During heavy lifting, a disk between spinal vertebrae is subjected to a 5000-N compressional force. (a) What pressure is created, assuming that the disk has a uniform circular cross section 2.00 cm in radius? (b) What deformation is produced if the disk is 0.800 cm thick and has a Young's modulus of 1.5×109 N/m2?

Answers

Answer 1

Answer:

[tex]3978873.58\ \text{Pa}[/tex]

[tex]0.00002122\ \text{m}[/tex]

Explanation:

F = Force = 5000 N

r = Radius of circular cross section = 2 cm

l = Length of disk = 0.8 cm

A = Area = [tex]\pi r^2[/tex]

Y = Young's modulus = [tex]1.5\times 10^9\ \text{N/m}^2[/tex]

Pressure is given by

[tex]P=\dfrac{F}{A}\\\Rightarrow P=\dfrac{5000}{\pi (2\times 10^{-2})^2}\\\Rightarrow P=3978873.58\ \text{Pa}[/tex]

The pressure on the cross section is [tex]3978873.58\ \text{Pa}[/tex]

The change in length of the cross section is given by

[tex]\Delta L=\dfrac{PL}{Y}\\\Rightarrow \Delta L=\dfrac{3978873.58\times 0.8\times 10^{-2}}{1.5\times 10^9}\\\Rightarrow \Delta L=0.00002122\ \text{m}[/tex]

The deformation produced is [tex]0.00002122\ \text{m}[/tex]


Related Questions

An electromagnet needs a magnetic metal core. To produce a magnetic field,
what else is required?

A second metal core?
A solenoid with current or no current running through it? Or a permanent magnet?

Answers

A second metal core i think

Answer:solenoid with current running through it

Explanation:just took the test

A motorcycle moving at a constant velcoity suddenly accelerates at a rate of 4.0 m/s/s to a speed of 35 m/s in 5.0 s. What was the initial speed of the motorcycle?

Answers

Answer:

15 m/s

Explanation:

v = u+ at

35 = u + 20

35-20 = u

u= 15 m/s

A 7.50 kg bowling ball has 70.4
kg•m/s of momentum. What is its
velocity?

Answers

Answer:

9.39 m/s

Explanation:

The velocity of the bowling ball can be found by using the formula

[tex]v = \frac{p}{m} \\ [/tex]

p is the momentum

m is the mass

From the question we have

[tex]v = \frac{70.4}{7.5} \\ = 9.38666666..[/tex]

We have the final answer as

9.39 m/s

Hope this helps you

A spaceship of mass mm circles a planet of mass M in an orbit of radius R. How much energy is required to transfer the spaceship to a circular orbit of radius 3R?

Answers

Answer:

ΔE = GmM/3R

Explanation:

The absolute potential energy of an object in a planet's field is given as:

E = -GmM/2r

where,

E = Potential Energy

G = Universal Gravitational Constant

m = mass of spaceship

M = Mass of Planet

r = distance from surface of planet

Therefore, for initial state:

E = E₁ and r = R

E₁ = - GmM/2R

and for final state:

E = E₂ and r = 3R

E₂ = - GmM/6R

So, the required energy will be:

ΔE = E₂ - E₁ = - GmM/6R + GmM/2R

ΔE = GmM(- 1/6R + 1/2R)

ΔE = GmM/3R

1000 kg car takes travels on a circular track having radius 100 m with speed 10 m/s. What is the direction of the acceleration of the car? *
A- outside track, and normal to track
B- towards the center and normal to the track
C- up
D- down

Answers

Answer:

B.

Explanation:

find the fundamental units involved in derived units
newton
watt
joule
pascal
cubic meter ​

Answers

9ycy8c8t 7f fixfuozofuxt8lsrupsurpaurae6pUeoUe6eoUeFipzuroz6d0, 7d0z6e0z7e0zurpz6e0z

Explanation:

force newton N - m·kg·s-2

pressure, stress pascal Pa N/m2 m-1·kg·s-2

energy, work, quantity of heat joule J N·m m2·kg·s-2

power, radiant flux watt W J/s m2·kg·s-3

volume cubic meter m3

A 1150 kg car is on a 8.70° hill.
Using X-Y axes tilted down the
plane, what is the y-component
of the weight?

Answers

Answer:

Explanation:

y-component = - mgcos(8.7)

= - (1150)(9.81)cos(8.7)

= - 11151.69378

= - 11151.69 N

The weight of the y-component is 11140.33N.

How to find the weight of the y-component?

