D.
The gravitational of a body is possessed by the body due to the virtue of its position.
The formula for gravitational potential energy is,
P.E = mgh joules
Substituting the values
196*9.8*250= 49k
a car is traveling along a straight road at a velocity of 30m/s when its engine cuts off. For the next ten seconds, the car slows down, and its average acceleration is a1. For the next five seconds, the car slows down further at a velocity of 24m/s, and its average acceleration is a2. The ratio of the average acceleration values is a1/a2=1.5. Find the velocity of the car at the end of the initial ten-second interval.
Answer:
See below....the question is unclear....pick the answer you think they want
Explanation:
Car goes from 30 m/s to 24 m/s in 10 seconds
acceleration = change in velocity/change in time
= -6 m/s / 10 s = - 3/5 m/s^2
The wording of the question is not very good.....perhaps they meant to say that after the 5 seconds the velocity is now 24 m/s ?
a1 *10 + a2 *5 = -6 m/s
( 6 m/s is the change in velocity from 30 to 24 m/s)
a1/a2 = 1.5 so a2 = a1/1.5 (substitute this in)
10 a1 + 5 ( a1/1.5) =-6
15 a1 + 5 a1 = -9
a1 = - .45 m/s^2
The velocity of the car at the end of the initial ten-second interval, given that the car further slows down for the next five seconds, is 25.5 m/s
How to determine the velocity of the car at the end of the 10 s interval?From the first statement, we have:
Initial velocity (u₁) = 30 m/sTime (t₁) = 10 secondsVelocity at the end of 10 s (v₁) = ?v₁ = u₁ + a₁t₁
v₁ = 30 + (a₁ × 10)
v₁ = 30 + 10a₁ ......(1)
From the second statement, we have:
Time (t₂) = 5 sFinal velocity (v₂) = 24 m/sRation of acceleration (a₁/a₂) = 1.5Initial velocity (u₂) = Velocity at the end of 10 sv₂ = u₂ + a₂t₂
24 = u₂ + (a₂ × 5)
24 = u₂ + 5a₂ .......(2)
But,
u₂ = v₁ = 30 + 10a₁
Thus, we have:
24 = 30 + 10a₁ + 5a₂
But,
a₁/a₂ = 1.5
a₁ = 1.5a₂ ..... (3)
Thus, we have:
24 = 30 + 10a₁ + 5a₂
24 = 30 + 10(1.5a₂) + 5a₂
24 = 30 + 15a₂ + 5a₂
24 = 30 + 20a₂
Collect like terms
24 - 30 = 20a₂
-6 = 20a₂
Divide both sides by 20
a₂ = -6 / 20
= -0.3 m/s²
Substituting the value of a₂ into equation 3, we have
a₁ = 1.5a₂
= 1.5 × -0.3
= -0.45 m/s²
Substitute the value of a₁ into equation 1 to obtain the velocity, v₁ at the end of the initial to 10 s
v₁ = 30 + 10a₁
= 30 + (10 × -0.45)
= 30 - 4.5
= 25.5 m/s
Thus, velocity of the car at the end of the initial ten-second interval is 25.5 m/s
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In a hydraulic lift whose input line has a cross-sectional area of 1.00 cm² and whose output line has
a cross-sectional area of 22.0 cm², what is the largest mass (kg) that can be lifted by an input force
of 200. N?
By using Newton's Second law of motion, the largest mass that can be lifted by this input force is equal to 2,244 kg.
How to calculate the largest mass?Based on the information provided, we can logically deduce that the output to input force ratio is equal to 22:1. This ultimately implies that, the input force is equal to 1000 N, which would result in an output force of 22,000 N.
Next, we would calculate the largest mass by using Newton's Second law of motion:
Mass = Force/Acceleration
Mass = 22,000/9.8
Mass = 2,244 kg.
Note: Acceleration due to gravity is equal to 9.8 m/s².
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the 3kg object in figure is released from rest at height of on a curved frictionless ramp. at the foot of the ramp is a spring of force constant 400N/m. the object slides down the ramp and into the spring,compressing it a distance x before coming momentarily to rest.
a) find x
b) describe the motion of the object (if any) after the block momentarily comes to rest?
(a) The compression of the spring is 7.4 cm.
(b) When the object comes to rest its potential energy from the given height will be converted into elastic potential energy of the spring.
