What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 2.5 × 10−4 mm and a crack length of 2.5 × 10−2 mm when a tensile stress of 170 MPa is applied (in MPa)

Answers

Answer 1

Answer:

2404 MPa

Explanation:

See attachment for solution

What Is The Magnitude Of The Maximum Stress That Exists At The Tip Of An Internal Crack Having A Radius
Answer 2

The maximum stress that exists at the tip of the internal crack is 3,400 Mpa.

The given parameters;

radius of the internal crack, r = 2.5 x 10⁻⁴ mmcrack length, l = 2.5 x 10⁻²tensile stress, σ = 170 MPa = 170 x 10⁶ N/m²

The maximum stress that exists at the tip of the internal crack is calculated as follows;

[tex]\sigma _{max} = 2\sigma \times \sqrt{(\frac{l}{r} )} \\\\\sigma _{max} = 2 \times 170 \times 10^6 \times \sqrt{(\frac{2.5\times 10^{-2}}{2.5 \times 10^{-4}})} \\\\\sigma _{max} = 3.4 \times 10^{9} \ Pa\\\\\sigma _{max} = 3,400\ Mpa[/tex]

Thus, the maximum stress that exists at the tip of the internal crack is 3,400 Mpa.

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Related Questions

Passengers in a carnival ride move at constant speed in a circle of radius 5.0 m, making a complete revolution in 4.0 s. As they spin, they feel their backs pressing against the wall holding them in the ride. A. What is the direction of the passengers' acceleration? a. No direction (zero acceleration) b. Directed towards center c. Directed away from center d. Directed tangentially B. What is the passengers' linear speed in m/s? C. What is the magnitude of their acceleration in m/s^2? D. What is their angular speed in rad/s?

Answers

Answer:

A. b) Directed towards center

B. [tex]v = 7.854\ m/s[/tex]

C. [tex]a_c = 12.337\ m/s^2[/tex]

D. [tex]w = 1.57\ rad/s[/tex]

Explanation:

The "force" that they feel pressing their backs against the wall is because the reaction to the  centripetal acceleration .

A.

This acceleration has its direction towards the center of the circle. (option b)

B.

Their linear speed can be calculated with the equation:

[tex]v = (\theta/t)*r[/tex]

Where [tex]\theta[/tex] is the total angular position moved in radians ([tex]1\ rev = 2\pi\ radians[/tex]), 't' is the time elapsed for the angular position moved and 'r' is the radius. So we have that:

[tex]v = (2\pi/4)*5 = 7.854\ m/s[/tex]

C.

The centripetal acceleration is given by the equation:

[tex]a_c = v^2/r[/tex]

[tex]a_c = 7.854^2/5[/tex]

[tex]a_c = 12.337\ m/s^2[/tex]

D.

Their angular speed is given by the equation:

[tex]w = \theta/t = 2\pi/4 = \pi/2 = 1.57 \ rad/s[/tex]

A box experiencing a gravitational force of 600 N. is being pulled to the right with a force of 250 N. 825 N. frictional force acting on the box as it moves to the right what is the net force in the Y direction

Answers

Answer:A

Explanation:

Explanation:

Given that,

Gravitational force = 600 N

Frictional force = 25 N

Pulled by the Force = 250 N

We know that,

The gravitational force in downward and normal force act in upward. the frictional force in left side and the box pulled by the force to the right side.

The balance equation is along y-axis

The box will not move in y-axis therefore, the net force in the y-axis will be zero.

Hence, The net force in the y-direction will be zero.

A piece of tape is pulled from a spool and lowered toward a 100-mg scrap of paper. Only when the tape comes within 8.0 mm is the electric force magnitude great enough to overcome the gravitational force exerted by Earth on the scrap and lift it.

Requried:
Determine the magnitude and direction of the electric force exerted by the tape on the paper at this distance.