To find the weight of the y-component:

Given,

Car weight = 1150 kg

Anfle = 8.70 degree

weight = mg = 1150 * 9.8

= 11270 N

Y-component =  mg cos∅

= 11270 * cos(8.70)

= 11140.33N

The aspect that pushes proper or left is referred to as the x-factor, and the element that pushes up or down is known as the y-component.

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The time constant for RC circuit with the values of R1 and C1 is 5ms. What will be the time constant for a new RC circuit with the values: R2=10R1 and C2 = 0.5C1.

a. 2.5ms
b. 15.5ms
c. 50ms
d. 25ms
e. 15ms

Answers

Answer:

d. 25 ms

Explanation:

In a RC circuit we call time constant to the product of the resistance times the capacitance, which represents the time when the charge reaches to the 63% of the final value, as follows:

       [tex]\tau_{1} = R_{1} *C_{1} = 5 ms (1)[/tex]

If we have a new circuit with new values for R and C, the time constant will be defined in the same way, as follows:

       [tex]\tau_{2} =10* R_{1} *0.5*C_{1} = 5*(R_{1}* C_{1}) = 5* \tau_{1} = 5* 5 ms = 25 ms (2)[/tex]

Question 1 of 5
In which way are electromagnetic waves different from mechanical waves?

Answers

Answer:

Electromagnetic waves differ from mechanical waves in that they do not require a medium to propagate. This means that electromagnetic waves can travel not only through air and solid materials, but also through the vacuum of space.

Explanation:

Answer:

In which way are electromagnetic waves different from mechanical waves?Electromagnetic waves can travel through empty space

Explanation:

I took A P E X Quiz.

a vertical solid steel post 29cm in diameter and 2.0m long is required to support a load of 8200kg, ignore the weight of the post. determine the stress in the post​

Answers

Answer:

The stress is  [tex]\sigma = 1.218*10^{6} \ N/m^2[/tex]

Explanation:

From the question we are told that

   The diameter of the post is  [tex]d = 29 \ cm = 0.29 \ m[/tex]

   The length is [tex]L = 2.0 \ m[/tex]

    The weight of the loading mass

Generally  the  radius of the post is mathematically represented as

     [tex]r = \frac{0.29}{2}[/tex]

=>   [tex]r = 0.145 \ m[/tex]

Generally the area of the post is  

       [tex]A = \pi r^2[/tex]

=>     [tex]A = 3.14 * 0.145 ^2[/tex]

=>     [tex]A = 0.066 \ m^2[/tex]

Generally the weight exerted by the load is mathematically represented as

        [tex]F = m * g[/tex]

=>      [tex]F = 8200 * 9.8[/tex]

=>      [tex]F = 80360 \ N[/tex]

Generally the stress is mathematically represented as

         [tex]\sigma = \frac{F}{A}[/tex]

=>      [tex]\sigma = \frac{80360 }{0.066}[/tex]

=>      [tex]\sigma = 1.218*10^{6} \ N/m^2[/tex]

You are lifting a 10 kg block straight up at a constant speed of 10 m/s. How much force are you exerting on the block?

Answers

Answer:

The force exerted is [tex]F = 100 \ N[/tex]

Explanation:

From the question we are told that

   The mass of the block is [tex]m_b = 10 \ kg[/tex]

    The speed is  [tex]v = 10 \ m/s[/tex]

Generally the force exerted to lift the object at constant speed is equivalent to the wight of the ball, this is mathematically represented as

       [tex]F = m * g[/tex]      Here  [tex]g = 10 \ m/s^2[/tex]

=>    [tex]F = 10 * 10[/tex]

=>    [tex]F = 100 \ N[/tex]

The force are you exerting on the block when the block is lifting straight up with constant speed is 98 N and this can be determined by using the given data.

Given :

You are lifting a 10 kg block straight up at a constant speed of 10 m/s.

The following steps can be used in order to determine the force are you exerting on the block:

Step 1 - According to the given data, the block is lifting straight up at a constant speed. So, the acceleration is zero.

Step 2 - So, the only force exerted on the block is the weight of the block.

Step 3 - So, the force are you exerting on the block is given by:

F = mg

F = 10 [tex]\times[/tex] 9.8

F = 98 N

For more information, refer to the link given below:

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For satellite travelling on circular orbit if radius of the orbit increased 4 times then the period of the satellite increased *


2

4

8

none of the above

Answers

Answer:

8

Explanation:

A kangaroo jumps straight up to a vertical height of 1.45 m. How long was it in the air before returning to Earth?