Compression of the spring [ x ]
The compression of the spring is calculated by applying Hooke's law as follows;
F = kx
x = F/k
x = (mg)/k
x = (3 x 9.8)/400
x = 0.074 m
x = 7.4 cm
When the object comes to rest its potential energy from the given height will be converted into elastic potential energy of the spring.
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the temprerture of an object is 40°c.what is this tempreture in farhenhite scale?
This is an exercise of thermal scales.
Data:T = 40 °C
T = °F ¿?
To convert degrees Celsius to degrees Fahrenheit, we have the formula:
[tex]\bf{^{\circ}F=\dfrac{9}{5} \ ^{\circ}C+32 \ \ \to \ \ \ Formula }[/tex]
We substitute our data into the formula.
[tex]\bf{^{\circ}F=\dfrac{9}{5} \times 40+32}[/tex]
First we divide 9/5 multiplied by 40.
[tex]\bf{=75+32}[/tex]
Add 75 plus 32
[tex]\bf{=104 \ ^{\circ}F}[/tex]
Therefore, the temperature change of 40 °C on the °F scale is equal to 104.
[tex]\huge \red{\boxed{\green{\boxed{\boldsymbol{\purple{Piscis04}}}}}}[/tex]
When a cold alcohol thermometer is placed in a hot liquid, the column of alcohol goes down slightly before going up. Explain why.
Answer:
See below
Explanation:
When a cold alcohol thermometer is placed in a hot liquid, the following takes place.The glass of thermometer expands because of the heat so the column of alcohol goes down.After some time, the heat travels to alcohol and then alcohol expands.So, Reason why alcohol goes down first:Alcohol first goes slightly down because of the expansion of the glass of thermometer before doing up.
[tex]\rule[225]{225}{2}[/tex]
During World War II, it was found that r, the radius of the shockwave produced during an atomic bomb explosion, depends on the energy released, W, the elapsed time, t, and the air density, .
The atomic bomb explosion, depends on the energy released, W, the elapsed time, t, and the air density which altogether lead to the vaporisation of the contents within the mushroom cloud created.
What is an atomic bomb?An atomic bomb is type of nuclear weapon that releases a vast amount of energy upon explosion in the form of fission reaction.
During an explosion of atomic bomb, a cloud of mushroom fire is formed which vaporises anything found within it. The extent of destruction depends on:
energy released, W, the elapsed time, t, and the air density.Therefore, the atomic bomb explosion, depends on the energy released, W, the elapsed time, t, and the air density which altogether lead to the vaporisation of the contents within the mushroom cloud created.
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Metals are used in many products because of the characteristic properties that most metals have.
Which product
requires the high luster of metals in order to work?
O mirrors
O pots and pans
O food wrap
O circuit boards
a) Mirrors
Mirror requires the high luster of metals in order to work.
What causes the highest luster of a metal?When light shine on to the surface of a metal, its electrons absorb small amounts of energy and become excited into one of its many empty orbitals.
The electrons immediately fall back down to lower energy levels and emit light.
This process is responsible for the high luster of metals.
Here,
Mirrors have metal coating on its back side, because of high reflective property , high lustrous mirror is preferred . When metal is more lustrous, the less light reflect into it and bounces back.
Coating one side of a piece of glass with shiny metals can turn it into a mirror, reflecting light coming toward it.
Window glass can reflect only eight percent of light hitting it, while mirrors can reflect 95 percent of light hitting them.
The glass in a mirror is usually coated with a layer of silver or aluminum.
Mirrors are shiny because when photons (rays of light) coming from an object , it strike the smooth surface of a mirror, then they bounce back at the same angle. Our eyes see these reflected photons as a mirror image.
Therefore,
Mirror requires the high luster of metals in order to work.
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Answer:A) mirrors
Explanation:
Aude slipped on ice going 7.6 m/s .What was her velocity before she hit the wal?
Answer:
Define universal gravitational constant. Why is newton's law of gravitation called universal law?
A hydrogen bond forms by the electrostatic interaction of opposite charges in two molecules. If
the bond length is 2x10 -10 m and the magnitude of the charges involved is
approximately 1.60x10 -20C, what is the force between the molecules involved in the bond?
