Answer:
Avoidant attachment pattern
Explanation:
The Attachment theory proposes that there are 4 attachment styles:
Secure attachment
Anxious attachment
Avoidant attachment
Anxious-Avoidant attachment
A child who does not seek their mother when they leave and are unbothered by the fact that the mother is not around shows the avoidant attachment pattern. The child is independent and self-directed, thus does not seek intimacy. These types of people are assumed to be fearful or avoidant when it comes to intimacy and commitment, as the theory suggests.
Differentiating it from the other style:
Secure attachment pattern show comfort in intimacy. They easily display interest to other parties, but at the same time, they do not necessarily seek intimacy as they are capable of being alone as well.
Anxious attachment on the other hand, is more like a needy and clingy type of behavior. They seek companionship more desperately and always needs reassurance.
Anxious-avoidant attachment is a more aggressive (in a defensive sense) type. They totally avoid commitment or any form of intimacy. What makes them different is the distrust. They do not trust anyone that tries to get close to them and can sometimes emotionally vent out, (verbally or physically) if anyone attempts to get to know them.
1. A racecar accelerates uniformly from 20.0 m/s to 50.0 m/s in 30 seconds
Determine the acceleration of the car and the distance traveled.
Answer: v² = u² + 2as
2as = v² - u²
a = (v² - u²) / 2s
a = (20.0² m²/s² - 6.00² m²/s²) / (2 * 50.0 m)
a = (400 m²/s² - 36 m²/s²) / (100 m)
a = (364 m²/s²) / (100 m)
a = 3.64 m/s²
Explanation:
Answer:1m/s^-2
Explanation:V=u+at where;
V(final velocity)=50m/s
U(Initial velocity)=20m/s
a(acceleration)=?
t(time)=30seconds
Therefore; 50=20+a(30)
Collect like terms
50-20=30a
30=30a
Divide both side by 30
a=1ms^-2
You are holding a positive charge and there are positive charges of equal magnitude 1 m to your north and 1 m to your east. What is the direction of the force on the charge you are holding?
1. to the east
2. to the southwest
3. to the south
4. to the northeast
Answer:
southwest
Explanation:
Here, all the particles have same charge this means that only repulsive force is acting on the particles according to Coulomb's laws.
One particle is 1 m to my north and another particle is to my east.
The particle to my north will push the positive charge I am holding to the south and the particle to my east will push the positive charge I am holding to the west.
Hence, the resultant direction of the force will be southwest.
A tiny 0.0250 -microgram oil drop containing 15 excess electrons is suspended between to horizontally closely-spaced metal plates that carry equal but opposite charges on their facing surfaces. The plates are both circular with a radius of 6.50cm. (k = 1/4pi(epsilon)0 = 9.0 x 10^9 N x m^2/C^2, e = 1.6 x 10^-19 C)a) How much excess charge must be on each plate to hold the oil drop steady? (b) Which plate must be positive, the upper one or the lower one?
To hold the oil drop steady, the electrical force must match its weight. The voltage required to generate this force can be determined, and hence the charge on each plate, using given constants and the weight of the drop. The upper plate should be positive as the drop has excess electrons.
Explanation:Firstly, we need to find the weight of the oil drop because it will be opposing the electrical force to keep the drop suspended. The mass can be found by converting the micrograms into kilograms (1 microgram = 10^-9 kg). So, the weight (W) of the oil drop will be the mass times gravity. W = 0.025 * 10^-9 *9.8 = 2.45 * 10^-10 N.
The electrical force that will hold the drop steady will exactly equal this weight. Since F = qE where F is the force, q is the charge, and E is the field strength, and E = V/d (where V is potential difference and d is distance), we can put these together to find the voltage that will create this field. V = Fd/q = Wd/q . All that's left is to plug all the given values in: the number of excess electrons in the oil drop times the elementary charge (1.6*10^-19 C), which gives us the total charge on the drop.
As for which plate must be positive, it would be the one opposing the negative charge of the oil drop to suspend it, therefore, the upper plate must be positive.
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A swimmer is determined to cross a river that flows due south with a strong current. Initially, the swimmer is on the west bank desiring to reach a camp directly across the river on the opposite bank. In which direction shoud the swimmer head?
A. The swimmer should swim southeast.
B. The swimmer should swim south.
C. The swimmer shuold swim northeast.
D. The swimmer should swim due north.
E. The swimmer should swim due east.
Answer:
C. The swimmer should swim northeast.
Explanation:
A swimmer expects to reach to the eastern bank of a river by swimming whose stream flows from north to south. Then in order to reach the directly opposite bank of the river when standing on the western bank, the swimmer must swim towards northeast direction so that the resultant of the velocity of flow and the velocity of swimming results in a net velocity towards the east direction.
