A simple watermelon launcher is designed as a spring with a light platform for the watermelon. When an 8.00 kg watermelon is put on the launcher, the launcher spring compresses by 10.0 cm. The watermelon is then pushed down by an additional 30.0 cm and it’s ready to go. Just before the launch, how much energy is stored in the spring?

Answer:296.76 N

Explanation:

Given

mass of box [tex]m=100 kg[/tex]

Let [tex]T_1[/tex] be the Tension in left side and [tex]T_2[/tex] be the Tension in the right side

From diagram

[tex]T_1\cos 80=T_2\cos 65[/tex]

[tex]T_1=T_2\cdot \frac{\cos 65}{\cos 80}[/tex]

and

[tex]T_1\sin 80+T_2\sin 65=100\cdot g[/tex]

[tex]T_2\left [ \tan 80\cdot \cos 65+\sin 65\right ]=100\cdot g[/tex]

[tex]T_2=\frac{100\cdot g}{\left [ \tan 80\cdot \cos 65+\sin 65\right ]}[/tex]

[tex]T_2=\frac{980}{3.3023}=296.76 N[/tex]

Answer:

Temperature = 20.35 °C

Explanation:

Arrhenius equation is as follows:

k = A*exp(-Ea/(R*T)) , where

k = rate of chirps

Ea = Activation Energy

R = Universal Gas Constant

T = Temperature (in Kelvin)

A = Constant

Given Data

Ea = 53.9*10^3 J/mol

R = 8.3145  J/(mol.K)

T = 273.15 + 25  K

k = 178  chirps per minutes

Calculation

Using the Arrhenius equation, we can find A,

A= 4.935x10^11

Now we can apply the same equation with the data below to find T at k=126,

k = A*exp(-Ea/(R*T))

Ea = 53.9*10^3

R = 8.3145

k = 126

T = 20.35 °C

Answer:

a. 0 W

b. 0.4375 J

c. 2 W

d. 2 W

Explanation:

a.

Recall that the power is directly proportional to the current in the circuit. (P=I*V) First, determine the current at t=0. Note that inductor prevents an instantaneous build-up of current. Therefore, initial power is 0 W.

P initial = 0 W

b.

given on the APPC equation sheet (last equation bottom right):

U = (1/2)*L*(I^2)

we know I = V/R

U = (1/2)*3.5*(4/8)^2

U = .4375 J

c.

P = R*I^2

I=V/R

P = 8*(4/8)^2

P=2 W

d. (same as part c.)

P = R*I^2

I=V/R

P = 8*(4/8)^2

P=2 W

a) The battery supplies electrical energy at a certain rate just after the circuit is completed. b) When the current has reached its steady-state value, the energy stored in the inductor can be calculated. c) The rate at which electrical energy is being dissipated in the resistance of the inductor can be found using the power dissipated formula. d) The rate at which the battery is supplying electrical energy to the circuit is the same as the power delivered by it.

a) Just after the circuit is completed, at what rate is the battery supplying electrical energy to the circuit?

To determine the rate at which the battery is supplying electrical energy, we need to calculate the power delivered by the battery. Power is given by the formula P = IV, where P is power in watts, I is current in amperes, and V is voltage in volts. Since we know the voltage (4.00 V) and the resistance (8.00 Ω), we can use Ohm's Law (V = IR) to find the current. Once we have the current, we can calculate the power delivered by the battery.

b) When the current has reached its final steady-state value, how much energy is stored in the inductor?

When the current reaches its final steady-state value, the inductor acts like a short circuit and has no effect on the circuit. Therefore, all the energy stored in the inductor is dissipated as heat due to the resistance. Since we know the inductance (3.50 H) and the resistance (8.00 Ω), we can calculate the energy stored in the inductor using the formula E = 0.5LI^2, where E is energy in joules, L is inductance in henries, and I is current in amperes.

c) What is the rate at which electrical energy is being dissipated in the resistance of the inductor?

The rate at which electrical energy is being dissipated in the resistance of the inductor is equal to the power dissipated. Power dissipated is given by the formula P = I^2R, where P is power in watts, I is current in amperes, and R is resistance in ohms. Since we know the resistance (8.00 Ω) and the current, we can calculate the power.

d) What is the rate at which the battery is supplying electrical energy to the circuit?

The rate at which the battery is supplying electrical energy to the circuit is the same as the power delivered by the battery, which we calculated in part a using the formula P = IV.

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Answer:

Explanation:

The case relates to interference in thin films , in which we study interference of light waves reflected by upper and lower surface of a medium or glass.

For constructive interference , the condition is

2μt = ( 2n+1)λ/2

μ is refractive index of glass , t is thickness , λ is wavelength of light.

putting the given values

2 x 1.53 x 350 x 10⁻⁹ = ( 2n+1) λ/2

λ = 2142nm /  ( 2n+1)

For n = 2

λ = 428 nm

This wave length will have constructive interference making this light brightest of all .