The force between the molecules involved in the bond is 6. 426 *10^-11 Newton
How to determine the force
Using the formula:
F = K[q1 x q2]/D^2
where K is coulombs constant =9 *10 ^9 Nm^2/C^2.
q1 and q2 = charges = 1.60x10 -20C
d = distance between the charges = 2x10 -10 m
Substitute the values into the formula
F = [tex]9 * 10^9\frac{ 1.60*10^ -20 * 1.60 *10^ -20}{2x10^ -10^{2} }[/tex]
F = [tex]9 *10^9\frac{2. 856* 10^-40}{4* 10^-20}[/tex]
F = [tex]9* 10^9 * 7. 14* 10^-21[/tex]
F = [tex]6. 426 * 10^-11[/tex] Newton
Thus, the force between the molecules involved in the bond is 6. 426 *10^-11 Newton
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A skier is standing on a horizontal patch of snow. She is holding onto a horizontal tow rope, which is about to pull her forward. The skier’s mass is 59 kg, and the coefficient of static friction between the skies and the snow is 0.14. What is the magnitude of the maximum force that the tow rope can apply to the skier without causing her to move?
The magnitude of the maximum force that the tow rope can apply to the skier without causing her to move will be 81.03 N,
What is the friction force?It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).
It is defined as the product of the coefficient of friction and normal reaction.
On resolving the given force and acceleration. Mathematically aion in the different components and balancing the equation gets. Components in the x-direction.
The normal force is balanced by weight;
N = mg
The magnitude of the maximum force that the tow rope can apply to the skier without causing her to move is;
F = μN
F= μmg
F=0.14 × 59 kg × 9.81 m/s²
F = 81.03 N
Hence, the magnitude of the maximum force will be 81.03 N,
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what is the radius of the 5th orbital in hydrogen?
Answer: 1.32 x 10^-9 m
Explanation:
The formula for the radius of nth orbital is:
[tex]r_{n}=\frac{n^{2}}{Z} \times 0.528 \text { Angstorms }[/tex]
For hydrogen, Z= 1, so we have:
[tex]$$\begin{aligned}r_{n} &=5^{2} \times 0.528 \text { Angstorm } \\&=13.2 \text { Angstorms } \\&=13.2 \times 10^{-10} \mathrm{~m} \\&=1.32 \times 10^{-9} \mathrm{~m}\end{aligned}$$[/tex]
Answer: 1.32*10^-9
Explanation: I had the same question and here's what I got:
rn = n2(r1)
n = 5
r = 5.29*10^-11
rn = n^2(r1)
= 5^2(5.29*10^-11)
r = 1.32*10^-9
Lesson 1, What is Physical Science.
Write an expository essay explain the difference between scientific knowledge and other kinds of science, include ideas regarding:
1) the scientific method.
2) true science versus false science.
3) examples of what scientists study and do not study.
4) what is required for a scientist to consider an idea to be true.
Physical science explains the physical world and not supernatural events and an idea is considered true by science if it can be observed, explained and reproduced by others.
What is physical science?
Physical science is a systematic study of the of the physical world and phenomena based on observation and experiments.
The scientific method involves:
observation and questionshypothesisexperimentsresults and further experimentstheories and lawsScience explains the physical world and not supernatural events.
Science studies physical phenomena such as the stars, the planets, living things, etc.
Science do not study about angels or demons or supernatural events.
An idea is considered true by science if it can be observed, explained and reproduced by others.
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5. A uniform solid ring of mass M = 10 kg and radius R = 2 m is initially rotating in mid-air with angular speed 20 rad/s [the Before state in figure (I) below]. It is then placed in contact with a surface (assume that the surface immediately feels the full weight of the ring). (1) M Before Wi (II) wf V Ө ə Ө After Ө wf H ? (a) What kind of frictional force acts on the ring upon contact with the surface? (b) If the ring begins to roll smoothly to the right with speed v = 10 m/s after 5 seconds of contact (the After state), find the coefficient of friction corresponding to the frictional force you mentioned in (a). (c) What is the increase in the thermal energy of the ring/surface system from right before the ring contacts the surface, to when the ring starts to roll smoothly? The ring then rolls smoothly up a ramp of 0 = π/6 rad and H = 5 m [see figure (II)] (d) What is the horizontal distance, from the end of the ramp, at which the ring lands?
(a) The kind of frictional force acts on the ring upon contact with the surface is kinetic friction before rolling and afterwards it is static friction.