A jet fighter pilot wishes to accelerate from rest at 5 ggg to reach Mach-3 (three times the speed of sound) as quickly as possible. Experimental tests reveal that he will black out if this acceleration lasts for more than 5.2 ss . Use 331 m/sm/s for the speed of sound
Answer
given,
acceleration, a = 5 g = 5 x 9.8 = 49 m/s²
speed of sound, v = 331 m/s
speed , v = 3 Mach = 3 x 331 m/s = 993 m/s
time, t = 5.2 s
a) Calculating the time period of black out.
time required by the aircraft to black out
using equation of motion
v = u + a t
initial velocity = 0 m/s
993 = 0 + 49 t
t = 20.26 s
the required time to reach 3 Mach speed is more than the given time hence, pilot will black out.
b) now, calculating the maximum speed it can reach in 5.2 s
using equation of motion
v = u + a t
initial velocity = 0 m/s
v = 0 + 49 x 5.2
v = 254.8 m/s
The worked examples of charged-particle motion are relevant to. The worked examples of charged-particle motion are relevant to. a. a transistor. b. a cathode-ray tube. c. magnetic resonance imaging cosmic rays. d. lasers.
Answer:
b. a cathode-ray tube.
the law of conversation of energy and describe the energy transformations that occur as you coast down a long hill on a bicycle and then apply the brakes to make the bike stop at the bottom.
Answer:
Potential energy is transformed into kinetic energy
friction work decreases kinetic energy
Explanation:
The law of conservation of the mechanical energy is the sum of kinetic energy plus the different forms of potential energy, this energy is constant throughout the trajectory if the dissipative force (friction) is zero.
Let us apply this to our case, in the upper part of the trajectory almost all the mechanical energy is potential, and a very small part is kinetic, the bicycle goes very slowly, as it descends without pedaling the speed increases so that the kinetic energy it increases and the height decreases therefore the potential energy decreases, but the sum of the two energies remains constant.
Potential energy is transformed into kinetic energy
When the brakes are applied, a dissipative force enters the system that causes part of the energy to be transformed into heat and part into work of this dissipative force against the wheel, two resulting in a net decrease in mechanical energy and therefore a decrease in the speed of the bicycle, the value of this decrease is given by
W = DK
A gas enters a compressor that provides a pressure ratio (exit pressure to inlet pressure) equal to 8. If a gage indicates the gas pressure at the inlet is 5.5 psig, what is the absolute pressure, in psia, of the gas at the exit?
Answer:
[tex]P_2=160\ psia[/tex]
Explanation:
Given that
Pressure ratio ,r= 8
Gauge pressure at inlet = 5.5 psig
Lets take atmospheric pressure = 14.5 lbf/in²
The absolute presure at inlet
P₁ =5.5 + 14.5 psia
P₁= 20 psia
Lets take absolute pressure at the exit =P₂
[tex]r=\dfrac{P_2}{P_1}[/tex]
[tex]P_2=8\times 20\ psia[/tex]
[tex]P_2=160\ psia[/tex]
Therefore the absolute pressure at the exit will be 160 psia.
If the speed of a car is increased by 80%, by what factor will its minimum braking distance be increased, assuming all else is the same? Ignore the driver's reaction time.
Answer:
[tex]KE_2=3.24 \times KE_1[/tex]
Explanation:
given,
Speed of car = v
new speed of car v'= 1.8 v
Kinetic energy of the car
[tex]KE_1 =\dfrac{1}{2}mv^2[/tex]
new Kinetic energy of energy
[tex]KE_2 =\dfrac{1}{2}mv'^2[/tex]
[tex]KE_2=\dfrac{1}{2}m(1.8v)^2[/tex]
[tex]KE_2=3.24 \times \dfrac{1}{2}mv^2[/tex]
[tex]KE_2=3.24 \times KE_1[/tex]
hence, the kinetic energy is increased by the factor of 3.24.
The momentum of an object is determined to be 7.2 ×× 10-3 kg⋅m/skg⋅m/s. Express this quantity as provided or use any equivalent unit. (Note: 1 kgkg = 1000 gg).