Answers

Answer:

1.08 s

Explanation:

From the question given above, the following data were obtained:

Height (h) reached = 1.45 m

Time of flight (T) =?

Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:

Height (h) = 1.45 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

1.45 = ½ × 9.8 × t²

1.45 = 4.9 × t²

Divide both side by 4.9

t² = 1.45/4.9

Take the square root of both side

t = √(1.45/4.9)

t = 0.54 s

Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).

Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:

Time (t) taken to reach the height = 0.54 s

Time of flight (T) =?

T = 2t

T = 2 × 0.54

T = 1.08 s

Therefore, it will take the kangaroo 1.08 s to return to the earth.

Which environment is least likely to support protists
A soil
B open ocean
C shallow pod
D organisms blood

Answers

Answer:

A: Soil

Explanation:

Protists need a moist environment to survive, and shallow ponds, oceans, and blood is all moist. So, the answer would be the soil, because that is the least moist environment out of these options.

A locomotive creates a
59,400 N force, which creates
an acceleration of 0.145 m/s.
What is the mass of the
locomotive?
Unit=kg

Answers

Answer:

410,000 kg

Explanation:

Use Newton's second law

F = ma

m = F/a

m = 59,400 N/(.145 m/s) = 410,000 kg

A 6.0-kilogram cart initially traveling at 4.0 meters per second east accelerates uniformly at 0.50 meter per second squared east for 3.0 seconds. What is the speed of the cart at the end of this 3.0 second interval? A. 1.5 m/s B. 5.5 m/s G. 3.0 m/s D. 7.0 m/s​

Answers

Answer: 5.5m/s

Explanation:

vf=vi+at

vf= 4.0m/s + (0.50m/s^2)(3.0s)

The speed of the cart at the end of this 3.0 second interval is 5.5 meter per seconds.

Given the following data:

Initial velocity = 4 m/sMass of cart = 6 KgAcceleration = 0.5 [tex]m/s^2[/tex]Time = 3 seconds

To find the speed of the cart at the end of this 3.0 second interval, we would use the first equation of motion;

[tex]V = U + at[/tex]

Where:

U is the initial velocity.V is the final velocity. a is the acceleration. t is the time measured in seconds.

Substituting the given parameters into the formula, we have;

[tex]V = 4 + 0.5(3)\\\\V = 4 + 1.5[/tex]

Final velocity, V = 5.5 m/s.

Therefore, the speed of the cart at the end of this 3.0 second interval is 5.5 meter per seconds.

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An object that moves in uniform circular motion has a centripetal acceleration of 13 m/s^2 . If the radius of the motion is 0.02m, what is the frequency of motion?

Answers

Answer:

f = 3.97 Hz

Explanation:

Given that,

Centripetal acceleration, [tex]a=13\ m/s^2[/tex]

The radius of motion is 0.02 m

The formula for the centripetal acceleration is given by :

[tex]a=\dfrac{v^2}{r}\\\\v=\sqrt{ar} \\\\v=\sqrt{13\times 0.02} \\\\v=0.5\ m/s[/tex]

The speed of an object in a circular path is given by :

[tex]v=\dfrac{2\pi r}{t}[/tex]

t is time period

Also, f=1/t (f is frequency)

[tex]f=\dfrac{v}{2\pi r}\\\\f=\dfrac{0.5}{2\pi \times 0.02}\\\\f=3.97\ Hz[/tex]

Hence, the frequency of motion s 3.97 Hz.

The frequency of the motion is 4.1 Hz.

Linear velocity?

The linear velocity of the of the object is calculated as follows;

[tex]a = \frac{v^2}{r} \\\\v^2 = ar\\\\v = \sqrt{ar} \\\\v = \sqrt{13 \times 0.02} \\\\v = 0.51 \ m/s[/tex]

Angular speed of the object

The angular speed of the object is calculated as follows;

[tex]\omega =\frac{v}{r} \\\\\omega = \frac{0.51}{0.02} \\\\\omega = 25.5 \ rad/s[/tex]

Frequency of motion

The frequency of the motion is calculated as follows;

[tex]\omega = 2\pi f\\\\f = \frac{\omega }{2\pi} \\\\f = \frac{25.5}{2\pi } \\\\f = 4.1 \ Hz[/tex]

Thus, the frequency of the motion is 4.1 Hz.

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1. When asteroids collided some of the broken materials fall into Earth's orbit. What do
astronomers call the debris when it hits planet Earth?