(b) If the ring begins to roll smoothly to the right with speed v = 10 m/s after 5 seconds of contact (the After state), the coefficient of friction corresponding to the frictional force is 0.612
(c) The increase in the thermal energy of the ring/surface system from right before the ring contacts the surface, to when the ring starts to roll smoothly is 7500 J.
(d) The horizontal distance, from the end of the ramp, at which the ring lands is 7.78 m
What is frictional force?When an object is moving on a rough surface, it experiences opposition. This opposing force is called the friction force.
Given is a uniform solid ring of mass M = 10 kg and radius R = 2 m is initially rotating in mid-air with angular speed 20 rad/s [the Before state in figure (I) below]. It is then placed in contact with a surface (assume that the surface immediately feels the full weight of the ring).
(a) The kind of frictional force acts on the ring upon contact with the surface is kinetic friction before rolling and afterwards it is static friction.
(b) angular frequency = velocity / radius
ωf = 10/2 = 5 rad/s and ωi = 20 rad/s
Angular acceleration, α = ωf - ωi /t
Put the values, we get
α = -15/5 = -3 rad/s²
Coefficient of friction, μ = a/g = rα/g
Plug the values, we get
μ = 0.612
Thus, the coefficient of friction corresponding to the frictional force is 0.612.
(c) The energy lost = heat generated
energy lost = 1/2 Iω² + 1/2 Mv²
energy lost = 1/2 MR²ω² + 1/2 Mv²
Plug the values, we get
energy lost = 7500 J
Thus, the increase in the thermal energy is 7500 J.
(d) The horizontal distance, from the end of the ramp, at which the ring lands
s= (v- 2gH) sin2θ /g
s = (10 - 2x (-9.81)x5 ) sin (2x π/6) / 9.81
s = 7.78m
Thus, the horizontal distance is 7.78 m
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What is the monthly (30 days) energy cost (in
$) of leaving a 40-W bulb on continuously if
electricity costs 21¢ per kWh? Round final
answer to the nearest cent.
Explanation:
Power= 40 W
cost=21¢/kWh
t=1 month=30 days = 720h
Energy= power*time
E= 40W* 720h = 28800kWh
Cost= Energy*rate
cost= 28800kWh*21¢/kWh
cost=604800¢
11. Trait theory claims that
O A. people from the same locations
share the same personality type.
B. you always behave the way your
personality type says you will.
O C. your personality is made up of a
number of traits.
O D. you have one characteristic that
defines your entire personality.
A model rocket with a mass of 5 kg has a downward thrust of 100 newtons. Assuming there are no other forces acting on the rocket, what is the rocket’s acceleration upward?
a. 20 m/s2
b. 100 m/s2
c. 5 m/s2
d. 105 m/s2
Answer:
A
Explanation:
F = m * a
100 = 5 * a
a = 100/5 = 20 m/s^2
How many seconds are in 28 hours?
Answer:
1680 seconds
Explanation:
[tex]28 hrs * \frac{60 s}{1 hr} =1680s[/tex]
The position of given object is s(t)= t^5 + cos 4t
Find the velocity and acceleration functions
Answer:
v(t)=s'(t)=5t⁴-sin(4).4, for velocity function at time t
a(t)=v'(t)=20t³, for acceleration function at time t
A cup of water is warmed from 21 °C to 85 °C. What is the difference between these two temperatures, in kelvins?
Answer:
337k
Explanation:
First, let us find the difference between the given two temperatures.
Difference = 85°C - 21°C
= 64°C
And now we have to write the temperature in kelvins.
To convert Celcius to Kelvins you can add 273 to the temperature in Celcius.
Let us find it now.
64°C + 273 = 337k
Therefore,
64°C ⇒ 337k
A local municipality placed an advert in the local newspaper for a competition of local high schools to participate to build a security system in the community hall to avoid vandalism and break ins
We will use the design process to meet the requirements of the competition by the given explained in the main answer.
What is the competition?The initiative was taken by two or more parties operating independently to win the business of a third party by competing to hire contractors on the best terms possible.
The complete question is
"A local businessperson placed an advert in the local newspaper for a competition for girls only from different schools to participate in the competition. The competition is about drawing ideas for a shopping complex for the local community. Girls from the different schools are excited about the competition. you are a group of boys in the local schools who saw the advert and feel it is unfair for you to be excluded from the competition.
consider yourself as one of the learners in the local schools.