For n = 1

λ = 714  nm

So second largest  brightness will belong to 700 nm wavelength.

Answer:

Completing the operation will require 9.765625 hours.

Explanation:

First, let's check which of the stages will be the process limitation:

Regarding the pump, it can perform up to 120 cy/h. The trucks will deliver 8 cy every 15 min, i.e., 32 cy/h, then this is the smaller capacity within the whole process.

Now, let's establish a simple relation:

Placement speed x Efficiency = Placed quantity / time,

solving for time. we have

[tex]t = \frac{q}{\epsilon s} = \frac{250 cy}{0.8\times 32\frac{cy}{h}} = \mathbf{9.765625 h}.[/tex]

The operation will require approximately 2.6 hours, taking into account the efficiency factor.

To determine how many hours the operation will require, we need to calculate the total time it takes for all the trucks to arrive at the site and unload the concrete. Since each truck delivers 8 cy of concrete every 15 mins, we can calculate the time it takes for all the trucks to deliver 250 cy of concrete: 250 cy / (8 cy/truck * 15 mins/truck) = 250 cy / (120 mins) = 2.08 hours.

However, we need to account for the 80% efficiency factor, which means the pump is only operating at 80% of its maximum output. Therefore, the total time required would be 2.08 hours / 0.8 = 2.6 hours.

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Answer:

L=500 mH

Explanation:

Here di/dt = 4A/s,    ε= -2V

Inductance of inductor, induced emf and rate of change of current have the following relation.

ε= [tex] - L\frac{di}{dt}[/tex]

⇒L= - ε/[tex]\frac{di}{dt}[/tex]

⇒L= -(-2)/ 4

⇒ L= 0.5 H   or

⇒ L= 500 mH

Answer:

final velocity  = - 1.75 mph

Explanation:

given data

mass m1 = 1 ton

mass m2 = 3 ton

velocity v = 41 mph

velocity u = 16 mph

to find out

what is final velocity  V

solution

we will apply here Conservation of momentum that is express as

mv + Mu = (m + M) × V    ...........................1

put here value we get

1 × 41 - 3 × 16 = (1 + 3 ) × V

solve it we get

41 - 48 = 4 V

V = [tex]\frac{-7}{4}[/tex]

final velocity  = - 1.75 mph

Final answer:

The final velocity of the SUV/car entanglement is 9.9 m/s.

Explanation:

In order to determine the final velocity of the SUV/car entanglement, we need to first calculate the momentum of each vehicle before the collision.

Momentum is determined by the product of an object's mass and velocity. So, the momentum of the 1 ton car before the collision is 1 ton (or 1000 kg) multiplied by its velocity of 41 mph (which is equivalent to 18.3 m/s). Therefore, the momentum of the car is 1000 kg  imes 18.3 m/s = 18300 kg*m/s.

Similarly, the momentum of the 3 ton SUV before the collision is 3 ton (or 3000 kg) multiplied by its velocity of 16 mph (which is equivalent to 7.1 m/s). Therefore, the momentum of the SUV is 3000 kg  imes 7.1 m/s = 21300 kg*m/s.

Since momentum is conserved in collisions, the total momentum before the collision is equal to the total momentum after the collision. This means that the final velocity of the SUV/car entanglement can be calculated by dividing the total momentum by the total mass. The total momentum is 18300 kg*m/s + 21300 kg*m/s = 39600 kg*m/s. The total mass is 1000 kg + 3000 kg = 4000 kg. So, the final velocity of the SUV/car entanglement is 39600 kg*m/s / 4000 kg = 9.9 m/s.

Answer:

B

Explanation:

B. Women have straighter teeth.

The subject of this question is Physics and it pertains to the concept of a ballistic pendulum.

The subject of the question is Physics. Specifically, it is related to the concept of a ballistic pendulum.

A ballistic pendulum consists of a block of dense material suspended from a cord. When a bullet is fired into the block, it lodges itself into the block in a specific manner, either elastically or by sticking together. The (block+bullet) swings upward to some maximum height due to the acquired velocity.

For example, if the block and bullet stick together after the collision, the system forms a ballistic pendulum. The bullet's initial kinetic energy is converted into potential energy as the (block+bullet) rises.

The ballistic pendulum consists of a block of dense material suspended from a cord and allowed to swing freely. A bullet is fired into the block, lodging itself into the block in a sticky collision. The (block+bullet) having now acquired a velocity, swings upward to some maximum height. 

A ballistic pendulum is a device that measures the speed of a projectile, such as a bullet. When a bullet of mass m is fired into a block of mass M suspended from a cord, it lodges itself into the block in a sticky collision. The combined mass then swings to a maximum height h, where the kinetic energy is converted into potential energy.