Answer:
Momentum, p = 720 g-cm/s
Explanation:
The momentum of an object is determined to be,
[tex]p=7.2\times 10^{-3}\ kg-m/s[/tex]
We need to express this quantity in any equivalent units. We know that the conversions are as follows :
1 kg = 1000 g
and 1 m = 100 cm
[tex]p=7.2\times 10^{-3}\ kg-m/s=7.2\times 10^{-3}\times (1000\ g)\times (100\ cm)/s[/tex]
p = 720 g-cm/s
So, the momentum of an object in any equivalent unit is 720 g-cm/s. Hence, this is the required solution.
The momentum of the object is 7.2 × 10^-3 kg·m/s, which is equivalent to 7.2 g·m/s when the mass is converted from kilograms to grams.
Explanation:The momentum of an object is defined as the product of its mass and velocity. It is a vector quantity, meaning it has both magnitude and direction. The magnitude of the given object's momentum is 7.2 × 10-3 kg·m/s. To express this quantity in equivalent units, we can convert the mass from kilograms to grams by noting that 1 kg equals 1000 grams. Thus, the momentum in grams·m/s can be calculated as (7.2 × 10-3 kg) × (1000 g/kg) = 7.2 g·m/s.
A 123 kg box is resting on the ground. The coefficient of static friction is 0.34. What force must be applied to the box to start it moving?
Answer:
410.254 N
Explanation:
Force: This can be defined as the product of the mass of a body. The Unit of force is Newton (N)
Deduced from the question,
Force applied to the box to start it moving = Force of friction.
Fₐ = F
Where Fₐ = Force applied to the box to start it moving, F = Force of friction.
But,
F = μR.......................... Equation 1
Where R = normal reaction, μ = coefficient of static friction.
R = W = mg ( on a level surface)
Where m = mass of the box = 123 kg, g = acceleration due to gravity = 9.81 m/s²
R = 123×9.81
R = 1206.63 N.
Also, μ = 0.34
Substituting into equation 1
F = 1206.63×0.34
F = 410.254 N.
Thus the force applied to the box to start it moving = 410.254 N
Aaden has decided to break up with his significant other. He has gone over what he is going to say, but can not decide where to have the conversation. After some serious thought, Aaden decided to have this discussion in his home because it was quiet and they would be able to have a better discussion face-to-face. Which element of the Communication Process Model is this?
Answer:
The correct answer is: b. Channel
Explanation:
The Communication Process Model (CPM) is a model that describes the steps through which effective exchange of information takes place from the sender of a message to the receiver(s) of the message. (Refer to attached image)
The CPM includes the following steps:
1. Noise- reducing physical/ physiological distractions that can impede clear communication.
2. Communicator- the person who initiates the communication
3. Message- the information exchanged between the sender and receiver of the message
4. Feedback- How the information from the sender is received by the recipient of the communication
5. Channel- The medium by which the information is delivered from the sender to the receiver.
In this instance, when Aaden decided to have a discussion with his significant other in his home using face-to-face communication, he is deciding on the channel or medium through which the communication will this place. By choosing face-to-face communication at his home, Aaden has determined what he believes is the most appropriate context for the conversation with his significant other, given the sensitive nature of the topic. In this way, Aaden has selected a channel of communication.
An empty graduated cylinder weighs 55.26 g. When filled with 48.1 mL of an unknown liquid, it weighs 92.39 g. The density of the unknown liquid is?
The density of the unknown liquid can be calculated by finding the mass of the liquid and dividing it by the volume. The mass of the liquid is 37.13 g and the volume is 48.1 mL. Thus, the density is 0.772 g/mL.
Explanation:The density of a substance is calculated by dividing its mass by its volume. Here, the mass of the unknown liquid can be calculated by subtracting the weight of the empty graduated cylinder from the weight of it filled with the liquid. That is, mass = 92.39 g - 55.26 g = 37.13 g. The volume of the liquid is given as 48.1 mL. Therefore, the density can be calculated as follows:
Density = mass/volume
= 37.13 g / 48.1 mL = 0.772 g/mL
This means the density of the unknown liquid is 0.772 g/mL.
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What is the equations to prove the answer is A?
Explanation:
Let's say your mass is M, the ball's mass is m, and the ball's speed is v. Let's say your final speed is V.
Momentum is conserved. If you catch the ball, your speed is:
mv = (m + M)V
V = mv / (m + M)
If you deflect the ball, your speed is:
mv = MV + m(-v)
V = 2mv / M
To compare these, we need the common denominator.
If you catch the ball:
V = mMv / (M (m + M))
If you throw the ball:
V = 2m(m + M)v / (M (m + M))
Comparing the numerators, we can see M < 2(m + M), so catching the ball results in a lower speed.
