Answer:
average pressure = 8.314 ×[tex]10^{-5}[/tex] pascal
density = 1.66 ×[tex]10^{-12}[/tex] kg/m³
Explanation:
given data
temperature of sun = 6000 K
concentration of atoms = [tex]10^{15}[/tex] particles/m³
to find out
average pressure and density of its atmosphere
solution
first we get here average pressure of atmosphere that is express as
average pressure = [tex]\frac{nRT}{V}[/tex] .............1
put here value we get
average pressure = [tex]\frac{10^15*8.314*6000}{6*10^{23}}[/tex]
average pressure = 8.314 ×[tex]10^{-5}[/tex] pascal
and
density of atmosphere will be
density = [tex]\frac{nM}{V}[/tex] .............2
density = [tex]\frac{10^{15}*1.66*10^{-27}}{1}[/tex]
density = 1.66 ×[tex]10^{-12}[/tex] kg/m³
8. A 40.0kg block of metal is suspended from a scale and is immersed in water. The dimensions of the block are 12.0cm x 10.0cm x 10.0cm. The 12.0cm dimension is vertical, and the top of the block is 5.00cm below the surface of the water. (a) What are the forces exerted by the water on the top and bottom of the block? (Take P0 = 1.0130 X 105 N/m2 .) (b) What is the reading of the spring scale? (c) Show that the buoyant force equals the difference between the forces at the top and bottom of the block.
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.
You are lowering two boxes, one on top of the other, down a ramp by pulling on a rope parallel to the surface of the ramp. Both boxes move together at a constant speed of 15.0 cm/s. The coefficient of kinetic friction between the ramp and the lower box is 0.444, and the coefficient of static friction between the two boxes is 0.800.
(a) What force do you need to exert to accomplish this? (b) What are the magnitude and direc- tion of the friction force on the upper box?
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.
Explanation: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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An inductor with an inductance of 3.50 H and a resistance of 8.00 Ω is connected to the terminals of a battery with an emf of 4.00 V and negligible internal resistance.a)
Just after the circuit is completed, at what rate is the battery supplying electrical energy to the circuit?Express your answer with the appropriate units.b)When the current has reached its final steady-state value, how much energy is stored in the inductor?Express your answer with the appropriate units.c)What is the rate at which electrical energy is being dissipated in the resistance of the inductor?Express your answer with the appropriate units.d)What is the rate at which the battery is supplying electrical energy to the circuit?
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.
Explanation: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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There is a skier at the top of a ski slope. The skier has potential energy. What gives the skier his potential energy? A. his speed B. friction C. gravity D. his efficiency
Answer:C
Explanation:
Skier at the top of a ski has Potential Energy due to gravity.
Potential Energy is the Energy Possessed by an object when it attains a height concerning some zero level Position.
During the process of attaining the height, some work has to be done against gravity and this energy stored within the object after attaining some height w.r.t relative zero position.
The answer is C: gravity
The roads are icy, and you observe a head-on collision on Summit, at the corner with Rhodes: a 1ton car swerves out of his lane and slides through a stop sign at 41 mph straight into a 3 ton SUV traveling at 16mph in the other direction. The car and the SUV crumple from the collision, and stick together.
1. What is the final velocity, in MPH (you don't need to enter MPH in your answer) , of the SUV/car entanglement (the positive direction is the direction the car was initially going)
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.
A continuous and aligned fiber reinforced composite having a cross-sectional area of 1130 mm^2 (1.75 in.^2) is subjected to an external tensile load. If the stresses sustained by the fiber and matrix phases are 156 MPa (22,600 psi) and 2.75 MPa (400 psi), respectively; the force sustained by the fiber phase is 74,000 N (16,600 lbf); and the total longitudinal strain is 1.25 10^ -3, determine: (a)the force sustained by the matrix phase, (b)the modulus of elasticity of the compos- ite material in the longitudinal direction, and (c) the moduli of elasticity for fiber and matrix phases.
