wind (b), also some others are:
- steam
- gas
- liquid fules
Consider a concave spherical mirr or that has focal length f = +19.5 cm.
a) What is the distance of an object from the mirror's vertex if the image is real and has the same height as the object? Follow the sign rules.
The distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.
What is concave mirror?A concave mirror has a reflective surface that is curved inward and away from the light source.
Concave mirrors reflect light inward to one focal point and it usually form real and virtual images.
Object distance of the concave mirrorApply mirrors formula as shown below;
1/f = 1/v + 1/u
where;
f is the focal length of the mirrorv is the object distanceu is the image distancewhen image height = object height, magnification = 1
u/v = 1
v = u
Substitute the given parameters and solve for the distance of the object from the mirror's vertex
1/f = 1/v + 1/v
1/f = 2/v
v = 2f
v = 2(19.5 cm)
v = 39 cm
Thus, the distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.
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a man carries a hand bag by hanging on his hand moves horizantaly wher the bag does not up or down what is the work done on the bag
Since the displacement is completely perpendicular to the direction of the applied force, the work done on the bag is zero.
When is the Work done on an object ?The work is done on an object when the force applied is multiply by the distance moved by the object in the direction of the force applied.
Given that a man carries a hand bag by hanging on his hand moves horizontally where the bag does not up or down.
What is work if the displacement is not in the direction of force ?
The work done can only be zero if the displacement is perpendicular to the direction of force. otherwise, it will not be equal to zero.
Also, the work done will be zero, if the displacement is zero.
In the question above, the displacement is completely perpendicular to the direction of the applied force.
Therefore, the work done on the bag is zero.
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ASAP NEED HELP !! :(
Why are temperatures more moderate around the fall and spring equinoxes?
C The angle at which Earth's axis tilts changes.
Neither end of Earth's axis is tilted toward the Sun.
The north end of Earth's axis is tilted toward the Sun.
C
The Earth briefly wobbles on its axis.
Answer:C is the answer
Explanation:
It is the most reasonable and the answer that makes most sense
Answer:
The angle at which Earth's axis tilts changes.
Based on the information provided, how confident can you be that this exoplanet is able to support life as we know it?
The information is missing here but data that may support the existence of life in an exoplanet may include the discovery of atmospheric biological gases.
What is an exoplanet?An exoplanet is any planet that surrounds another star beyond the sun, and therefore they are discovered by analyzing other galaxies in the Universe.
The existence of life in these exoplanets can be indirectly supported by the presence of gases that are only produced by organisms, which are detected by suitable telescopes that analyze emission spectra.
In conclusion, the information is missing here but data that may support the existence of life in an exoplanet include biological gases.
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Question 5 & 6 plissssssss
Question: 5
The length of the pendulum is 7.6 m.
What is the expression of length of a pendulum in term of time period?Time period of the pendulum (T) = 2π×√(L/g)L= length of pendulum, g = acceleration due to gravity on earthSo, L = T²g/4π²
What is the length of the pendulum, if the time period is 3.20 s and acceleration due to gravity becomes 3×g?T= 3.20 sL = (3.2²×3×9.8)/4π²= 7.6 m
Thus, we can conclude that the length of the pendulum is 7.6 m i.e option C is correct.
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Question: 6
The object takes 2.55 seconds to reach the ground.
What is the expression of time taken to reach the earth surface by an object?From the conversation of energy, (1/2)mv²=mghSo, v=√(2gh)From Newtown's equation of motion, v=u+atHere, a= acceleration due to gravity which is gSo, √(2gh)=gtt= √(2h/g)
What is the time taken by an object dropped from 31 m to reach the ground?t= √(2×31/9.8)
= 2.55s
Thus, we can conclude that the option A is correct.
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*According to Bohr's Theory, what is the maximum number of electron orbital layers?*
Four are known: s, p, d, and f.
Thank you,
Eddie
3 A rocket of mass 1200 kg is travelling at 2000 m/s. It fires its engine for 1 minute. The forward thrust provided by the rocket engines is 10 kN (10 000 N).