1. Conservation of momentum: m[tex]v_{i}[/tex] = (m + M)[tex]v_{f}[/tex]

2. Conservation of energy: 1/2(m + M)[tex]v_{f}^{2}[/tex] = (m + M)gh.

From these equations, [tex]v_{i}[/tex] can be derived:

[tex]v_{i}[/tex] = (1 + M/m) √(2gh)

Thus, the ballistic pendulum helps measure the speed of the bullet before the collision.

Answer:

Velocity of the ping pong ball must be = V2= 6,035.34m/s

Explanation:

M1= momentum of the bowling ball

m1 = mass of the bowling ball= 5.8kg

v1= velocity of the bowling ball= 1.59m/s

M2= momentum of the ping pong ball

m2= mass of the ping pong ball= 1.528 g/1000=  0.001528kg

v2= velocity of the ping pong ball

Momentum of the bowling ball= M1= m1v1= 5.8* 1.59= 9.222 kg-m/s

Momentum of the ping pong ball = M2= M1= m2v2

= 0.001528 *v2= 9.222

v2= 9.222/0.001528= 6,035.34 m/s

Answer:

C) false. The image is formed by the prolongation of the rays, so it is VIRTUAL

Explanation:

Let's review each of the statements

A) True. The image is the same size as the object in a flat mirror, m = 1

B) True. The rays are not inverted, so the right images

C) false. The image is formed by the prolongation of the rays, so it is VIRTUAL

D) True. Flat mirrors reverse left and right

E) True. When using trigonometry the angles are equal, therefore two triangles formed have the same leg, and the distance to the object and the image are equal

Answer:

The image is real is not a characteristics of a plane mirror.

Explanation:

A plane mirror is a reflector with a flat reflective covering. For light beams hitting a plane mirror, the angle of reflection matches the angle of incidence. The angle of the incidence is the angle within the incident beam and the surface normal flat mirror.

Reason for Image is not real in the plane mirror:

Thus we can say, real image is not a characteristic of a plane mirror.

To learn more about real images, refer:

Final answer:

To find the maximum shearing force in the pegs during the truck's acceleration, Newton's second law is applied, and the weight and acceleration of the gate are considered. The final shear force for each peg is 7.7021 lbs.

Explanation:

To find the maximum shearing force in each peg during the acceleration of the pickup truck with the trailer, we need to apply Newton's second law.

First, let's convert the speed from miles per hour to feet per second (1 mi/hr = 1.46667 ft/s):

37 mi/hr × 1.46667 ft/s/mi/hr = 54.2667 ft/s

Using the kinematic equation v^2 = u^2 + 2as (where v is final velocity, u is initial velocity, s is distance, and a is acceleration), we can solve for acceleration (a) since the truck starts from rest (u = 0):

54.2667^2 = 0 + 2 × a × 215

a = × 54.2667^2 / (2 × 215) = 6.88776 ft/s^2

The force affecting the gate pegs comes from the horizontal component of the gravitational force and the force due to acceleration. Assume that the mass of the gate is 72 lbs; the gravitational force is:

F_gravity = m × g = 72 lbs (Since g in lbs already accounts for Earth's gravitational acceleration)

To find the force due to acceleration, we use F = m × a:

F_acceleration = 72 lbs × 6.88776 ft/s^2

However, since lbs already include g, we convert lbs to mass in slugs:

m = 72 lbs / 32.2 ft/s^2 = 2.23602 slugs

F_acceleration = 2.23602 slugs × 6.88776 ft/s^2 = 15.4042 lbs

The total force on each peg is thus the shearing force due to the acceleration, but as there are two pegs, we divide the total force by 2, assuming the load is distributed evenly:

Shear force per peg = F_acceleration / 2 = 15.4042 lbs / 2 = 7.7021 lbs (rounded to four significant figures)

Answer:

1.  2176 kilograms of plastic  2. $15,232  3. $397 (U)     4. $947 (F)   $1,344  (U)

Explanation:

Generally, cost variance analysis can be used to estimate the difference between the actual cost and the expected costs. However, if the actual cost is more than the expected cost, then the variance is said to be unfavorable and vice versa.

1. The standard quantity of kilograms of plastic (SQ) that is allowed to make 3,200 helmets?

We know that each helmet requires 0.68 kilograms of plastic. Thus, to make 3200 helmets, we will need 0.68*3200 = 2176 kilograms of plastic

2. The standard materials cost allowed (SQ × SP) to make 3,200 helmets.

We also know that each helmet costs $7.00 per kilogram. Therefore, to make 3200 helmets, the standard materials cost = $7.00*2176 = $15,232

3. The material spending variance = difference between the actual cost and the standard cost = $15,629 - $15,232 = $397  U

4.  The materials price variance and the materials quantity variance?

The materials price variance  is the actual cost- (standard cost per kilogram x actual number of plastic used) . Therefore:

Materials price variance = $15,629 - ($7 x 2,368 kg)

Materials price variance = $15,629 - $16,576 = ($947) F

Since the budgeted cost is relatively higher than the actual cost, the materials price variance is favorable (F) by $947.