A rubber ball is dropped from rest from a height h. The ball bounces off the floor and reaches a height of 2h/3. How can we use the principle of the conservation of mechanical energy to interpret this observation?
a) During the collision with the floor, the floor did not push hard enough on the ball for it to reach its original height.
b) Some of the ball’s potential energy was lost in accelerating it toward the floor.
c) The force of the earth’s gravity on the ball prevented it from returning to its original height.
d) Work was done on the ball by the gravitational force that reduced the ball’s kinetic energy.
e) Work was done on the ball by non-conservative forces that resulted in the ball having less total mechanical energy after the bounce.
Please explain how to figure this out
Answer:
Option e.
In this case the work is done on the ball by nonconservative forces that resulted in the ball having less total mechanical energy after the bounce.
Explanation:
This is the type of nonelastic collision when a moving ball hits the ground.Although the conservation of mechanical energy possessed by the ball which is the sum of P.E and K.E., but kinetic energy is not conserved.The non conservative force did the work on the ball that after bouncing lost some of the mechanical energy of that ball. The kinetic energy in the beginning is converted in some other energy like friction and air resistance in this case.A force of 14 N acts on a 5 kg object for 3 seconds. a. What is the object’s change in momentum? b. What is the object’s change in velocity?
Answer:The answer can be calculated by doing thefollowing steps;
Explanation:
A typical neutron star may have a mass equal to that of the Sun but a radius of only 20 km.(a) What is the gravitational acceleration at the surface of such a star?_______m/s2(b) How fast would an object be moving if it fell from rest through a distance of 16 m on such a star? (Assume the star does not rotate.)_______m/s
Answer:
[tex]331665750000\ m/s^2[/tex]
3257806.62409 m/s
Explanation:
G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²
M = Mass of Sun = [tex]1.989\times 10^{30}\ kg[/tex]
r = Radius of Star = 20 km
u = Initial velocity = 0
v = Final velocity
s = Displacement = 16 m
a = Acceleration
Gravitational acceleration is given by
[tex]g=\dfrac{GM}{r^2}\\\Rightarrow g=\dfrac{6.67\times 10^{-11}\times 1.989\times 10^{30}}{20000^2}\\\Rightarrow g=331665750000\ m/s^2[/tex]
The gravitational acceleration at the surface of such a star is [tex]331665750000\ m/s^2[/tex]
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 331665750000\times 16+0^2}\\\Rightarrow v=3257806.62409\ m/s[/tex]
The velocity of the object would be 3257806.62409 m/s
(a) the acceleration due to gravity is 3.3 × 10¹¹ m/s²
(b) the final velocity of the object is 3.26 × 10⁶ m/s
Gravitational force:Given that the neutron star has a mass equal to Sun, M = 2 × 10³⁰ kg
Radius of the star is R = 20 km = 20 × 10³
The weight on the surface of a neutron star that has the same mass as our Sun and a diameter of 21.0 km will be 8.29 × 10¹³ N
(a) The gravitational force on the surface of the neutron star is given by:
F = GMm/R²
mg = GMm/R²
where G is the gravitational constant
M is the mass of the body
m is the mass of the person
and, R is the radius of the body
So,
g = GM/R²
g = (6.67 × 10⁻¹¹)( 2 × 10³⁰)/ (210 × 10³)²
g = 3.3 × 10¹¹ m/s²
(b) According to the third equation of motion:
v² = u² + 2gh
v² = 2gh , since u = 0 as the object is at rest initially
v = [tex]\sqrt{2\times3.3\times10^{11}\times16}[/tex]
v = 3.26 × 10⁶ m/s
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Suppose the fetus's ventricular wall moves back and forth in a pattern approximating simple harmonic motion with an amplitude of 1.7 mm and a frequency of 3.0 Hz. Find the maximum speed of the heart wall (in m/s) during this motion. Be careful of units!
Answer:
The maximum speed of the heart wall during this motion is 0.032 m/s.
Explanation:
Given that,
Amplitude of the simple harmonic motion, A = 1.7 mm = 0.0017 m
Frequency of the fetus's ventricular wall, f = 3 Hz
We need to find the maximum speed of the heart wall during this motion. The maximum speed of the object that is executing SHM is given by :
[tex]v_{max}=A\omega[/tex]
[tex]v_{max}=2\pi f A[/tex]
[tex]v_{max}=2\pi \times 3\times 0.0017[/tex]
[tex]v_{max}=0.032\ m/s[/tex]
So, the maximum speed of the heart wall during this motion is 0.032 m/s. Hence, this is the required solution.