(i) Use increase in momentum = F x t to calculate the increase in momentum of the rocket.
(ii) Use your answer to a to calculate the increase in velocity of the rocket and its new velocity after firing the engines.
The impulse shared by the object equals the difference in momentum of the object. In equation form,
F • t = m • Δ v. In a collision, objects experience an impulse; the impulse causes and is equal to the difference in momentum.
How to calculate thrust provided by the rocket engines is 10 kN (10 000 N).?
a)There is this impulse-momentum change equation.
[tex]where m$ is the mass of a body, $F$ is a force acting to the body, $t$ is time and $D E L A T A N\}=V_{2}-V_{1}$ is the change of velocity.We consider everything is happen along a straight line, and gravitation does not participate.So, the increase of momentum is $\mathrm{F}^{*} \mathrm{t}=10000 \mathrm{~N} * 60$ seconds $=600000 \mathrm{~N}^{*} \mathrm{~s}=600000\left(\mathrm{~kg}^{*} \mathrm{~m}\right)^{*} \mathrm{~s} / \mathrm{s}^{\wedge} 2=600000 \mathrm{~kg}{ }^{*} \mathrm{~m} / \mathrm{s}$.[/tex]
We consider everything exits happen along a straight line, and gravitation does not participate.
So, the increase of momentum is F×t = 10000 N × 60 seconds = 600000 N*s = 600000 (kg*m)*s/s^2 = 600000 kg*m/s.
[tex]$$\Delta(\mathrm{V})=\frac{\mathrm{F.t}}{\mathrm{m}}=\frac{600000}{1200}=500 \mathrm{~m} / \mathrm{s} .$$[/tex]
New velocity after engine was firing during 60 seconds is 2000 + 500 = 2500 m/s.
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Three ropes A, B and C are tied together in one single knot K. (See figure.)
If the tension in rope A is 50.5 N, then what is the tension in rope B?
Assuming point K is held in equilibrium, by Newton's second law we have
• net horizontal force
[tex]F_C \cos\left(\tan^{-1}\left(\dfrac57\right)\right) - F_A = 0[/tex]
• net vertical force
[tex]F_C \sin\left(\tan^{-1}\left(\dfrac57\right)\right) - F_B = 0[/tex]
where the angle [tex]\theta[/tex] that rope C makes with the horizontal axis satisfies
[tex]\tan(\theta) = \dfrac{9-4}{11-4} = \dfrac57[/tex]
Solve the first equation for [tex]F_C[/tex].
[tex]F_C = F_A \sec\left(\tan^{-1}\left(\dfrac57\right)\right)[/tex]
(Recall that [tex]\sec(x)=\frac1{\cos(x)}[/tex].)
Substitute this into the second equation and solve for [tex]F_B[/tex].
[tex]F_B = F_C \sin\left(\tan^{-1}\left(\dfrac57\right)\right)[/tex]
[tex]F_B = F_A \sec\left(\tan^{-1}\left(\dfrac57\right)\right) \sin\left(\tan^{-1}\left(\dfrac57\right)\right)[/tex]
[tex]F_B = F_A \tan\left(\tan^{-1}\left(\dfrac57\right)\right)[/tex]
(Recall that [tex]\tan(x)=\frac{\sin(x)}{\cos(x)}[/tex].)
[tex]F_B = \dfrac57 F_A[/tex]
[tex]\boxed{F_B \approx 36.1\,\rm N}[/tex]
PLEASE HELP 20 POINTS
Scientists often need to look for patterns that occur in the data they collect and analyze. Explain why identifying patterns is important, using an example from your investigation about habitable worlds.
Answer:
Patterns in science are a little different. Data doesn't have to follow a trend, always going up or down over time. A pattern is a when data repeats in a predictable way. A good example of a pattern in science comes from the father of genetics, Gregor Mendel.After data is collected, it can be analyzed by looking for trends, patterns, and relationships. Trends are general directions of data, such as an overall increase in global temperature. Patterns don't necessarily involve data going one way or the other, but rather describe a repeating observation.In order to interpret and understand scientific data, one must be able to identify the trends, patterns, and relationships in it. Examine the importance of scientific data and recognize how understanding its trends, patterns, and relationships can lead a researcher to support or refute a hypothesis. Updated: 01/06/2022 What Is Scientific Data?