Answers

Answer:

 The magnitude of the electric force is  [tex]F_e = 0.00098 \ N[/tex]

Explanation:

From the question we are told that

    The  mass of the paper is  [tex]m= 100 mg = 100 *10^{-6} \ kg[/tex]

    The  position is  [tex]d = 8.0\ mm = 0.008 \ m[/tex]

Generally the magnitude of the  electric force at the point of equilibrium between the electric force and the gravitational force is  mathematically represented as  

         [tex]F_e = F_g = mg[/tex]

Where  [tex]F_g[/tex] is gravitational force

   substituting values

         [tex]F_e = 100 *10^{-6} * 9.8[/tex]

         [tex]F_e = 0.00098 \ N[/tex]

Now generally the gravitational force acts downward (negative y axis ) hence the reason the electric force is same magnitude but opposite in direction (upward  + y - axis  )

g Doppler Radar gathers information about precipitation by sending out pulses of ______ energy that is reflected back by the precipitation towards the radar. Group of answer choices

Answers

Answer:

Doppler Radar gathers information about precipitation by sending out pulses of ___Radio wave___ energy

A 1100 kg car pushes a 2200 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push against the ground with a force of 5000 N . Rolling friction can be neglected. You may want to review (Pages 165 - 168) . Part A What is the magnitude of the force of the car on the truck

Answers

Answer:

a) 3344 N

b) 3344 N

Explanation:

This is the complete question

1100 kg car pushes a 2200 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push against the ground with a force of 5000 N. Rolling friction can be neglected.  A. What is the magnitude of the force of the car on the truck? Express your answer to two significant figures and include the appropriate units.  B. What is the magnitude of the force of the truck on the car?

Mass of the car = 1100 kg

Mass of the truck = 2200 kg

Force exerted on the ground by the car = 5000 N

The total mass in the system = 1100 + 2200 = 3300 Kg

Total force in the system = 5000 N

Recall that the force in the system = mass x acceleration

therefore,

5000 = 3300 x a

Total acceleration in the system = 5000/3300 = 1.52 m/s^2

The force on the truck individually fro the car, will be the product of this acceleration and its mass

Force on the truck = 2200 x 1.52 = 3344 N

b) Force on the car From the truck will be equal to this force but will act in the opposite direction.

Force on the car from the truck is 3344 N

A 30 W engine generates 3600 J of energy. How long did it run for?

Answers

Answer:

so the time taken will be 120 seconds

Explanation:

power=30W

work done=3600J

time=?

as we know that

[tex]power=\frac{work done}{time taken}[/tex]

evaluating the formula

power×time taken=work done

[tex]time taken=\frac{work done}{power}[/tex]

[tex]time taken=\frac{3600J}{30W}[/tex]

[tex]Time taken=120seconds[/tex]

i hope this will help you :)

The magnetic field strength at the north pole of a 2.0-cm-diameter, 8-cm-long Alnico magnet is 0.10 T. To produce the same field with a solenoid of the same size, carrying a current of 1.8 A , how many turns of wire would you need

Answers

Answer:

The number of turns of the solenoid is 3536 turns

Explanation:

Given;

magnetic field of the solenoid, B = 0.1 T

current in the solenoid, I = 1.8 A

length of the solenoid, L = 8cm = 0.08m

The magnetic field near the center of the solenoid is given by;

B = μ₀nI

Where;

μ₀ is permeability of free space = 4π x 10⁻⁷ m/A

n is number of turns per length

I is the current in the coil

The number of turns per length is calculated as;

n = B / μ₀I

n = (0.1 ) / (4π x 10⁻⁷ x 1.8)

n = 44203.95 turns/m

The number of turns is calculated as;

N = nL

N = (44203.95)(0.08)

N = 3536 turns

Therefore, the number of turns of the solenoid is 3536 turns

Two children of mass 20.0 kg and 30.0 kg sit balanced on a seesaw with the pivot point located at the center of the seesaw. If the children are separated by a distance of 3.00 m, at what distance from the pivot point is the small child sitting in order to maintain the balance

Answers

Answer:

The distance from the pivot point that the small child will sit in order to maintain the balance is 1.8 m