Explain why the project should be given to boys also who have better skills for this project than the girls do.
1.2 think about the design process skills, investigates, design, make, evaluate and communicate."
Explain how you will use the design process to meet the requirements of the competition.
[ 1.1 ]Boys should also be given the project because some of them have more artistic talent than girls do when it comes to sketching designs for local shopping centers.
Boys can also present a clear picture of how local communities can expand their shopping options.
For instance, they can encourage local merchants to offer seasonal items because doing so will benefit their businesses.
They can also suggest that retailers encourage customers to bring a friend to participate in competitions, which will be highly regarded.
[1.2] In order to fulfill the needs of the competition, I would make sure to explore how a shopping center for the neighborhood functions, for instance, by making sure I am aware of the strategies one may use to enhance his or her firm in the neighborhood.
In order to win the competition, I would make sure that I was confident in what I was saying and that I was audible enough to be heard during the competition.
Finally, I would make sure I met the competition's conditions by staying on-topic and solely addressing the issues raised in the advertisement
Hence will use the design process to meet the requirements of the competition.
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a generator uses loops of area 0.239 m^2, rotating 373 rad/s in a 0.0639 t magnetic field. how many loops must the coil have to generate a maximum emf of 2500 v?
The loops must the coil have to generate a maximum emf of 2500 will be 439.
What is the faraday law of electromagnetic induction?According to Faraday's law of electromagnetic induction, the rate of change of magnetic flux linked with the coil is responsible for generating emf in the coil resulting in the flow of amount of current.
Given data;
Area,A = 0.239 m²
Angular velocity,ω=373 rad/sec
Magnetic field,B=0.0639 T
Maximum emf,E= 2500V
The formula for the maximum induced voltage is;
E{max} = N × B × A × ω
2500 = N × 0.639 × 0.0239 × 373
N = 438.66
N = 439 \ turns
Hence, loops must the coil have to generate a maximum emf of 2500 will be 439.
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Answer: 11392.99
Explanation: I had the same problem and here are the steps:
Values:
N (Number of loops) = 2000
A (area) = 0.239
ω (angular velocity) = 373
B (magnitude) = 0.0639
ε = NABω
= (2000)(0.239)(373)(0.0639)
ε = 11392.99 (FINAL ANSWER)
A small sphere of mass 10 kg
is released from rest at a height of
15.0 m above the ground level.
The sphere experiences a constant
resistive force (due to air
resistance) of magnitude R = 10.0
N.
a) Calculate the speed of the
sphere after it has fallen
through a distance of 5.00 m
bCalculate the speed of the ball just before a it hits the gound.
Answer:
Approximately [tex]9.39 \; {\rm m\cdot s^{-1}}[/tex] after the sphere has travelled a distance of [tex]5\; {\rm m}[/tex].
Approximately [tex]16.3\; {\rm m\cdot s^{-1}}[/tex] right before touching the ground (a distance of [tex]15\; {\rm m}[/tex].)
Assumption: [tex]g = 9.81\; {\rm N\cdot kg^{-1}}[/tex].
Explanation:
Weight of the sphere: [tex]m\, g = 9.81\; {\rm N \cdot kg^{-1}} \times 10\; {\rm kg} = 98.1\; {\rm N}[/tex], downwards.
Drag on the sphere: [tex]10.0\; {\rm N}[/tex] upwards.
Net force on the sphere: [tex]98.1\; {\rm N} - 10\; {\rm N} = 88.1\; {\rm N}[/tex] downwards.
Acceleration of the sphere: [tex]a = F_\text{net} / m = 88.1\; {\rm N} / (10\; {\rm kg}) = 8.81\; {\rm m\cdot s^{-2}}[/tex].
Apply the SUVAT equation [tex]v^{2} - u^{2} = 2\, a\, x[/tex], where [tex]v[/tex] is the final velocity, [tex]u[/tex] is the initial velocity ([tex]0[/tex] in this case, as the sphere was released from rest,) and [tex]x[/tex] is the distance (displacement) that the sphere has travelled so far.
Rearrange this equation to obtain an expression for [tex]v[/tex]:
[tex]\displaystyle v = \sqrt{2\, a\, x + u^{2}}[/tex].