The materials quantity variance = (Actual number of plastic used x Standard cost per kilogram) - Standard cost

Materials quantity variance = ($7 x 2,368 kg) - $15,232

Materials quantity variance = $16,576 - $15,232 = $1,344  U

Since the budgeted cost is relatively higher than the standard cost, the materials quantity variance is unfavorable (U) by $1,344.

To find the standard quantity of plastic and standard materials cost allowed, multiply the standard quantity per helmet by the number of helmets produced and multiply the standard quantity by the standard price per kilogram, respectively. The materials spending variance is the difference between the actual cost and the standard cost that should have been incurred. The materials price variance is the difference between the actual quantity of plastic used multiplied by the standard price per kilogram and the actual cost, while the materials quantity variance is the difference between the standard quantity of plastic allowed multiplied by the standard price per kilogram and the actual cost.

To calculate the standard quantity of plastic (SQ) allowed to make 3,200 helmets, multiply the standard quantity per helmet by the number of helmets produced. In this case, SQ = 0.68 kg/helmet * 3,200 helmets = 2,176 kg. To calculate the standard materials cost allowed (SQ × SP), multiply the standard quantity by the standard price per kilogram. In this case, the materials cost allowed is $15,232 (2,176 kg * $7.00/kg).

To calculate the materials spending variance, subtract the actual cost from the standard cost that should have been incurred. In this case, the materials spending variance is $397 (Standard Cost - Actual Cost = $15,232 - $15,629). The materials price variance is calculated by subtracting the actual quantity of plastic used multiplied by the standard price per kilogram from the actual cost.

The materials quantity variance is calculated by subtracting the standard quantity of plastic allowed multiplied by the standard price per kilogram from the actual cost. In this case, the materials price variance is $7,774 (Actual Cost - Actual Quantity * Standard Price = $15,629 - 2,368 kg * $7.00/kg) and the materials quantity variance is -$2,145 (Actual Quantity * Standard Price - Standard Cost = 2,368 kg * $7.00/kg - $15,232).

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Explanation:

We need to find how many calories is 1 BTU.

Given

          1 BTU = 1054 J

          1 calorie = 4.186 J

So we have

          1 BTU = 4.186 x 251.79 J

          1 BTU =251.79 calorie

          1 BTU = 252 calorie.

Option C is the correct answer.

Answer:

"when an infection has cleared, a small number of B and T cells are present in the blood with memory of the virus, allowing them to activate and destroy viruses more quick and fast next time when they enter the body."which is known as immune system."

Explanation:

Our immune system is not only designed to destroy disease-causing codes but remembers codes it's encountered so they are able to fight against them with when they return. when we all come across these types of invaders our layer of protection stops it from entering our body. When body senses a virus or other infections, like bacteria, the body tends to try and destroy the foreign invaders. The first step is the macrophages is activated. Macrophages destroy the foreign invaders.  If the viruses penetrates deep into the body it result into infection. Where the T cells and B cells are activated to fight against these viruses.T & B cells contains antibodies that attach to the virus and label it as foreign for other cells to destroy.

Final answer:

The forces exerted by the water on the submerged block are due to pressure differences and can be calculated using the dimensions of the block, the density of water, and the depth of immersion. The scale reading reflects the difference between the block's weight and the buoyant force. Lastly, the buoyyant force is confirmed to be the difference between the top and bottom forces on the block, in alignment with Archimedes' principle.

Explanation:

The student's question involves calculating the forces exerted by water on a submerged block and the reading of a scale when the block is immersed in water. To calculate these forces, one must consider the forces exerted by the water on the top and bottom of the block, which are due to the differences in pressure at these points, as well as the buoyant force acting on the block. Using the principle of Archimedes, which states that the upthrust or buoyant force on a submerged object is equal to the weight of the fluid it displaces, we can find the buoyant force which in turn will help to determine the scale reading.

The weight of the water displaced can be calculated using the volume of the block and the density of the water. To answer part (a), the pressure difference at the top and bottom of the block due to the water column can be calculated using the given atmospheric pressure and depth of the block in the water. For part (b), the scale reading would be the weight of the block minus the buoyant force exerted by the water. Part (c) involves showing that the buoyant force is indeed the difference in forces at the top and bottom of the block, confirming Archimedes' principle.

Answer:

5. 7.452 min

Explanation:

Momentum is conserved, so:

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

0 = (0.695 kg) (-12 m/s) + (59 kg) v

v = 0.1414 m/s

She is 63.2 m away, so the time it takes to reach the shuttle is:

t = 63.2 m / 0.1414 m/s

t = 447.1 s

t = 7.452 min

To find how long it will take for the astronaut to reach the shuttle, the conservation of momentum is used to calculate her velocity after she throws the camera. She will travel at a velocity of 0.141 m/s. It will take her approximately 7.47 minutes to cover the 63.2-meter distance to the shuttle.