Explanation:
If A = 6î - 8ĵ, B = −8î + 3ĵandC = 26î + 19ĵ. Find a and b Such that aẢ +bB + C = 0
Sol^n :
aA=6ai-8aj
bB=-8bi-3bj
C=26i-19j
Given aA+bB+C=0i+0j
(6a-8b+26)i-(8a+3b+19)j=0i+0j
comparing coefficient of i and j
6a-8b=-26
2a-4b=-13.....(1)
8a+3b=-19......(2)
making the coefficient of a similar
8a-16b=-52....(3)
8a+3b=-19....(4)
equation 4-3 implies
19b=33
b=33/19
from (1)
2a=-13+4(33/19)
2a=-115/19
a=-115/38
A stone of weight 10N falls from the top of a 250m high cliff. a) Calculate how much work is done by the force of gravity in pulling the stone to the foot of the cliff. b) How much energy is transferred to the stone?
Answer:
work done = ( force × displacement)
(a)The force acting on the block is it's self weight and displacement is equal to height of the tower.
work done by gravity = (250 × 10) = 2500 joule
(b) The work done by gravity 2500 joule is transferred to the object in the form of it's kinetic energy.
"The acceleration due to gravity on the surface of the earth is 9.8 m/s ²." What does it mean?
Answer: It simply means that a freely falling object would increase its velocity by 9.8 m/s per second
Answer:
Hello!
"The acceleration due to gravity on the surface of the earth is 9.8 m/s ²." What does it means that
That every second an object is in free fall, gravity will cause the velocity of the object to increase 9.8 m/s. So, after one second, the object is traveling at 9.8 m/s.
In 1656, the Burgmeister (mayor) of the town of Magdeburg, Germany, Otto Von Guericke, carried out a dramatic demonstration of the effect resulting from evacuating air from a container. It is the basis for this problem. Two steel hemispheres of radius 0.430 m (1.41 feet) with a rubber seal in between are placed together and air pumped out so that the pressure inside is 15.00 millibar. The atmospheric pressure outside is 940 millibar.
1. Calculate the force required to pull the two hemispheres apart. [Note: 1 millibar=100 N/m2. One atmosphere is 1013 millibar = 1.013×105 N/m2 ]
2. Two equal teams of horses, are attached to the hemispheres to pull it apart. If each horse can pull with a force of 1450N (i.e., about 326 lbs), what is the minimum number of horses required?
The values of the required solutions are
F= 53731 N N=37.05What is the minimum number of horses required?What is Force?
In the field of physics, an influence that can alter the motion of an object is referred to as a force. An object having mass can experience a change in its velocity, often known as an acceleration, when subjected to a force. Intuitively, force can also be conceptualized as either a push or a pull. Because it may be measured in both magnitude and direction, a force is considered a vector quantity.
What is atmospheric pressure?
The pressure that is exerted within the atmosphere of the Earth is referred to as barometric pressure as well as atmospheric pressure. As a unit of pressure, the standard atmosphere is defined as having a value of 101,325 Pa. This value is equivalent to 1013.25 millibars, 760 mm Hg, 29.9212 inches Hg, or 14.696 psi.
In most situations, the equation for force can be expressed numerically as
Force = pressure * A
Therefore
[tex]F= (970 - 15 B)(\pi * (0.430 )^2)[/tex]
F= 53731 N
In conclusion, If each horse can pull with a force of 1450N
The number of horses required is
[tex]N=\frac{60754 }{1450}[/tex]
N=37.05
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The figure illustrates flow through a pipe with diameters of 1.0 mm and 2.0 mm and with different elevations. Px is the pressure in the pipe, and Vx is the speed of a non-viscous incompressible fluid at locations x = Q,R,S,T, or U. Options are: Greater than, Less than, Equal to
PU is ... PQ
VU is ... 2VT
PR is ... PU.