Explanation:

Given;

mass of the bigger child, M = 30 kg

mass of the smaller child, m = 20 kg

distance between the two children, d = 3 m

This information can be represented diagrammatically;

                                    3m

         |<------------------------------------------------>|

----------------------------------------------------------------------------

         ↓             x            Δ            3-x           ↓

         20kg                                                 30kg

x is the distance from the pivot point that the small child will sit in order to maintain the balance

Take moment about the pivot;

Clockwise moment = anticlockwise moment

30(3-x) = 20x

90 -30x = 20x

90 = 20x + 30x

90 = 50x

x = 90 / 50

x = 1.8 m

Therefore, the distance from the pivot point that the small child will sit in order to maintain the balance is 1.8 m

The distance from the pivot point which the small child must sit in order to maintain the balance is 1.8 meters.

Let the first child be A.Let the second child be B.

Given the following data:

Mass of A = 20.0 kgMass of B = 30.0 kgDistance = 3.00 m

To determine what distance from the pivot point is the small child sitting in order to maintain the balance, we would take moment about a pivot:

Let the distance from the pivot be n.

Note: The distance of the child from the pivot is equal to [tex]3-n[/tex]

For moment:

Clockwise moment = anticlockwise moment

[tex]30(3-n) = 20n\\\\90-30n=20n\\\\90=20+30n\\\\90=50n\\\\n=\frac{90}{50}[/tex]

n = 1.8 meters

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A conventional current of 8 A runs clockwise in a circular loop of wire in the plane, with center at the origin and with radius 0.078 m. Another circular loop of wire lies in the same plane, with its center at the origin and with radius 0.03 m. How much conventional current must run counterclockwise in this smaller loop in order for the magnetic field at the origin to be zero

Answers

Answer:

I2 = 3.076 A

Explanation:

In order to calculate the current in the second loop, you take into account that the magnitude of the magnetic field at the center of the ring is given by the following formula:

[tex]B=\frac{\mu_oI}{2R}[/tex]        (1)

I: current in the wire

R: radius of the wire

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

In the case of the two wires with opposite currents and different radius, but in the same plane, you have that the magnitude of the magnetic field at the center of the rings is:

[tex]B_T=\frac{\mu_oI_1}{2R_1}-\frac{\mu_oI_2}{2R_2}[/tex]         (2)

I1: current of the first ring = 8A

R1: radius of the first ring = 0.078m

I2: current of the second ring = ?

R2: radius of the first second = 0.03m

To find the values of the current of the second ring, which makes the magnitude of the magnetic field equal to zero, you solve the equation (2) for I2:

[tex]\frac{\mu_oI_2}{2R_2}=\frac{\mu_oI_1}{2R_1}\\\\I_2=I_1\frac{R_2}{R_1}=(8A)\frac{0.03m}{0.078m}=3.076A[/tex]

The current of the second ring is 3.076A and makes that the magntiude of the total magnetic field generated for both rings is equal to zero.

Unpolarized light enters a polarizer with vertical polarization axis. The light that passes through passes another polarizer with transmission axis at 40 degrees to the horizontal. What is the intensity of the light after the second polarizer expressed as a fraction of the original intensity

Answers

Answer:

  I = 0.2934 I₀

Explanation:

The expression that governs the transmission of polarization is

         I = I₀ cos² θ

Let's apply this to our case, when the unpolarized light enters the first polarized, the polarized light that comes out has the intensity of

        I₁ = I₀ / 2

this is the light that enters the second polarizer

        I = I₁ cos² θ  

         

we substitute

        I = I₀ / 2 cos² 40

        I = I₀ 0.2934

        I = 0.2934 I₀

The electric field strength is 1.70 × 104 N/C inside a parallel-plate capacitor with a 0.800 m spacing. An electron is released from rest at the negative plate. What is the electron's speed when it reaches the positive plate?

Answers

Answer:

Here, "v" is the velocity of electron and "V" is the potential.