The given scenario can be solved using the principle of conservation of momentum. The momentum of the system (the astronaut and the camera) before and after the camera is thrown must be equal since there are no external forces acting on the system. The momentum of the astronaut can be determined using the equation p = m × v, where p is momentum, m is mass, and v is velocity.

To find the velocity of the astronaut after throwing the camera, we use the following relationship:

Solving for the velocity of the astronaut, we get:

0 = (59 kg × velocity of astronaut) + (0.695 kg × (-12 m/s))

Velocity of astronaut = (0.695 kg × 12 m/s) / 59 kg = 0.141 m/s

To calculate how long it will take for the astronaut to reach the shuttle, we divide the distance by her velocity:

Time = Distance / Velocity = 63.2 m / 0.141 m/s = 448.227 s

Convert seconds to minutes:

Time in minutes = 448.227 s / 60 = 7.47 minutes

The closest answer from the options provided is 7.452 min, but rounding should be done judiciously, and the calculated value is between 7.452 min and 7.824 min, so it's important to check the context in which the answer will be used, as the rounding might affect the selection of the correct option.

For the calculation of resistance there are generally two paths. The first is through Ohm's law and the second is through the relationship

[tex]R = \frac{pL}{A}[/tex]

Where

p = Specific resistance of material

L = Length

A = Area

The area of nerve axon is given as

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

[tex]A = \pi (5*10^{-6})^2[/tex]

[tex]A = 7.854*10^{-11}m^2[/tex]

The rest of values are given as

[tex]p= 2 \Omega\cdot m[/tex]

[tex]L = 2cm = 0.02m[/tex]

Therefore the resistance is

[tex]R = \frac{pL}{A}[/tex]

[tex]R = \frac{2*0.02}{7.854*10^{-11}}[/tex]

[tex]R = 509.3*10^6\Omega[/tex]

[tex]R = 509.3M\Omega[/tex]

Answer:

Pnitrogen=3.18 kPa, Poxygen=1.62 kPa , Ptotal= 4.80 kPa

Explanation:

partial pressure equation becomes Ptotal = Pnitrogen + Poxygen

Partial pressure of Nitrogen

Pnitrogen= nRT/V

n=no of moles =mass/molar mass

mass of nitrogen=0.6kg

Molar mass of nitrogen gas=28gmol^-1

n=0.6/28=0.0214moles

R=0.2968 kPa·m3/kg·K

T=300k

V=0.6m^3

Pnitrogen=(0.0214 * 0.2968 * 300)/0.6

Pnitrogen=3.18 kPa

Likewise

Poxygen=nRT/V

n=0.4/32=0.0125moles

R=0.2598 kPa·m3/kg·K

T=300k

V=0.6m^3

Poxygen=(0.0125 * 0.2598 * 300)/0.6

Poxygen=1.62 kPa

Ptotal= 3.18+1.62= 4.80 kPa

Using the ideal gas law, the partial pressure for Nitrogen ([tex]N_{2}[/tex]) and Oxygen ([tex]O_{2}[/tex]) is 148.4 kPa and 86.6 kPa respectively. The total pressure of the mixture is the sum of these two partial pressures, equaling 235 kPa.

The pressure exerted by individual gases in a mixture is known as partial pressure. The calculation of the partial pressure of each gas, and the total pressure of the mixture involves using the ideal gas law. In this case, the ideal gas equation is P = mRT/V, where P represents the pressure, m is the mass of gas, R is the gas constant, T is the temperature and V is the volume. Thus, for Nitrogen ([tex]N_{2}[/tex]), the partial pressure is (0.6 kg × 0.2968 kPa·m3/kg·K × 300K) / 0.6 m3 = 148.4 kPa, and for Oxygen ([tex]O_{2}[/tex]), the partial pressure is (0.4 kg× 0.2598 kPa·m3/kg·K × 300K) / 0.6 m3 = 86.6 kPa. The total pressure, then, is the sum of these partial pressures, which equals 148.4 kPa + 86.6 kPa = 235 kPa.

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Answer:

a) [tex]F_m=1803.013\ N[/tex]

b) [tex]E=53665.84\ MPa[/tex]

c) [tex]E_f=124800\ MPa[/tex]

   [tex]E_m=2200\ MPa[/tex]

Explanation:

Given:

Now, the area of fiber phase:

[tex]A_f=\frac{F_f}{\sigma_f}[/tex]

[tex]A_f=\frac{74000}{156}[/tex]

[tex]A_f=474.359\ mm^2[/tex]

∴Area of matrix phase:

[tex]A_m=A-A_f[/tex]

[tex]A_m=1130-474.359[/tex]

[tex]A_m=655.641\ mm^2[/tex]

(a)

Now the force sustained by the matrix phase:

[tex]F_m=\sigma_m\times A_m[/tex]

[tex]F_m=2.75\times 655.641[/tex]

[tex]F_m=1803.013\ N[/tex]

(b)

Total stress on the composite:

[tex]\sigma=\frac{(F_f+F_m)}{A}[/tex]

[tex]\sigma=\frac{(74000+1803.013)}{1130}[/tex]

[tex]\sigma=67.082\ MPa[/tex]

Now,Modulus of elasticity of the composite:

[tex]E=\frac{\sigma}{\epsilon}[/tex]

[tex]E=\frac{67.082}{1.25\times 10^{-3}}[/tex]

[tex]E=53665.84\ MPa[/tex]

(c)

Since, strain will be same in this case throughout the material.