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:
cross-sectional area of reinforced composite, [tex]A=1130\ mm^2[/tex]stress sustained by the fiber phase, [tex]\sigma_f=156\ MPa[/tex]force sustained by the fiber phase, [tex]F_f=74000\ N[/tex]Total strain on the composite, [tex]\epsilon=1.25\times 10^{-3}[/tex]stress sustained in the matrix phase, [tex]\sigma_m=2.75\ MPa[/tex]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.
Explanation: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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How many calories are equal to one BTU? (One calorie = 4.186 J, one BTU = 1 054 J.)
a. 0.252b. 3.97c. 252d. 397
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.
Acetylene torches are used for welding. These torches use a mixture of acetylene gas, C2H2, and oxygen gas, O2 to produce the following combustion reaction: 2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g) Imagine that you have a 6.50 L gas tank and a 4.00 L gas tank. You need to fill one tank with oxygen and the other with acetylene to use in conjunction with your welding torch. If you fill the larger tank with oxygen to a pressure of 115 atm , to what pressure should you fill the acetylene tank to ensure that you run out of each gas at the same time? Assume ideal behavior for all gases.
Answer:
Pressure in acetylene gas tank will be 74.8atm
Explanation:
Step 1: Using the ideal gas equation, determine the number of moles of oxygen
[tex]n=\frac{PV}{RT}[/tex]
[tex]P_O=115atm[/tex]
[tex]V_O=6.50L[/tex]
[tex]n_O=\frac{P_OV_O}{RT}[/tex]
[tex]n_O=\frac{115*6.50}{RT}[/tex]
[tex]n_O=\frac{747.5}{RT}[/tex]
As temperature is unknown and assumed to be the same for both gases, and the ideal gas constant will be the same for both cases, these values are left as constants 'T' and 'R'.
Step 2: Determine the proportionate number of moles of acetylene required based on the chemical equation
2 moles of acetylene require 5 moles of oxygen for complete combustion
Thus, 0.4 moles of acetylene are required per mole of oxygen
[tex]n_A=0.4n_O[/tex]
[tex]n_A=0.4*\frac{747.5}{RT}[/tex]
[tex]n_A=\frac{299}{RT}[/tex]
Step 3: Determine the pressure of acetylene tank required in 4.00L tank
[tex]n_A=\frac{P_AV_A}{RT}[/tex]
[tex]\frac{299}{RT}=\frac{P_A*4.00}{RT}[/tex]
[tex]P_A=\frac{299}{4.00}[/tex]
[tex]P_A=74.75 atm[/tex]
Assumptions:
Temperature is the same for both gases and constant
The ideal gas constant is the same for both gases
The combustion reaction is complete and there are no limiting factors
A row of five 1.2 N wooden blocks is being pushed across a tabletop at a constant speed by a toy tractor that exerts a force of 1.7 N on the row. What is the coefficient of kinetic friction between the wooden blocks and the tabletop?
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
what consistent physiological pattern are more common in women's teeth? A. Women's canines are generally sharper. B. Women have straighter teeth. C. Women have 1 less set of molar teeth. D. Women normally have longer pre-molars.
Answer:
B
Explanation:
B. Women have straighter teeth.
A hypodermic needle can be thought of as a large area pipe leading to a small pipe that opens into the vein. The diameter of the plunger is 1.2 cm and the needle diameter is 0.15 cm. If the gauge pressure in the vein is 25 mmHg, what minimum is needed on the plunger in order for a fluid flow into the vein to occur
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
A concrete pump of 120-cy/hr max. output is used to place 250 cy of concrete. It is supplied by 8 cy transit-mix trucks arriving to the site every 15 mins. Barring interruptions how many hours will the operation require, considering 80% efficiency factor.
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.
Explanation: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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A bowling ball of mass 5.8 kg moves in a
straight line at 1.59 m/s.
How fast must a Ping-Pong ball of mass
1.528 g move in a straight line so that the two
balls have the same momentum?
Answer in units of m/s.
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
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 __________________. The (block+bullet) having now acquired a velocity, swings upward to some maximum height. (See lab manual)
The subject of this question is Physics and it pertains to the concept of a ballistic pendulum.
Explanation: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.