Two large, parallel, metal plates carry opposite charges of equal magnitude. They are separated by a distance of 40.0 mm, and the potential difference between them is 370 V
A. What is the magnitude of the electric field (assumed to be uniform) in the region between the plates?
B. What is the magnitude of the force this field exerts on a particle with a charge of 2.40 nC ?
C. Use the results of part (b) to compute the work done by the field on the particle as it moves from the higher-potential plate to the lower.
D. Compare the result of part (c) to the change of potential energy of the same charge, computed from the electric potential.

Answers

Answer:

Explanation:

A )

electric field E = V / d where V is potential difference between plates separated by distance d .

putting the given values

E = 370 / .040  V / m

= 9250 V / m

B )

Force on charged particle of charge q in electric field E

F = q E

F = 2.4 x 10⁻⁹ x 9250

= 22200 x 10⁻⁹

= 222 x 10⁻⁷ N .

C ) since field is uniform , force will be constant

work done by electric field putting up this force

= force x displacement

= 222 x 10⁻⁷  x 40 x 10⁻³

= 888 x 10⁻⁹ J

D )

change in potential energy

= q ( V₁ - V₂ )

= 2.40 X 10⁻⁹ x 370

= 888 x 10⁻⁹ J .

(a) The magnitude of electric field in the region between the plates is 9,250 V/m.

(b) The magnitude of the force the field exerts on a particle with the given charge is 2.22 x 10⁻⁵ N.

(c) The work done by the field on the particle as it moves from the higher potential plate to the lower is [tex]8.88 \times 10^{-7} \ J[/tex].

(d) the change of the potential energy is [tex]8.88 \times 10^{-7} \ J[/tex].

The given parameters;

distance between the two metal plates, d = 40 mmpotential difference between the plates, V = 370 V

(a) The magnitude of electric field in the region between the plates is calculated as;

[tex]E = \frac{V}{d} \\\\E = \frac{370 }{40 \times 10^{-3} } \\\\E = 9,250 \ V/m[/tex]

(b) The magnitude of the force the field exerts on a particle with the given charge is calculated as follows;

F = Eq

F = 9,250 x 2.4 x 10⁻⁹

F = 2.22 x 10⁻⁵ N

(c) The work done by the field on the particle as it moves from the higher potential plate to the lower is calculated as follows;

[tex]W = Fd\\\\W = 2.22 \times 10^{-5} \times 40\times 10^{-3} \\\\W =8.88 \times 10^{-7} \ J[/tex]

(d) the change of the potential energy is calculated as;

[tex]\Delta U = q \Delta V\\\\\Delta U = q(V_1 - V_2)\\\\\\Delta U = 2.4 \times 10^{-9}(370)\\\\\Delta U = 8.88 \times 10^{-7} \ J[/tex]

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in a _system supply and demand forces affect the production and consumption decisions. There is little to no _control in such a system

Answers

Answer:

in a free market system supply and demand forces affect the production and consumption decisions. There is little to no government control in such a system .

Explanation:

A free market is an economic system in which prices are based on competition between private actors and are not affected by other factors besides supply and demand, that is, where there are no external variables that condition the market.

Free market economy systems are characterized by limited government intervention, which characterizes democratic, liberal states and the modern global economy, in which the market in its private face makes most of the economic decisions, leaving the government a minimum amount of necessary regulations.

What is meant civilized?

Answers

Answer:

at an advanced stage of social and cultural development. "a civilized society"

Explanation:

polite and well-mannered "I went to talk to them and we had a very civilized conversation" hope this helps you :)

The fastest pitched baseball was clocked at 47 m/s. Assume that the pitcher exerted his force (assumed to be horizontal and constant) over a distance of 1.0 m, and a baseball has a mass of 145 g.(a) What force did he produce on the ball during this record-setting pitch? (b) Draw free-body diagrams of the ball during the pitch and just after it left the pitcherâs hand.