Now the modulus of elasticity of fiber phase:

[tex]E_f=\frac{\sigma_f}{\epsilon}[/tex]

[tex]E_f=\frac{156}{1.25\times 10^{-3}}[/tex]

[tex]E_f=124800\ MPa[/tex]

Now the modulus of elasticity of matrix phase:

[tex]E_m=\frac{\sigma_m}{\epsilon}[/tex]

[tex]E_m=\frac{2.75}{1.25\times 10^{-3}}[/tex]

[tex]E_m=2200\ MPa[/tex]

The force sustained by the matrix phase is 3125 N, the modulus of elasticity of the composite material in the longitudinal direction is 124800 MPa, and the moduli of elasticity for the fiber and matrix phases cannot be calculated without information about the volume fractions.

To determine the force sustained by the matrix phase, we can use the equation:

Force sustained by matrix phase = Stress in matrix phase * Cross-sectional area

Using given values, the force sustained by the matrix phase is:

Force sustained by matrix phase = 2.75 MPa * 1130 mm2 = 3125 N

To determine the modulus of elasticity of the composite material in the longitudinal direction, we can use the equation:

Modulus of elasticity = Stress / Strain

Using given values, the modulus of elasticity of the composite material in the longitudinal direction is:

Modulus of elasticity = 156 MPa / (1.25 x 10-3) = 124800 MPa

To determine the moduli of elasticity for the fiber and matrix phases, we can multiply the modulus of elasticity of the composite material by the volume fractions of the fiber and matrix phases. In this case, we don't have information about the volume fractions, so we cannot calculate the moduli of elasticity for the fiber and matrix phases.

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Answer:

The rate clock is about

F = 8 GHz

Explanation:

f₁ = 4 G Hz , t₁ = 10 s , t₂ = 6s , f₂ = 1.2 f₁

Can organize to find the rate clock the designer build to the target so

X / 4 Ghz = 10 s , 1.2 X /  Y = 6 s

X * Y = 10 s  ⇒ F = 10 s

1.2 * 4 G Hz = 6 s

F = 10 * ( 1.2 * 4 G Hz ) / 6

F = 10 * ( 1.2 * 4 x 10 ⁹ Hz )  /  6

F = 8 x 10 ⁹ Hz

F = 8 GHz

Final answer:

To run a program in 6 seconds on Machine B, which needs 1.2 times as many clock cycles as Machine A, the target clock rate should be 8 GHz.

Explanation:

The student asked for the target clock rate needed for computer B to run a program in 6 seconds, given that computer A runs it in 10 seconds with a 4 GHz clock and that machine B needs 1.2 times as many clock cycles as machine A for the same program. To solve this, we know that time (T) is equal to the number of cycles (N) divided by the clock rate (C), or T = N / C. For machine A, TA = NA / CA and for machine B, TB = (1.2 * NA) / CB. If machine A completes the program in 10 seconds, the number of cycles it uses is CA * TA, which is 4 GHz * 10 s, yielding 40 billion cycles.

Machine B needs to run these 40 billion cycles in 6 seconds. Also, machine B requires 1.2 times the number of cycles of machine A; thus we have (1.2 * 40 billion) / 6 s to find CB, the clock rate for machine B. This simplifies to 8 GHz. Thus, for machine B to run the program in 6 seconds, the target clock rate should be 8 GHz.

Answer:

(a) 57.17 N

(b) 146.21 N up the ramp

Explanation:

Assume the figure attached

(a)

The angle of ramp, [tex]\theta=tan ^{-1} \frac {2.5}{4.75}=27.75854^{\circ}\approx 27.76^{\circ}[/tex]

[tex]N=(32+48)*9.81*cos 27.76^{\circ}=694.4838 N[/tex]

m=32+48=80 kg

[tex]T+ \mu N= mg sin \theta[/tex]

[tex]T=mg sin \theta - \mu N= mg sin \theta- \mu mg cos \theta= mg (sin \theta - \mu cos \theta) [/tex] where [tex]\mu[/tex] is coefficient of kinetic friction

[tex]T=80*9.81(sin 27.76^{\circ} -0.444 cos 27.76^{\circ})=57.16698 N \approx 57.17 N[/tex]

(b)

Upper box doesn’t accelerate

[tex]F_r= mgsin\theta= 32*9.81sin 27.76^{\circ}=146.2071\approx 146.21 N[/tex]  

The direction will be up the ramp

To lower the boxes down the ramp, a force is needed that is equal to the force of kinetic friction on the lower box. The magnitude and direction of the friction force on the upper box is the same as the force of kinetic friction on the lower box.