You and your dog go for a walk to the park. On the way, your dog takes many side trips to chase squirrels or examine fire hydrants. When you arrive at the park, do you and your dog have the same displacement?
Answer:
yes. We will have the same displacement.
Explanation:
It's in the definition of a displacement that " displacement is the change in position and if the initial and final position is same then the displacement will be zero". So in the park from starting position if we end up walking at the same position then our displacement will be zero because our initial and final positions are same.
An isotope of Oxygen has 8 protons, 10 neutrons, and 8 electrons. What is the atomic mass of this isotope?
Answer:
18
Explanation:
An isotope an atom of a particular element is another atom having same number of protons or electrons but different number of neutrons.
For example oxygen
element proton neutron
O-16 8 8
O-17 8 9
O-18 8 10
Now atomic mass of atom = no. of protons +no. of neutrons
therefore, mass number of Oxygen has 8 protons, 10 neutrons, and 8 electrons. = 8+10 =18.
Answer:
The atomic mass of this isotope is 18.
Explanation:
Given that,
Number of proton = 8
Number of neutron =10
Number of electron = 8
Atomic mass :
Atomic mass is the addition of number of neutron and number of proton or electron.
We need to calculate the atomic mass of this isotope
Using formula of atomic mass
[tex]A=N+P[/tex]
Where, N = number of neutron
P = number of proton
Put the value into the formula
[tex]A=10+8[/tex]
[tex]A=18[/tex]
Hence, The atomic mass of this isotope is 18.
A piano tuner stretches a steel piano wire with a tension of 800 N. The steel wire is 0.400 m long and has a mass of 3.00 g. What is the frequency of its fundamental mode of vibration?
Answer:
408.25 Hz.
Explanation:
The fundamental frequency of a stretched string is given as
f' = 1/2L√(T/m') .................... Equation 1
Note: The a steel piano wire is a string
Where f' = fundamental frequency of the wire, L = length of the wire, T = tension on the wire, m' = mass per unit length of the wire.
Given: L = 0.4 m, T = 800 N,
Also,
m' = m/L where m = mass of the steel wire = 3.00 g = 3/1000 = 0.003 kg.
L = 0.4 m
m' = 0.003/0.4 = 0.0075 kg/m.
Substituting into equation 1
f' = 1/(2×0.4)[√(800/0.0075)]
f' = 1/0.8[√(106666.67)]
f' = 326.599/0.8
f' = 408.25 Hz.
Hence the frequency of the fundamental mode of vibration = 408.25 Hz.
To find the frequency of the fundamental mode of vibration, use the equation f₁ = vₓ / 2L, where vₓ is the speed of waves in the string and L is the length of the string. The speed of waves can be calculated using vₓ = √(T/μ), where T is the tension in the string and μ is the linear mass density of the string. Plugging in the given values, the frequency of the fundamental mode is 1886.75 Hz.
Explanation:In order to find the frequency of the fundamental mode of vibration of the steel piano wire, we need to use the equation:
f1 = vw/2L
Where f1 is the frequency of the fundamental mode, vw is the speed of waves in the string, and L is the length of the string. The speed of waves in the string can be calculated using the equation:
vw = √(T/μ)
Where T is the tension in the string and μ is the linear mass density of the string.
Plugging in the given values, we have:
T = 800 N, L = 0.400 m, and μ = 3.00 g = 0.00300 kg
Converting the linear mass density to kg/m, we have:
μ = 0.00300 kg / 0.400 m = 0.00750 kg/m
Plugging these values into the equation for the speed of waves, we have:
vw = √(800 N / 0.00750 kg/m) = √106666.67 m/s
Finally, plugging the speed of waves and the length of the string into the equation for the frequency of the fundamental mode, we have:
f1 = (√106666.67 m/s) / (2 * 0.400 m)
Calculating this, we get:
f1 = 1886.75 Hz
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A multiparous client presents to the labor and delivery area in active labor. The initial vaginal examination reveals that the cervix is dilated 4 cm and 100% effaced. Two hours later the client experiences rectal pressure, followed by delivery 5 minutes later. How is this delivery best documented
Answer:
correct answer is Precipitous vaginal delivery
Explanation:
given data
cervix dilated = 4 cm
effaced = 100%
delivery = 5 minutes later
solution
correct answer is Precipitous vaginal delivery because precipitous take delivery time less than = 3 hours
A multipara progress at rate 1.5 cm of dilation per hour
and it is progress for 10 cm for the deliver and birth averages approx 20 minute
so here correct answer is Precipitous vaginal delivery
Explain why incremental development is the most effective approach for developing business software systems. Why is this model less appropriate for real-time systems engineering?