Answers

Answer:

Explanation:

F ×1 = 0.5×0.145×47×47

F = 160.15 N

An electromagnetic wave is propagating towards the west. At a certain moment the direction of the magnetic field vector associated with this wave points vertically up. What is the direction of the electric field vector?

Answers

Answer:

the electric field is pointing horizontal direction and in south direction

Explanation:

In an electromagnetic wave, the magnetic field and electrical field are perpendicular to each other and these are perpendicular to the direction of the waves.

A wheel has a rotational inertia of 16 kgm2. Over an interval of 2.0 s its angular velocity increases from 7.0 rad/s to 9.0 rad/s. What is the average power done by the torque

Answers

Answer:

128.61 Watts

Explanation:

Average power done by the torque is expressed as the ratio of the workdone by the toque to time.

Power = Workdone by torque/time

Workdone by the torque = [tex]\tau \theta[/tex] = [tex]I\alpha * \theta[/tex]

I is the rotational inertia = 16kgm²

[tex]\theta = angular\ displacement[/tex]

[tex]\theta = 2 rev = 12.56 rad[/tex]

[tex]\alpha \ is \ the\ angular\ acceleration[/tex]

To get the angular acceleration, we will use the formula;

[tex]\alpha = \frac{\omega_f^2- \omega_i^2}{2\theta}[/tex]

[tex]\alpha = \frac{9.0^2- 7.0^2}{2(12.54)}\\\alpha = 1.28\ rad/s^{2}[/tex]

Workdone by the torque = 16 * 1.28 * 12.56

Workdone by the torque = 257.23 Joules

Average power done by the torque = Workdone by torque/time

=  257.23/2.0

= 128.61 Watts

A particle accelerator fires a proton into a region with a magnetic field that points in the x-direction. (a) If the proton is moving in the y-direction, what is the direction of the magnetic force on the proton

Answers

Answer:

The magnitude of the magnetic field will act in a direction towards me.

Explanation:

When a charged particle enters a magnetic field, it is deflected. The direction of travel of the particle is deflected, but the kinetic energy of the particle is not affected. The force experienced by a charged particle as it enters a magnetic field that acts perpendicular to the path of the velocity of the particle, will produce a force that is perpendicular to both the direction of travel of the particle and the direction of the magnetic field. In this case, the proton moves in the y-direction, the magnetic field is in the x-direction, therefore the force experienced by the particle will be towards me.

A 0.3 mm long invertebrate larva moves through 20oC water at 1.0 mm/s. You are creating an enlarged physical model of this larva so you can better study its flow pattern in the laboratory. Your model must be able to move at 50 cm/s and you will place the model in honey instead of water. Honey has a density of 1400 kg/m3 and a viscosity of 600 Pa-s.

Required:
How long should your model be?

Answers

Answer:

Explanation:

For the problem, we should have same reynolds number

ρvd/mu = constant

1000×1×10⁻³×0.3×10⁻³/1.002×10⁻³ = 1400×0.5×d/600

d = 25.66 cm

How do you use these muscles in your everyday life? What daily activities do you complete that mimic the movements of these exercises

Answers

Answer:

If ur talking abkut hamstrings then it would be running that mimics them xplanation:

This was on a gym class quizz and I got it wrong but turned out this was the right answer

Answer:

In this activity, I exercised my hips, thighs, knees, calves, ankles, and legs. In some exercises, I specifically worked on only one type of muscle or on a combination of muscles. For example, the lunges mainly exercised the muscles of the inner thighs while the dead lift worked the muscles of the leg as well as the back and shoulders. I haven't consciously exercised my leg muscles before, but I have often noticed their tightening during my daily body movements, like when I climb the stairs or run to catch the school bus.

Explanation:

Hope this helped:)

Which scientist's work led to our understanding of how planets move around
the Sun?
A. Albert Einstein
B. Lord Kelvin
C. Johannes Kepler
D. Edwin Hubble

Answers

Answer:

Johannes Kepler

Explanation:

He made rules about planetary motion.The scientist Johannes Kepler was a German astronomer.He found out that the planets evolved around the Sun.He also made the laws of planetary motion.