VR is ... VS
VQ is ... VU
PR is ... PS
a.
i. PU is greater than PQ.ii. VU is Greater than 2VTb.
i. PR is Equal to PU.ii. VR is Equal to VSc.
i. VQ is Equal to VUii. PR is Greater than PS.What is pressure?Pressure is the force per unit area on a surface.
What is speed?Speed is the distance moved per unit time.
Pressure
Since pressure, P = hρg where
h = depth, ρ = density of liquid and g = acceleration due to gravity.Since ρ and g are constant
P ∝ h
So, we see that pressure is directly proportional to depth.
a. i. Pressure between R and USince U is lower than Q, Pressure at U is greater than pressure at Q.
So,PU is greater than PQ.
ii. Speed between U and TUsing the continuity equation
VUAU = VTAT where
VU = speed at U, AU = cross-sectional area at U = π(dU)² where dU = diameter at U = 1.0 mmVUT= speed at T, AT = cross-sectional area at T = π(dT)² where dT = diameter at T = 2.0 mmSo, VUAU = VTAT
VUπ(dU)² = VTπ(dT)²
VU = VT(dT)²/(dU)²
VU = VT(2.0)²/(1.0)²
VU = VT(4)
VU = 4VT
Since VU = 4VT,VU is Greater than 2VT
b i. Pressure between R and USince R is at the same depth as U, Pressure at R is equal to pressure at U.
So,PR is Equal to PU.
ii. Speed between R and SUsing the continuity equation
VRAR = VSAS where
VR = speed at R, AR = cross-sectional area at R = π(dR)² where dR = diameter at R = 2.0 mmVS= speed at S, AS = cross-sectional area at S = π(dS)² where dS = diameter at S = 2.0 mmSo, VRAR = VSAS
VRπ(dR)² = VSπ(dS)²
VR = VS(dS)²/(dS)²
VR = VS(2.0)²/(2.0)²
VR = VS(1)
VR = VS
Since VR = VS,VR is Equal to VS
c. i. Speed between Q and UUsing the continuity equation
VQAQ = VUAU where
VQ = speed at Q, AQ = cross-sectional area at Q = π(dQ)² where dQ = diameter at Q = 1.0 mmVU = speed at U, AU = cross-sectional area at U = π(dU)² where dU = diameter at U = 1.0 mmSo, VQAQ = VUAU
VQπ(dQ)² = VUπ(dU)²
VQ = VU(dU)²/(dQ)²
VQ = VU(1.0)²/(1.0)²
VQ = VU(1)
VQ = VU
Since VQ = VU, VQ is Equal to VU
Ii. Pressure between R and SSince R is lower than S, Pressure at R is greater than pressure at S.
So,PR is Greater than PS.
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Which of the following is correct concerning the uncontrolled burn phase?
Group of answer choices
The uncontrolled burn phase is characterized by uncontrolled combustion in a cylinder until fuel accumulated during ignition delay is burned.
This is also called the flame propagation phase.
Many points in the combustion chamber will simultaneously reach the threshold values required for ignition, and multiple flame fronts will move through the air-fuel mixture.
(All three statements are correct.)
The option that is the correct one concerning the uncontrolled burn phase is:
The uncontrolled burn phase is characterized by uncontrolled combustion in a cylinder until fuel accumulated during ignition delay is burned.What is uncontrolled combustion?
Uncontrolled Combustion is known to be the the time and place in which a kind of an ignition will stop and it is said to be never fixed by anything in regards to the compression ignition engine as seen in SI engines.
Note that the four Stages of combustion are:
1. Pre-flame combustion
2. Uncontrolled combustion
3. Controlled combustion and
4. After burning
Hence, The uncontrolled burn phase is characterized by uncontrolled combustion in a cylinder until fuel accumulated during ignition delay is burned as all the fuel need to burn out.
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In traveling to the Moon, astronauts aboard the Apollo spacecraft put spacecraft into a slow rotation to distribute the Sun's energy evenly (so one side would not become too hot). At the start of their trip, they accelerated from no rotation to 1.0 revolution every minute during a 12-min time interval. Think of the spacecraft as a cylinder with a diameter of 8.5 m rotating about its cylindrical axis.
a)Determine the angular acceleration of the ship.