To determine the force required to lower the two boxes down the ramp, we can use the equation: force = friction force + weight force. Since the boxes are moving at a constant speed, the force of kinetic friction on the lower box is equal to the force applied to it. Therefore, the force needed to lower the boxes is equal to the force of kinetic friction on the lower box which can be calculated using the equation: force = coefficient of kinetic friction × normal force.

For the magnitude and direction of the friction force on the upper box, we can use the equation: friction force = coefficient of static friction × normal force. Since the boxes are moving together, we can assume that the friction force on the upper box is equal to the force of kinetic friction on the lower box. Therefore, the magnitude and direction of the friction force on the upper box is the same as the force of kinetic friction on the lower box.

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The velocity of the collar at the instant the rod becomes horizontal can be determined using the conservation of angular momentum principle. By calculating the moment of inertia of the disk and equating the initial and final angular momenta, we can solve for the final angular velocity. Since the disk rolls without slipping, the final linear velocity of the disk can be determined using the equation v' = Rω'.

To determine the velocity of the collar at the instant the rod becomes horizontal, we can use conservation of angular momentum. Since the disk rolls without slipping, its angular momentum is conserved. The angular momentum is given by the product of the moment of inertia and the angular velocity. When the rod becomes horizontal, the angular velocity of the disk is equal to the velocity of the collar.

We can calculate the moment of inertia of the disk using the formula I = (1/2)MR^2, where M is the mass of the disk and R is its radius. Substituting the values, we get I = (1/2)(20 lb)(2 ft)^2. We also know that the angular momentum is conserved, so the initial angular momentum is equal to the final angular momentum. The initial momentum is Iω, where ω is the initial angular velocity. The final momentum is Iω', where ω' is the final angular velocity.

Since the disk rolls without slipping, the linear velocity of the disk is equal to the radius times the angular velocity, v = Rω. So, the final linear velocity of the disk is equal to the final angular velocity times the radius, v' = Rω'. Substituting the values, we can solve for ω'.

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Answer:246.92 kJ

Explanation:

Given

Initial Temperature [tex]T_i=30^{\circ}C[/tex]

Final Temperature [tex]T_f=129^{\circ}C[/tex]

no of moles [tex]n=0.3[/tex]

Pressure is constant [tex]P=1 atm [/tex]

As Pressure is constant therefore work done is given by

[tex]W=P\Delta V[/tex]

[tex]P\Delta V[/tex] can also be written as [tex]P\Delta V=nR\Delta T[/tex]

[tex]W=nR\Delta T[/tex]

[tex]W=0.3\times 8.314\times (129-30)[/tex]

[tex]W=246.92 kJ[/tex]                        

Answer:

angular acceleration = 209.44 [rad/s^2]

Explanation:

First we have to convert the velocities which are in revolutions per minute to radians on second.

where:

[tex]w_{0} = 1000 [\frac{rev}{min}]*[\frac{2\pi rad }{1rev}]  * \frac{1min}{60s} = 104.7[\frac{rad}{s} ]\\w = 2000 [\frac{rev}{min}]*[\frac{2\pi rad }{1rev}]  * \frac{1min}{60s} = 209.4[\frac{rad}{s} ][/tex]

Now we can find the angular acceleration:

[tex]w=w_{0} + \alpha *t\\\alpha =\frac{w-w_{0} }{t} \\\alpha =\frac{209.43-104.71}{\0.5 } \\\alpha = 209.44[\frac{rad}{s^{2} } ][/tex]

The angular acceleration of the crankshaft is [tex]\(209.44 \, \text{rad/s}^2\)[/tex].

To find the angular acceleration, we can use the following formula:

[tex]\[ \alpha = \frac{\Delta \omega}{\Delta t} \][/tex]

where:

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

[tex]\(\Delta \omega\)[/tex] is the change in angular velocity

[tex]\(\Delta t\)[/tex] is the change in time

First, we need to convert the angular velocities from revolutions per minute (rpm) to radians per second (rad/s).

1. Initial angular velocity [tex](\(\omega_i\))[/tex]:

[tex]\[ \omega_i = 1000 \, \text{rpm} \][/tex]

2. Final angular velocity [tex](\(\omega_f\))[/tex]:

[tex]\[ \omega_f = 2000 \, \text{rpm} \][/tex]

To convert rpm to rad/s:

[tex]\[ \omega (\text{rad/s}) = \omega (\text{rpm}) \times \frac{2\pi \, \text{rad}}{60 \, \text{s}} \][/tex]

[tex]\[ \omega_i = 1000 \times \frac{2\pi}{60} \, \text{rad/s} = \frac{2000\pi}{60} \, \text{rad/s} \approx 104.72 \, \text{rad/s} \][/tex]