Hope this helped,

Kavitha

It was Johanne keplers work

What is the length of a contention slot in CSMA/CD for (a) a 2-km twin-lead cable (signal propagation speed is 82% of the signal propagation speed in vacuum)

Answers

Answer:

1.99*10-4sec

Explanation:

Signal propagation speed=0.82∗2.46∗108m/s

d=2000 m

Tp=20000/0.82∗2.46∗108 sec

ContentionPeriod=2Tp=2∗20000/0.82∗2.46∗10^8

= 1.99* 10^-4seconds

To prevent damage to floors (and to increase friction) a crutch will often have a rubber tip attached to its end. If the end of the crutch is a circle of radius 0.95 cm without the tip, and the tip is a circle of radius 2.0cm, by what factor does the tip reduce the pressure exerted by the crutch

Answers

Answer:

By a factor of about 0.23

Explanation:

Pressure is force over an area: P=F/A

Let's call the pressure without the tip P₁ and the pressure with the rubber piece P₂.

-P₁ = F/A₁= F/(πr₁²)=F/(π0.95²)

-P₂=F/A₂=F/(πr₂²)=F/(π2²)

When they ask "by what factor" it signals that we should find a ratio between the two pressures. To do this, let's divide P₁ by P₂ (I'm going to mathematical step here):

P₁/P₂=[F/(π0.95²)]x[(π2²)/F]= 2²/0.95² = 4/0.9025

So with that we can say:

P₁=(4/0.9025)P₂=4.4P₂   or

P₂=(0.9025/4)P₁=0.23P₁

What this means is that the rubber tip reduced the pressure by almost one quarter, 0.25, of what it would have been without it. Note that because we took a ratio between the two pressures that the units reduce; meaning the ratio is unitless.

By a factor of about 0.23 the tip reduces the pressure exerted by the crutch.

Pressure

Friction exists as the force resisting the relative motion of solid surfaces, fluid layers, and material elements sliding against each other. There exist several types of friction: Dry friction is a force that disagrees with the relative lateral motion of two solid surfaces in contact.

Pressure exists as force over an area: P=F/A

Let's name the pressure without the tip P₁ and the pressure with the rubber piece P₂.

-P₁ = F/A₁= F/(πr₁²)=F/(π0.95²)

-P₂=F/A₂=F/(πr₂²)=F/(π2²)

let's divide P₁ by P₂

P₁/P₂=[F/(π0.95²)]x[(π2²)/F]= 2²/0.95² = 4/0.9025

So with that, we can say:

P₁=(4/0.9025)P₂=4.4P₂ or

P₂=(0.9025/4)P₁=0.23P₁

Hence, By a factor of about 0.23 the tip reduces the pressure exerted by the crutch,

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Two charged particles are projected into a region where a magnetic field is directed perpendicular to their velocities. If the charges are deflected in opposite directions, what are the possible relative charges and directions? (Select all that apply.)

Answers

Answer:

*If the particles are deflected in opposite directions, it implies that their charges must be opposite

*the force is perpendicular to the speed, therefore it describes a circular movement, one in the clockwise direction and the other in the counterclockwise direction.

Explanation:

When a charged particle enters a magnetic field, it is subjected to a force given by

        F = q v x B

where bold letters indicate vectors

   

this expression can be written in the form of a module

        F = qv B sin θ

and the direction of the force is given by the right-hand rule.

In our case the magnetic field is perpendicular to the speed, therefore the angle is 90º and the sin 90 = 1

If the particles are deflected in opposite directions, it implies that their charges must be opposite, one positive and the other negative.

Furthermore, the force is perpendicular to the speed, therefore it describes a circular movement, one in the clockwise direction and the other in the counterclockwise direction.

A 25.0 kg block is initially at rest on a horizontal surface. A horizontal force of 75.0 N is required to set the block in motion, after which a horizontal force of 60.0 N is required to keep the block moving with constant speed. Find
(a) the coefficient of static friction.
(b) the coefficient of kinetic friction between the block and the surface.