First, understand that the bullet embeds itself in the block. This scenario is a completely inelastic collision, also known as a sticky collision.Upon impact, the kinetic energy of the bullet and block system is high, which gets converted into potential energy at the maximum height of the swing.To find the initial speed of the bullet use the conservation of momentum and energy principles: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.
Bandar Industries Berhad of Malaysia manufactures sporting equipment. One of the company’s products, a football helmet for the North American market, requires a special plastic. During the quarter ending June 30, the company manufactured 3,200 helmets, using 2,368 kilograms of plastic. The plastic cost the company $15,629. According to the standard cost card, each helmet should require 0.68 kilograms of plastic, at a cost of $7.00 per kilogram. Required: 1. What is the standard quantity of kilograms of plastic (SQ) that is allowed to make 3,200 helmets? 2. What is the standard materials cost allowed (SQ × SP) to make 3,200 helmets? 3. What is the materials spending variance? 4. What is the materials price variance and the materials quantity variance? (For requirements 3 and 4, indicate the effect of each variance by selecting "F" for favorable, "U" for unfavorable, and "None" for no effect (i.e., zero variance). Input all amounts as positive values. Do not round intermediate calculations.)
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.
Explanation: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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A 100 kg box is hanging from two strings. String #1 pulls up and left, making an angle of 80o with the horizontal on the left, and string #2 pulls up and to the right, making an angle of 65o with the horizontal on the right. FInd the tension in string #2 (on the right side)
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]
Our favorite program runs in 10 seconds on computer A, which has a 4 GHz. clock. We are trying to help a computer designer build a new machine B, that will run this program in 6 seconds. The designer can use new (or perhaps more expensive) technology to substantially increase the clock rate, but has informed us that this increase will affect the rest of the CPU design, causing machine B to require 1.2 times as many clock cycles as machine A for the same program. What clock rate should we tell the designer to target?
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.
The system consists of a 20-lb disk A, 4-lb slender rod BC, and a 1-lb smooth collar C. If the disk rolls without slipping, determine the velocity of the collar at the instant the rod becomes horizontal. The system is released from rest when != 45°.
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'.
Explanation: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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We had a homework problem in which the Arrhenius equation was applied to the blinking of fireflies. Several other natural phenomena also obey that equation, including the temperature dependent chirping of crickets. A particular species, the snowy tree cricket, has been widely studied. These crickets chirp at a rate of 178 times per minute at 25.0°C, and the activation energy for the chirping process is 53.9 kJ/mol. What is the temperature if the crickets chirp at a rate of 126 times per minute?
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
Explain in detail why it may take longer for you to fight off the flu virus the first time this season than it would if you caught the same strain a second time.
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.
A(n) 59 kg astronaut becomes separated from
the shuttle, while on a spacewalk. She finds
herself 63.2 m away from the shuttle and moving with zero speed relative to the shuttle. She
has a(n) 0.695 kg camera in her hand and decides to get back to the shuttle by throwing
the camera at a speed of 12 m/s in the direction away from the shuttle.
How long will it take for her to reach the
shuttle? Answer in minutes.
1. 10.06 min
2. 9.687 min
3. 8.569 min
4. 8.197 min
5. 7.452 min
6. 7.824 min
7. 10.43 min
8. 8.942 min
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.
Explanation: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.
There was an accident at a uranium processing plant in Tokai, Japan, on September 30, 1999. Using your favorite search engine, find out what happened and compare this accident with the ones at Chernobyl and Three Mile Island. How could these accidents have been avoided? Do these incidents prove that nuclear power plants should be phased out? Why or why not?
For the three accidents mentioned there were human factors that caused the destabilization of the reactors and generated catastrophies.
In the case of Tokai, there was an excess of Uranium due to the fact that workers (Not qualified, since such work did not merit it) added to the containers, which generated an excess in the filling generating an emulsion of radiation to all the personnel.
In the case of Chernobyl, it was the experimentation through a series of tests to reduce the power, during which a series of imbalances occurred in the reactor 4 of this nuclear power plant, which led to the uncontrolled overheating of the reactor core nuclear-
In the case of Three Mile Island it was a design error in which the water level was underestimated, believing that the required level was available but in the end it was noted that said water level was not sufficient which caused the melting of a water dipstick.