For example, after the ball travelled a distance of [tex]5.00\; {\rm m}[/tex], [tex]x = 5.00 \; {\rm m}[/tex]:
[tex]\begin{aligned} v &= \sqrt{2\, a\, x + u^{2}} \\ &= \sqrt{2 \times 8.81\; {\rm m\cdot s^{-2}} \times 5.0\; {\rm m} + 0} \\ &\approx 9.39\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
Similarly, [tex]x = 15.0\; {\rm m}[/tex] right before landing, such that:
[tex]\begin{aligned} v &= \sqrt{2\, a\, x + u^{2}} \\ &= \sqrt{2 \times 8.81\; {\rm m\cdot s^{-2}} \times 15.0\; {\rm m} + 0} \\ &\approx 16.3\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
The following equation shows the position of a particle in time t, x=at2i + btj where t is in second and x is in meter. A=2m/s2, b=1m/s. Find
A, the average velocity of the particle in the time interval t₁=2sec and t₂=3sec
B, the velocity and acceleration at any time t.
C, the average acceleration in the time interval given in part (a)
will give brilliance
A. The average velocity of the particle in the time interval t₁=2sec and t₂=3sec is 4 i m/s.
B. The velocity and acceleration at any time t is 4t i +1 j and 4i.
C. The average acceleration in the time interval given in part (a) is 4
What is acceleration?Acceleration can be defined as the rate change of velocity with time.
acceleration a = (Δv) / (Δt)
Given the following equation shows the position of a particle in time t, x=at² i + bt j
where t is in second and x is in meter. a=2m/s², b=1m/s.
A. Velocity is time rate of change of displacement.
V = dx/dt
V = d/dt (2t² i + 1t j)
V = 4t i +1 j
Velocity in time interval,
At t₁=2sec , V₂= 4x2 i + 1 j = 8 i + 1 j
At t₂=3sec, V₃ = 4x3 i + 1 j = 12 i + 1 j
Average velocity in time interval t₁=2sec and t₂=3sec is
, V₃ - V₂ = (12 i + 1 j) - ( 8 i + 1 j )
Average velocity = 4 i
B. Velocity at time t, is
V(t) = V = 4t i +1 j
Acceleration a = dV/dt
a(t) = d/dt (4t i +1 j)
a(t) = 4 i +0 j
C. Average acceleration in time interval t₁=2sec and t₂=3sec is
a = 4 m/s²
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Estimate the gravitational force of attraction between two people sitting side by side on a park bench. How does this force compare with the gravitational force exerted on each of them by the Earth
Weight is compared with the gravitational force exerted on each of them by the Earth.
What is Newton's law of gravitation?Newton's law of gravity states that each particle having mass in the universe attracts each other particle with a force known as the gravitational force.
The gravitational force is proportional to the product of the masses of the two bodies and inversely proportional to the square of their distance.
When mass increases and distance reduces, gravity rises. Gravity also lowers when the distance between two points grows and the mass decreases.
The gravitational force is balanced by weight;
[tex]\rm F = W \\\\ G\frac{mM}{r^2} =mg[/tex]
Hence, weight is compared with the gravitational force exerted on each of them by the Earth.
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3. Sarah is a world-class diver in the women's 3-m springboard competition. Her height
h (in metres), above the water t seconds after she leaves the board is given by
h = -4.9t2 + 8.8t + 3. How long is Sarah in the air before she reaches the water?
Make sure to round your answer to the nearest tenth.
Answer:
2011, 2013, 2016 Alberta Education
Answer:
1.163 seconds
Explanation:
see the attached. solve the quadratic equation.
Three small but dense objects are located in the x-y plane as shown in the figure. The objects have the following masses: mA = 3.19 kg, mB = 2.55 kg and mC = 1.41 kg.
Determine the x and the y coordinates of the center of the mass of this system. The objects are small in size, they can be treated as point masses.
x coordinate: ?
y coordinate: ?
The x and the y coordinates of the center of the mass of this system will be 43.1 m and 3.12 m respectively.
What is the center of mass?A location is established in relation to an object or set of objects in the center of mass. It is the system's average position across all of its components.