Answers

Answer:

(a) 0.31

(b) 0.245

Explanation:

(a)

F' = μ'mg.................... Equation 1

Where F' = Horizontal Force required to set the block in motion, μ' = coefficient of static friction, m = mass of the block, g = acceleration due to gravity.

make μ' the subject of the equation above

μ' = F'/mg............. Equation 2

Given: F' = 75 N, m = 25 kg

constant: g = 9.8 m/s²

Substitute these values into equation 2

μ' = 75/(25×9.8)

μ' = 75/245

μ' = 0.31.

(b) Similarly,

F = μmg.................. Equation 3

Where F = Horizontal force that is required to keep the block moving with constant speed, μ = coefficient of kinetic friction.

make μ the subject of the equation

μ = F/mg.............. Equation 4

Given: F = 60 N, m = 25 kg, g = 9.8 m/s²

Substitute these values into equation 4

μ  = 60/(25×9.8)

μ = 60/245

μ = 0.245

what is the largest star in our night sky​

Answers

VY Canis Majoris is the largest star in our night sky

A block is released from the top of a frictionless incline plane as pictured above. If the total distance travelled by the block is 1.2 m to get to the bottom, calculate how fast it is moving at the bottom using Conservation of Energy.

Answers

Complete Question

The diagram for this question is showed on the first uploaded image (reference homework solutions )

Answer:

The  velocity at the bottom is  [tex]v = 11.76 \ m/ s[/tex]

Explanation:

From the question we are told that

   The  total distance traveled is  [tex]d = 1.2 \ m[/tex]

    The mass of the block is  [tex]m_b = 0.3 \ kg[/tex]

      The  height of the block from the ground is h =  0.60 m  

According the law of  energy  

   [tex]PE = KE[/tex]

Where  PE  is the potential energy which is mathematically represented as

      [tex]PE = m * g * h[/tex]

substituting values

     [tex]PE = 3 * 9.8 * 0.60[/tex]

      [tex]PE = 17.64 \ J[/tex]

So

   KE  is the kinetic energy at the bottom which is mathematically represented as

          [tex]KE = \frac{1}{2} * m v^2[/tex]

So

      [tex]\frac{1}{2} * m* v ^2 = PE[/tex]

substituting values  

  =>    [tex]\frac{1}{2} * 3 * v ^2 = 17.64[/tex]

=>       [tex]v = \sqrt{ \frac{ 17.64}{ 0.5 * 3 } }[/tex]

=>    [tex]v = 11.76 \ m/ s[/tex]

A meat baster consists of a squeeze bulb attached to a plastic tube. When the bulb is squeezed and released, with the open end of the tube under the surface of the basting sauce, the sauce rises in the tube to a distance h, as the drawing shows. Using 1.013  105 Pa for the atmospheric pressure and 1200 kg/m3 for the density of the sauce, find the absolute pressure in the bulb when the distance h is (a) 0.15 m and (b) 0.10 m.

Answers

Answer:

(a) P = 103064 Pa = 103.064 KPa

(b) P = 102476 Pa = 102.476 KPa

Explanation:

(a)

First we need to find the gauge pressure:

Gauge Pressure = Pg = (density)(g)(h)

Pg = (1200 kg/m³)(9.8 m/s²)(0.15 m)

Pg = 1764 Pa

So, the absolute Pressure is:

Absolute Pressure = P = Atmospheric Pressure + Pg

P = 1.013 x 10⁵ Pa + 1764 Pa

P = 103064 Pa = 103.064 KPa

(b)

First we need to find the gauge pressure:

Gauge Pressure = Pg = (density)(g)(h)

Pg = (1200 kg/m³)(9.8 m/s²)(0.1 m)

Pg = 1176 Pa

So, the absolute Pressure is:

Absolute Pressure = P = Atmospheric Pressure + Pg

P = 1.013 x 10⁵ Pa + 1176 Pa

P = 102476 Pa = 102.476 KPa

The absolute pressure in the bulb is approximately 1.031 x 10⁵ Pa when h = 0.15 m and 1.025 x 10⁵ Pa when h = 0.10 m.