Answers

Answer:

meteoroids

Explanation:

when an asteroid (or really anything else) falls to earth, it is called a meteoroid

In a simulation on earth, an astronaut in his space suit climbs up a vertical ladder. On the moon, the same astronaut makes the same climb. In which case does the gravitational potential energy of the astronaut change by a greater amount?

Answers

Answer:

Gravitational potential energy of the astronaut will change by a greater amount on the earth

Explanation:

Gravitational potential energy is expressed by the formula;

GPE = mgh

This means that the gravitational potential energy is directly proportional to the gravity(g)

Now, from constant values, gravity of moon is 1.62 m/s² while gravity of the earth is 9.81 m/s².

This means that if we plug in the values of g on the earth and g on the moon, the potential energy on the earth would be greater than that of the moon

Thus, gravitational potential energy of the astronaut will change by a greater amount on the earth

A 0.12-m-radius grinding wheel takes 5.5 s to speed up from 2.0 rad/s to 11.0 rad/s. What is the wheel's average angular acceleration?

Answers

Answer:

0.56rad/s²

Explanation:

Using the equation of motion

wf = wi + αt

wf is the final angular velocity

wi is the initial angular velocity

α is the angular acceleration

t is the time

Given

wf = 11.0rad/s

wi =2.0rad/s

t = 5.5secs

Substitute into the formula and get α

11.0 = 2.0+5α

11.0-2.0 = 5α

9.0 = 5α

α = 5/9.0

α ≈ 0.56rad/s²

Hence the wheel's average angular acceleration is 0.56rad/s²

The wheel's average angular acceleration is equal to 1.64 [tex]rad/s^2[/tex].

Given the following data:

Radius = 0.12 meterTime = 5.5 secondsInitial angular velocity = 2.0 rad/sFinal angular velocity = 11.0 rad/s

To determine the wheel's average angular acceleration, we would apply the first equation of kinematics:

Mathematically, the angular acceleration of an object is given by the formula:

[tex]\alpha = \frac{\omega_f - \omega_i}{t}[/tex]

Where:

[tex]\omega_i[/tex] is the initial angular velocity.[tex]\omega_f[/tex] is the final angular velocity.t is the time.

Substituting the given parameters into the formula, we have;

[tex]\alpha =\frac{11.0\;-\;2.0}{5.5} \\\\\alpha =\frac{9.0}{5.5}[/tex]

Angular acceleration = 1.64 [tex]rad/s^2[/tex]

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The following is the longitudinal characteristic equation for an F-89 flying at 20,000 feet at Mach 0.638. The Short Period natural frequency is _____.

Answers

Answer:

hello your question is incomplete  attached below is the missing part  

answer : short period oscillations frequency  = 0.063 rad / sec

              phugoid oscillations natural frequency ( [tex]w_{np}[/tex] ) = 4.27 rad/sec

Explanation:

first we have to state the general form of the equation

= [tex]( S^2 + 2\alpha _{p} w_{np} S + w^{2} _{np} ) (S^{2} + 2\alpha _{s} w_{ns}S + w^{2} _{ns} ) = 0[/tex]

where :

[tex]w_{np} = Natural frequency of plugiod oscillation[/tex]

[tex]\alpha _{p} = damping ratio of plugiod oscilations[/tex]

comparing the general form with the given equation

[tex]w^{2} _{np}[/tex]  = 18.2329

[tex]w^{2} _{ns} = 0.003969[/tex]

hence the short period oscillation frequency ( [tex]w_{ns}[/tex] ) =  0.063 rad/sec

phugoid oscillations natural frequency ( [tex]w_{np}[/tex] ) = 4.27 rad/sec

What specific changes in two climate variables are expected to lead to major decreases in soil moisture southern Africa and the Mediterranean region?

Answers

Answer:

Less precipitation, droughts9: How might agriculture in southern Europe change by the end of the century if conditions follow the RCP8.

Explanation:

Precipitation and droughts are the specific changes in two climate variables that are expected to lead to major decreases in soil moisture.

What is drought?

Drought is defined as a period of protracted water scarcity, whether it is due to atmospheric surface water, or groundwater constraints.

Droughts can last months or years, although they can be proclaimed in as little as 15 days.

It has the potential to have a significant influence on the afflicted region's ecology and agriculture as well as harm the local economy.

Precipitation and droughts are the specific changes in two climate variables that are expected to lead to major decreases in soil moisture in southern Africa and the Mediterranean region.

Hence Precipitation and droughts are the specific changes in two climate variables.