Given data;
There are three little objects that are densely spaced out in the x-y plane.;
The value of the masses
m₁=1.41 kg
m₂=2.55 kg
m₃=3.19 kg
[tex]\rm X_{cm} =\frac{ (W_1x_1 + W_2x_2 + W_3x_3)}{(W_1 + W_2 + W_3)} \\\\\ X_{cm} ==\frac{1.41 \times 2 + 2.55 \times 6 +3.19 \times 4}{1.41+2.55+3.19} \\\\ X_{cm} =4.31 \ m[/tex]
[tex]\rm Y_{cm} =\frac{ (W_1y_1 + W_2y_2 + W_3y_3)}{(W_1 + W_2 + W_3)} \\\\\ X_{cm} ==\frac{1.41 \times 2 + 2.55 \times 4 +3.19 \times 7}{1.41+2.55+3.19} \\\\ X_{cm} =3.12 \ m[/tex]
Hence, the x and the y coordinates of the center of the mass of this system will be 4.31 m and 3.12 m respectively.
To learn more about the center of mass, refer;
https://brainly.com/question/8662931
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Suppose the resistance of a copper wire is 1.05 N at 30°C.
Find the temperature at which the resistance becomes 1.50
Ω.
Answer:
42.9°C
Explanation:
Resistance of a conducting material is directly proportion to Temperature, that is, when being heated, their is more of collisions between the molecules of the material which as a result offers more opposition to the flow of current through it.
therefore R=kT
where R=resistance of the wire
T=Temperature
k= proportionality constant
when R=1.05 T=30°C
so k=1.05/30
k=0.035
when R=1.5 k=0.035
so T=1.5/0.035
T=42.9°C. or 43°C approximated
A box with a mass of 40 kg sits at rest on a frictionless tile floor. with your foot, you apply a 20 N force in a horizontal direction. what is the acceleration of the box?
Answer:
0.5 m/s²
Explanation:
according to Newton's second law, we are goven a relationship between force, mass and acceleration, with the formula:
F = m×a
F for force
m for mass
a for acceleration
we use the given data and get:
20 = 40×a
we find a=20/40=0.5m/s²
acceleration of the box is 20m/s^2
given
mass (m) = 40 kg
force (f) = 20n
acceleration (a) =?
we know,
f = m × a
20 = 40 × a
40 - 20 = a
20m/s^2 = a
If the distance between two objects is cut in half, what happens to the
gravitational force between them?
A. It decreases to 1/2 its original magnitude.
B. It decreases to 1/4 its original magnitude.
O
C. It increases to 4 times its original magnitude.
D. It increases to 2 times its original magnitude.
What’s the allowable ampacity of a No. 12 copper wire with type THHN insulation installed in a raceway with 35 wires? Question 3 options: A) 10 A B) 15 A C) 8 A D) 12 A
The allowable ampacity of a No. 12 copper wire with type THHN insulation installed in a raceway with 35 wires is approximately 12 amperes. This is best represented by option D.
According to the National Electrical Code Table (NEC), a derating factor must be applied for any raceway with more than three current-carrying conductors (in our case, copper wires). In a raceway with 31-40 current-carrying conductors, the derating factor is 50%. Because the ampacity of a No. 12 copper wire usually is 25 amperes (A), we can multiply this by our derating factor to find that:
25 • 0.50 = 12.5
The exact allowable ampacity of this copper wire is 12.5 amperes; however, we can round down to find that the closest answer is 12 A.
Hence, option D represents the best answer.
Answer:
D) 12 A
Explanation:
You want to know the allowable ampacity of AWG 12 type THHN copper wire in a raceway with 35 wires.
Nominal ampacityTable 310.15(B)(16) of the 2017 National Electrical Code gives the nominal rating of AWG 12 type THHN copper wire as 30 A when installed in an environment with a temperature of 30 °C or less.
DeratingTable 310.15(B)(3)(a) gives the adjustment factor when there are more than three conductors in a raceway or cable. For 31–40 conductors, the ampacity must be derated to 40% of its nominal value.
In a raceway with 35 wires, the allowable ampacity of THHN AWG 12 wire is ...
(30 A)(0.40) = 12 A
__
Additional comment
For environmental temperatures above 30 °C, Table 310.15(B)(2)(a) gives additional derating factors. The adjustment factor for temperatures up to 50 °C (122 °F) is 0.82, which would bring the allowable ampacity down to about 9.8 A.