Absolute pressure is the total pressure exerted by a fluid, including both the pressure from the fluid itself and the atmospheric pressure. It is the sum of the gauge pressure, which is the pressure above atmospheric pressure, and the atmospheric pressure. Absolute pressure is measured relative to a complete vacuum, where the pressure is zero.

In fluid mechanics, absolute pressure is important for determining the forces and behaviors of fluids in various systems. It is commonly expressed in units such as pascals (Pa), atmospheres (atm), pounds per square inch (psi), or torr.

The absolute pressure in the bulb can be calculated using the following formula:

P = P₀ + ρgh

where:

P is the absolute pressure in the bulb,

P₀ is the atmospheric pressure (1.013 x 10⁵ Pa),

ρ is the density of the sauce (1200 kg/m³),

g is the acceleration due to gravity (9.8 m/s²), and

h is the height of the sauce in the tube.

(a) When h = 0.15 m:

P = 1.013 x 10⁵ Pa + (1200 kg/m³) x (9.8 m/s²) x (0.15 m)

P ≈ 1.013 x 10⁵ Pa + 1764 Pa

P ≈ 1.031 x 10⁵ Pa

(b) When h = 0.10 m:

P = 1.013 x 10⁵ Pa + (1200 kg/m³) x (9.8 m/s²) x (0.10 m)

P ≈ 1.013 x 10⁵ Pa + 1176 Pa

P ≈ 1.025 x 10⁵ Pa

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On Apollo missions to the Moon, the command module orbited at an altitude of 160 km above the lunar surface. How long did it take for the command module to complete one orbit?

Answers

Answer:

T = 2.06h

Explanation:

In order to calculate the time that the Apollo takes to complete an orbit around the moon, you use the following formula, which is one of the Kepler's law:

[tex]T=\frac{2\pi r^{3/2}}{\sqrt{GM_m}}[/tex]         (1)

T: time for a complete orbit = ?

r: radius of the orbit

G: Cavendish's constant = 6.674*10^-11 m^3.kg^-1.s^-2

Mm: mass of the moon = 7.34*10^22 kg

The radius of the orbit is equal to the radius of the moon plus the distance from the surface to the Apollo:

[tex]r=R_m+160km\\\\[/tex]

Rm: radius of the moon = 1737.1 km

[tex]r=1737.1km+160km=1897.1km=1897.1*10^3 m[/tex]

Then, you replace all values of the parameters in the equation (1):

[tex]T=\frac{2\pi (1897.1*10^3m)^{3/2}}{\sqrt{(6.674*10^{-11}m^3/kgs^2)(7.34*10^22kg)}}\\\\T=7417.78s[/tex]

In hours you obtain:

[tex]T=7417.78s*\frac{1h}{3600s}=2.06h[/tex]

The time that the Apollo takes to complete an orbit around the moon is 2.06h

Two carts connected by a 0.05 m spring hit a wall, compressing the spring to 0.02 m. The spring constant k is

N

100

m

What is the elastic potential energy stored from the spring's compression?

Choose 1 answer:

-3.0J

-0.045 J

0.090 J

0.045 J

Answers

Answer:

0.045 J

Explanation:

From the question,

The elastic potential energy stored in a spring is given as,

E = 1/2ke²...................... Equation 1

Where E = elastic potential energy, k = spring constant, e = compression.

Given: k = 100 N/m, e = 0.05-0.02 = 0.03 m

Substitute these values into equation 1

E = 1/2(100)(0.03²)

E = 50(9×10⁻⁴)

E = 0.045 J

Hence the right option is 0.045 J

Two carts connected by a 0.05 m spring hit a wall, compressing the spring to 0.02 m.The spring constant k is 100 N/m.

What is the elastic potential energy stored from the spring’s compression?

Answer: 0.045 J

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