋅m/s just before it lands on the ground.
For what amount of time was the ball in the air?
The time spent in the air by the ball at the given momentum is 6.43 s.
The given parameters;
momentum of the ball, P = 0.9 kgm/sweight of the ball, W = 0.14 NThe impulse experienced by the ball is calculated as follows;
[tex]Ft = \Delta P[/tex]
where;
[tex]Ft[/tex] is impulse
[tex]\Delta P[/tex] is change in momentum
The time of motion of the ball is calculated as follows;
[tex]t = \frac{\Delta P}{F} \\\\t = \frac{0.9 - 0}{0.14} \\\\t = 6.43 \ s[/tex]
Thus, the time spent in the air by the ball at the given momentum is 6.43 s.
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