All this leaves us with reflections on the security protocols followed for the construction or management of these nuclear power plants, and calls into question the human capacity to react to one of these catastrophes. Today, these plants not only put human health at risk but also generate waste that pollutes the planet. Alternatives such as renewable and clean energy already become more popular every day and are about to leave nuclear energy in the past to give rise to a new human stage.
A 5 kg object near Earth's surface is released from rest such that it falls a distance of 10 m. After the object falls 10 m, it has a speed of 12 m/s. Which of the following correctly identifies whether the object-Earth system is open or closed and describes the net external force?
A. The system is closed, and the net external force is zero.
B. The system is open, and the net external force is zero.
C. The system is closed, and the net external force is nonzero.
D. The system is open, and the net external force is nonzero.
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 .
A particle of mass M moves along a straight line with initial speed vi. A force of magnitude F pushes the particle a distance D along the direction of its motion.
By what multiplicative factor RK does the initial kinetic energy increase, and by what multiplicative factor RWdoes the work done by the force increase (with respect to the case when the particle had a mass M)?
If one of the quantities doubles, for instance, it would increase by a factor of 2. If a quantity stays the same, then the multiplicative factor would be 1.
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
A circuit with current increasing at a rate of 4 A/s contains an inductor, L. If the induced emf is -2 V, what is the inductance of the inductor?
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
Which one of the following statements is not a characteristic of a plane mirror?
A) The image is the same size as the object.
B) The image is always upright.
C) The image is real.
D) The image is reversed left to right compared to the object.
E) The image is the same distance behind the mirror as the object is in front of the mirror.
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:
A real image is that picture that is created when the light beams originating from an object meet each other after reflection or refraction.A real image can be captured on the screen. The real image is constantly inverted. A common example of a real image is the image formed on the cinema screen.Thus we can say, real image is not a characteristic of a plane mirror.
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the crankshaft of an engine increases its spin from 1,000 to 2,000 rpm in 0.50 s what is its angular acceleration
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]
A cylinder contains 0.300 mol of carbon dioxide (CO2) gas at a temperature of 30.0 ∘C. The cylinder is provided with a frictionless piston, which maintains a constant pressure of 1.00 atm on the gas. The cylinder is placed on a hot plate and a 920 J of heat flows into the gas, thereby raising its temperature to 129 ∘C. Assume that the CO2 may be treated as an ideal gas. How much work is done by the gas in this process?
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]
The driver of a pickup truck accelerates from rest to a speed of 37 mi/hr over a horizontal distance of 215 ft with constant acceleration. The truck is hauling an empty 460-lb trailer with a uniform 72-lb gate hinged at O and held in the slightly tilted position by two pegs, one on each side of the trailer frame at A. Determine the maximum shearing force developed in each of the two pegs during the acceleration.
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)
NASA operates a 2.2-second drop tower at the Glenn Research Centre in Cleveland, Ohio. At this facility, experimental packages are dropped from the top of the tower, on the 8th floor of the building. During their 2.2 seconds of free fall, experiments experience a microgravity environment similar to that of a spacecraft in orbit,
(a) What is the drop distance of a 2.2-s tower?
(b) How fast are the experiments travelling when they hit the air bags at the bottom of the tower?
(c) If the experimental package comes to rest over a distance of 0.75 m upon hitting the air bags, what is the average stopping acceleration?
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.
To find the drop distance (d) for a 2.2-second fall, we use the equation d = 0.5 * g * t, where g is the acceleration due to gravity (9.8 m/s²) and t is time in seconds. This gives us d = 0.5 * 9.8 * 2.22 = 23.65 meters.To calculate the final velocity (v) when the experiments hit the air bags, we employ v = g * t, resulting in v = 9.8 * 2.2 = 21.56 m/s.The average stopping acceleration (a) can be derived from the equation v² = 2 * a * d, rearranged to a = v² / (2*d), where d is the stopping distance of 0.75 m, leading to a = (21.562) / (2*0.75) = 311.43 m/s².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.