The incremental development is an effective tool for business software applications while it cannot be applied on real-time systems engineering.
Business software technologies are complex. Business software applications are often upgraded with changes in requirments of business goals and procedures.
Real-time systems engineering require a lot of hardware components that are quite difficult to change easily.
Therefore, the incremental development is an effective tool for business software applications while it cannot be applied on real-time systems engineering.
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Incremental development is effective for business software due to its flexibility, reduced risk, and continuous improvement. However, it is less appropriate for real-time systems engineering because these systems require predictable, precise, and immediate responses that iterative changes can undermine.
Incremental development, often aligned with agile methodologies, is effective for business software systems because it emphasizes flexibility and continuous improvement. This approach allows teams to develop software in small, manageable increments, making it easier to adapt to changing business needs and customer feedback. Regular iterations ensure that the project delivers business value continuously, minimizing risks associated with large-scale overhauls.
Several advantages make incremental development suitable for business software:
Flexibility: Allows for adjustments based on ongoing customer feedback and changing market conditions. Reduced Risk: Smaller, manageable increments mean issues can be identified and addressed early.Continuous Improvement: Continuous iterations help in progressively improving functionality and user satisfaction.Enhanced Collaboration: Active customer involvement ensures that the end product aligns well with business needs.Limitations for Real-Time Systems Engineering:
Predictability: Real-time systems need predictable performance and behavior, which is challenging to maintain with iterative changes.Timing Constraints: Immediate, precise reactions are critical in real-time systems, and incremental updates can introduce unwanted delays or inconsistencies.Rigidity: Typically, real-time systems have rigid requirements that do not align well with the adaptable approach of incremental development.You pull your car into your driveway and stop. The drive shaft of your car engine, initially rotating at 2400 rpm, slows with a constant rotational acceleration of magnitude 30 rad/s2.
Complete question is: You pull your car into your driveway and stop. The drive shaft of your car engine, initially rotating at 2400 rpm,slows with a constant rotational acceleration of magnitude 30 rad/s². How long does it take for the drive shaft to stop turning?
Answer:
8.37 s
Explanation:
Initial rotational velocity, ω₀ = 2400 rpm = 2400 × 2π/60 = 251.2 rad/s
Final velocity, ω = 0
rotational deceleration, α =- 30 rad/s²
Use the first equation of rotational motion:
[tex] t = \frac{\omega - \omega_o}{\alpha}[/tex]
Substitute the values:
[tex] t = \frac{0 - 251.2}{-30}=8.37 s[/tex]
A playground toy has four seats, each 6.4kg , attached to very light rods of length r= 1.5m , as seen from below in the figure.
If two children, with masses of 16kg and 23kg , sit in seats opposite one another, what is the moment of inertia about the rotation axis?
Answer:
The moment of inertia about the rotation axis is 145.35 kg.m²
Explanation:
Given:
Mass of each seat = 6.4 kg
length of rods attached to the seat = radius of the seat from center = 1.5m
Moment of inertia = M₁r²
Moment of inertia about the first child:
total mass at that point = mass of the child + mass of the seat
= 16kg + 6.4 kg = 22.4 kg
Moment of inertia at the point where the first child sits = 22.4 *1.5²
= 50.4 kg.m²
Moment of inertia about the second child:
total mass at that point = mass of the child + mass of the seat
= 23kg + 6.4 kg = 29.4 kg
Moment of inertia at the point where the second child sits = 29.4 *1.5²
= 66.15 kg.m²
Moment of inertia about the two empty seats:
Moment of inertia about the two seats = 2(6.4*1.5²)
= 28.8 kg.m²
Moment of inertia about the rotation axis = Moment of inertia about the point of two children + moment of inertia about the two seat
= (50.4 + 66.15 + 28.8) kg.m²
= 145.35 kg.m²
Therefore, the moment of inertia about the rotation axis is 145.35 kg.m²
The moment of inertia about the rotation axis is 145.35 kgm2.
How do you calculate the moment of inertia?Given that the mass m_s of the seat is 6.4 kg, length r of the rod is 1.5 m. The masses of two children is m1 = 16 kg and m2 = 23 kg.
The moment of inertia about the first child is calculated given below.