[tex]\[ \omega_f = 2000 \times \frac{2\pi}{60} \, \text{rad/s} = \frac{4000\pi}{60} \, \text{rad/s} \approx 209.44 \, \text{rad/s} \][/tex]

Now, we can find the change in angular velocity:

[tex]\[ \Delta \omega = \omega_f - \omega_i \][/tex]

[tex]\[ \Delta \omega = 209.44 \, \text{rad/s} - 104.72 \, \text{rad/s} = 104.72 \, \text{rad/s} \][/tex]

Given that the change in time [tex](\(\Delta t\))[/tex] is 0.50 s:

[tex]\[ \alpha = \frac{104.72 \, \text{rad/s}}{0.50 \, \text{s}} \][/tex]

[tex]\[ \alpha = 209.44 \, \text{rad/s}^2 \][/tex]

Answer:

Explanation:

Initial kinetic energy of M = 1/2 M vi²

let final velocity be vf

v² = u² + 2a s

vf² =  vi² + 2 (F / M) x D

Kinetic energy

= 1/2 Mvf²

= 1/2 M ( vi² + 2 (F / M) x D

1/2 M vi² + FD

Ratio with initial value

1/2 M  vi² + FD) / 1/2 M  vi²

RK = 1 + FD / 2 M  vi²

To solve this problem we will apply the concepts related to the Kinetic Friction Force for which we define as

[tex]F = \mu_k N[/tex]

Where,

N = Normal Force (Mass for gravity)

[tex]\mu_k =[/tex] Kinetic frictional coefficient

The total force applied is 1.7N and the Force from the (normal) weight is equivalent to five times 1.2N, therefore:

[tex]F = 5(N)\mu_k[/tex]

[tex]1.7 = 5(1.2)(\mu_k)[/tex]

[tex]\mu_k = \frac{1.7}{5*1.2}[/tex]

[tex]\mu_k =0.283[/tex]

Therefore the coefficient of kinetic friction between the wooden blocks and the tabletop is 0.283

Answer:D

Explanation:

Given

mass of object [tex]m=5 kg[/tex]

Distance traveled [tex]h=10 m[/tex]

velocity acquired [tex]v=12 m/s[/tex]

conserving Energy at the moment when object start falling and when it gains 12 m/s velocity

Initial Energy[tex]=mgh=5\times 9.8\times 10=490 J[/tex]

Final Energy[tex]=\frac{1}{2}mv^2+W_{f}[/tex]

[tex]=\frac{1}{2}\cdot 5\cdot 12^2+W_{f}[/tex]

where [tex]W_{f}[/tex] is friction work if any

[tex]490=360+W_{f}[/tex]

[tex]W_{f}=130 J[/tex]

Since Friction is Present therefore it is a case of Open system and net external Force is zero

An open system is a system where exchange of energy and mass is allowed and Friction is acting on the object shows that system is Open .

Answer:

23.7402 m

21.582 m/s

310.521816 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

g = Acceleration due to gravity = 9.81 m/s² = a

Equation of motion

[tex]s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 2.2^2\\\Rightarrow s=23.7402\ m[/tex]

The drop distance is 23.7402 m

[tex]v=u+at\\\Rightarrow v=0+9.81\times 2.2\\\Rightarrow v=21.582\ m/s[/tex]

When they hit the air bags at the bottom of the tower the speed of the experiments is 21.582 m/s

The final speed of the fall will be the initial velocity of stopping

[tex]v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-21.582^2}{2\times 0.75}\\\Rightarrow a=-310.521816\ m/s^2[/tex]

The average stopping acceleration is 310.521816 m/s²

Final answer:

Calculations reveal a drop distance of 23.65 meters, a final velocity of 21.56 m/s, and an average stopping acceleration of 311.43 m/s² for the experimental package in the 2.2-second drop tower scenario.

Explanation:

The question encompasses the calculation of drop distance, final velocity, and average stopping acceleration of an experiment package in a microgravity test environment. To solve these, we apply the equations of motion under uniform acceleration.

This analysis presupposes a vacuum environment in the drop tower to negate air resistance, thereby approximating conditions of free fall and achieving microgravity for the duration of the drop.

According to the statement and the data presented, it is presumed that the variable to look for is force. To solve this problem it is necessary to apply the concepts related to the Force according to the pressure and the Area.

Mathematically the Force can be expressed as

[tex]F = PA[/tex]

Where,

P = Pressure (At this case, the plunger pressure)

A = Cross-sectional Area ( At this case the plunger area), defined for a circle.

Our values are given as,

[tex]r = 0.006m[/tex]

[tex]P = 25mmHg[/tex]

Replacing we have that

[tex]F = (25mmHg \big [\frac{133N/m^2}{1mm\cdot Hg} \big ])(\pi 0.006^2)[/tex]

[tex]F = 0.376N[/tex]

Therefore the minimum force needed on the plunger inorder for a fluid flow into the vein to occur is 0.376