To learn more about the drought refer to the link;

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when is thermal equilibrium achived between two identical objects



need help ASAP

Answers

Answer: When two objects in contact with each other are at different temperatures, they are said to be in thermal equilibrium.

Explanation: . When two objects not in contact with each other are at the same pressure, they are said to be in thermal equilibrium.

Which type of mass movement is likely to result in considerable property damage, but rarely causes loss of life? a. debris avalanche b. rock fall c. mudflow d. creep

Answers

Answer:

The correct answer is D. Creep.

Explanation:

Ground creep is a slow downward movement of a hill or mountain slope without the formation of demolition forms. The decisive factor for this is the continuous flow of movement of the soil.

The main driver of collapse is the movement of the surface layer particles during expansion in a direction perpendicular to the slope, followed by vertical collapse on contraction. The visible effect of the collapsing is the inclination of fences and poles, as well as trees that grow out of the ground towards the slope and have trunks curved vertically, in more extreme cases it may be cracks on the walls of buildings.

3. Sarah drops a 10.0 kg objed from a 200 m high bridge over a river. a) What work is done by the objec as a result of its fall? ​

Answers

work done = force x distance
10kg is 98.1 newtons
98.1x200=19620
i’m not sure though sorry if this is wrong :)

What is the volume of a box if he has Length=7 cm Width=5cm , Height=10cm ?

Answers

Answer:

Volume of Cuboid = Height*Width*Length

Explanation:

Volume of Cuboid = 10*5*7

= 350 cu² cm

Answer:

Diagram:-

[tex]\setlength{\unitlength}{0.74 cm}\begin{picture}\thicklines\put(5.6,5.4){\bf A}\put(11.1,5.4){\bf B}\put(11.2,9){\bf C}\put(5.3,8.6){\bf D}\put(3.3,10.2){\bf E}\put(3.3,7){\bf F}\put(9.25,10.35){\bf H}\put(9.35,7.35){\bf G}\put(3.5,6.1){\sf 5cm}\put(7.7,6.3){\sf 7cm}\put(11.3,7.45){\sf 10cm}\put(6,6){\line(1,0){5}}\put(6,9){\line(1,0){5}}\put(11,9){\line(0,-1){3}}\put(6,6){\line(0,1){3}}\put(4,7.3){\line(1,0){5}}\put(4,10.3){\line(1,0){5}}\put(9,10.3){\line(0,-1){3}}\put(4,7.3){\line(0,1){3}}\put(6,6){\line(-3,2){2}}\put(6,9){\line(-3,2){2}}\put(11,9){\line(-3,2){2}}\put(11,6){\line(-3,2){2}}\end{picture}[/tex]

Required Answer:-

It is a cuboid

where

length =l=7cmwidth=b=5cmheight =h=10cm

As we know that in a cuboid

[tex]{\boxed{\sf Volume=lbh}}[/tex]

Substitute the values

[tex]{:}\longrightarrow[/tex][tex]\sf Volume =7×5×10 [/tex]

[tex]{:}\longrightarrow[/tex][tex]\sf Volume=35×10 [/tex]

[tex]{:}\longrightarrow[/tex][tex]\sf Volume=350cm^3 [/tex]

Why are the coral reefs suffering? (site 2) explain

Answers

Answer:

bcuz ov yo fat mamma

Explanation:

Answer:

Water pollution

Explanation:

The spring constant, k, for a 22cm spring is 50N/m. A force is used to stretch the spring and when it is measured again it is 32cm long. Work out the size of this force

Answers

Answer:

5N

Explanation:

Given parameters:

Original length = 22cm

Spring constant, K  = 50N/m

New length = 32cm

Unknown

Force applied  = ?

Solution:

The force applied on a spring can be derived using the expression below;

   Force  = KE

 k is the spring constant

 E is the extension

  extension = new length - original length

  extension  = 32cm  - 22cm  = 10cm

convert the extension from cm to m;  

   100cm  = 1m;

    10cm will give 0.1m

So;

  Force  = 50N/m x 0.1m  = 5N

When a mass of a cart is 10 kg, and an applied force is 5 N, The acceleration of the cart is

5.0 m/s2
2.0 m/s2
0.5 m/s2
0.2 m/s2

Answers

Answer:

0.5 m/s2

Explanation:

F = ma

5 = 10a

a = 5/10

a =0.5

A medicine ball has a mass of 5kg and is thrown with a speed of 3 m/sec what is it's kinetic energy

Answers

KE = 1/2mv^2

KE = 1/2 (5kg)(3m/s)

KE = 22.5 J
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