[tex]MI _1 = (m_s+m_1)r^2[/tex]
[tex]MI_1 = (6.4+16)\times 1.5^2[/tex]
[tex]MI_1 = 50.4 \;\rm kg m^2[/tex]
The moment of inertia about the second child is calculated given below.
[tex]MI _2= (m_s+m_2)r^2[/tex]
[tex]MI_2 = (6.4+23)\times 1.5^2[/tex]
[tex]MI_2 = 66.15 \;\rm kg m^2[/tex]
The moment of inertia about the two empty seats is calculated given below.
[tex]MI_s = 2(m_sr^2)[/tex]
[tex]MI_s = 2 ( 6.4\times 1.5^2)[/tex]
[tex]MI_s = 28.8 \;\rm kgm^2[/tex]
The moment of inertia about the rotation axis is the total sum of the moment of inertia about the point of two children and the moment of inertia about the two seats.
[tex]MI = MI_1 +MI_2+MI_s[/tex]
[tex]MI = 50.4 + 66.15 + 28.8[/tex]
[tex]MI = 145.35 \;\rm kgm^2[/tex]
Therefore, the moment of inertia about the rotation axis is 145.35 kgm2.
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The Chartered Institute of Management Accountants defines activity-based cost management as follows: "An approach to the costing and monitoring of activities, which involves tracing resource consumption and costing final outputs. Resources are assigned to activities, and activities to cost objects based on consumption estimates."
Answer:
ABC is the activity based accounting. It is the costing done for each separate activity which maybe unit level, batch level, product sustaining or facility sustaining.
Explanation:
The basic difference between ABC and traditional costing systems can be explained with the help of the following diagram.
Traditional Costing System
Overhead Cost Accounts (For each individual expense e.g. tax)
First Stage Allocation
Cost Centers ( normally Departments)Cost Centers ( normally Departments)N. Cost Centers ( normally Departments)
Second Stage Allocation ( Direct Labor Or Machine Hours)
Cost Objects ( products, services and customers)
ACTIVITY BASED COSTING SYSTEMS
Overhead Cost Accounts (For each individual expense e.g. tax)
First Stage Allocation ( resource cost drivers)
Activity Cost Centers Activity Cost CentersN. Activity Cost Centers
Second Stage Allocation ( activity cost drivers)
Cost Objects ( products, services and customers) (Direct Costs)
Four steps are involved in the design of ABC systems.
identifying the major activities that take place in an organizationassigning cost to cost pools / cost centers for each activitydetermining the cost for each major activityassigning the cost of activities to products according to the product's demand for activities.The first two steps relate to the first stage and final two steps to the second stage of the two - stage allocation process shown above.
You and a friend work in buildings six equal-length blocks apart, and you plan to meet for lunch. Your friend strolls leisurely at 1.2 m/s, while you like a brisker pace of 1.7 m/s . Knowing this, you pick a restaurant between the two buildings at which you and your friend will arrive at the same instant if both of you leave your respective buildings at the same instant?
Answer:
3.51724 blocks
2.48275 blocks
Explanation:
L = Length of one block
t = Time taken by both person
[tex]Distance=Speed\times Time[/tex]
[tex]1.2t+1.7t=6L\\\Rightarrow 2.9t=6L\\\Rightarrow t=\dfrac{6}{2.9}L[/tex]
Distance that I walk
[tex]s=1.7t\\\Rightarrow s=1.7\times \dfrac{6}{2.9}L\\\Rightarrow s=3.51724L[/tex]
The Distance that I will walk is 3.51724 blocks
Distance my friend will walk
[tex]s=1.2t\\\Rightarrow s=1.2\times \dfrac{6}{2.9}L\\\Rightarrow s=2.48275L[/tex]
Distance my friend will walk is 2.48275 blocks
To determine the restaurant where you and your friend will arrive at the same instant, we can use the equation distance = speed × time.
Explanation:To determine the restaurant where you and your friend will arrive at the same instant, we can use the equation distance = speed × time. Let's assume the distance between the buildings is 6 blocks, which is the same for both of you. We'll call the time it takes for both of you to reach the restaurant 't'. For you, the distance covered will be 6 blocks at a speed of 1.7 m/s, so your equation will be 6 = 1.7t. For your friend, the equation will be 6 = 1.2t.
Solving these equations, we find that t = 3.53 seconds for you and t = 5 seconds for your friend. Since you want to arrive at the same instant, you need to use the higher time, which is 5 seconds. Thus, the restaurant where you and your friend can meet at the same instant is located 5 seconds away from both of your buildings.
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In science, a broad idea that has been repeatedly verified so as to give scientists great confidence that it represents reality is called:____________
In science, a broad idea that has been repeatedly verified so as to give scientists great confidence that it represents reality is called "a theory".
Explanation:
In science, the interpretation of a feature of the organic world that can be tested in repeat manner and analysed by applying agreed tests validation methods, calculation and observation in according to the scientific method, such process is called as a theory in science.
The difference lie between a theory and a hypothesis. Because hypothesis is an "educated guess". Overall it is either a proposed interpretation of an observed phenomenon, or a logical inference of a possible causal association between several phenomena.
It requires 0.30 kJ of work to fully drive a stake into the ground. If the average resistive force on the stake by the ground is 828 N, how long is the stake?
Answer:
Length of the stake will be 0.3623 m
Explanation:
We have given energy required to fully drive a stake into ground = 0.30 KJ = 300 J
Average resistive force acting on the floor is equal to F = 828 N
We have to find the length of the stake
We know that work done is given by
W = Fd, here W is work done , F is average force and d is the length of the stake
So 300 = 828×d
d = 0.3623 m
So length of the stake will be 0.3623 m
The length of the stake is approximately 0.362 meters.
To solve this problem, we can use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy. In this case, the work done to drive the stake into the ground is equal to the potential energy gained by the stake when it is fully driven into the ground.
The work done (W) is given as 0.30 kJ, which is equal to 0.30 * 10^3 J.
The resistive force (F) is given as 828 N.
The distance (d) over which the work is done is the length of the stake.
We can use the formula for work:
[tex]\[W = F \cdot d\][/tex]
Rearranging the formula to solve for \(d\):
[tex]\[d = \frac{W}{F}\][/tex]
Substituting the given values:
[tex]\[d = \frac{0.30 \times 10^3 \, J}{828 \, N}\][/tex]
[tex]\[d = 0.362 \, m\][/tex]
What type of hybridization is exhibited by the nitrogen atom in the following substance pairs are present on the nitrogen?: and how many lone
a. sp hybridization and 2 lone pairs
b. sp hybridization and 1 lone pair
c. sp hybridization and 2 lone pairs
d. sp hybridization and I lone pair
e. sp hybridization and 1 lone pair
f. sp hybridization and 2 lone pair
[tex]sp^3[/tex] hybridization and 1 lone pair is exhibited by the nitrogen atom in the following substance pairs are present on the nitrogen.
b. [tex]sp^3[/tex] hybridization and 1 lone pair
Explanation:
The Nitrogen particle is [tex]sp^3[/tex] hybridized with one crossover orbital involved by the solitary pair. Likewise, nitrogen is [tex]sp^3[/tex] hybridized which implies that it has four [tex]sp^3[/tex] half and half orbitals. The sub-atomic structure of water is predictable with a tetrahedral game plan of two solitary sets and two holding sets of electrons. Two of the [tex]sp^3[/tex] hybridized orbitals cover with s orbitals from hydrogens to frame the two N-H sigma bonds.
Nitrogen utilizes [tex]sp^3[/tex] orbitals to accomplish this geometry. Three of the mixtures are utilized to frame bonds to hydrogen and the fourth contains the solitary pair. Whereas lone pairs are the pairs of electron on atoms that don't take an interest in the holding bonding of two atoms. To distinguish solitary matches in a particle, make sense of the number of valence electrons of the molecule and subtract the number of electrons that have partaken in the holding.
Hybridization of nitrogen depends on its bonding context; sp hybridization leads to a linear geometry, whereas sp³ hybridization results in a trigonal pyramidal geometry with one lone pair, as seen in ammonia.
Explanation:The question is asking about the hybridization of nitrogen in various substances. Hybridization describes the mixing of atomic orbitals to form new hybrid orbitals that can accommodate bonding and lone pairs in molecules. For a nitrogen atom with sp hybridization, the molecule usually has a linear geometry, as in the case of hydrogen cyanide (HCN). Here, nitrogen has one sp hybrid orbital with a lone pair and one with a sigma bond, while the two p orbitals form pi bonds, resulting in a triple bond.
In cases where nitrogen is sp³ hybridized, the nitrogen atom can form three sigma bonds with its sp³ hybrid orbitals and retain one lone pair, giving a trigonal pyramidal geometry. An example of this is in ammonia (NH₃), where the nitrogen atom is bonded to three hydrogen atoms and has one lone pair.