Express your answer using two significant figures.
b)Determine the radial component of the linear acceleration of a point on the skin of the ship 9.5 min after it started this acceleration.
Express your answer to two significant figures and include the appropriate units.
c)Determine the tangential component of the linear acceleration of a point on the skin of the ship 9.5 min after it started this acceleration.
Express your answer to two significant figures and include the appropriate units.
The angular acceleration is 4.44*[tex]10^{-5} rev/s^{2}[/tex] , radial component is 0.016 m/[tex]s^{2}[/tex], tangential component is 0.9347*[tex]10^{-5} m/s^{2}[/tex].
Angular acceleration,
ω=1/60=0.016 rev/s
[tex]\alpha[/tex]=0.016/(6*60)=4.44*[tex]10^{-5} rev/s^{2}[/tex]
The angular acceleration is 4.44*[tex]10^{-5} rev/s^{2}[/tex]
Radial component of the linear acceleration=[tex]\alpha_{r}[/tex]
v=ωr=0.016*4.75=0.076 m/s
[tex]\alpha_{r}[/tex]=0.076 /4.75=0.016 m/[tex]s^{2}[/tex]
The tangential component = [tex]\alpha_{t}[/tex]=4.44*[tex]10^{-5} rev/s^{2}[/tex]/4.75=0.9347*[tex]10^{-5} m/s^{2}[/tex]
Angular accelerationThe temporal rate at which angular velocity changes is known as angular acceleration. Due to the fact that there are two different types of angular velocity—spin angular velocity and orbital angular velocity—there are also two different types of angular acceleration, referred to as spin angular acceleration and orbital angular acceleration, respectively. The terms "orbital angular acceleration" and "spin angular acceleration" describe the angular acceleration of a point particle about a fixed origin and, respectively, the angular acceleration of a rigid body about its center of rotation.
The unit of measurement for angular acceleration is the angle per unit time squared, or, in SI units, radians per second squared. It is typically denoted by the symbol alpha ().
In traveling to the Moon, astronauts aboard the Apollo spacecraft put spacecraft into a slow rotation to distribute the Sun's energy evenly (so one side would not become too hot). At the start of their trip, they accelerated from no rotation to 1.0 revolution every minute during a 12-min time interval. Think of the spacecraft as a cylinder with a diameter of 8.5 m rotating about its cylindrical axis.
a)Determine the angular acceleration of the ship.
Express your answer using two significant figures.
b)Determine the radial component of the linear acceleration of a point on the skin of the ship 9.5 min after it started this acceleration.
Express your answer to two significant figures and include the appropriate units.
c)Determine the tangential component of the linear acceleration of a point on the skin of the ship 9.5 min after it started this acceleration.
Express your answer to two significant figures and include the appropriate units.
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1. The diagram shows a satellite traveling in uniform circular motion around the Earth.
(a) Give the relation between radius of the orbit and the velocity of the satellite.
(b ) The satellite is kept in orbit by a force. On the diagram draw an arrow to show the
direction of this force.
Answer:
M V R = constant angular momentum is constant because no forces act in the direction of V
Since M (mass) = constant
V R = constant
The force is directed along the gravitational force vector (towards the center of rotation)
why is the bulb of hydrometer is made heavier give two reasons
The reason behind the heavier hydrometer bulb is the sinking of hydrometer is inversely proportional to the density of hydrometer, hence hydrometers is made heavier.
We all know that hydrometers float in liquid hence to maintain the centre of gravity while floating the hydrometer is made heavier using lead shots.
A vector in the xy plane has components -14.0 units in the x-direction and 30.0 units in the y-direction. What is the magnitude of the vector? What is the angle between the vector and the positive x-axis?
[tex]\huge\underline{\underline{\boxed{\mathbb {SOLUTION:}}}}[/tex]
We would calculate the magnitude by applying pythagorean theorem:
[tex]\longrightarrow \sf{Magnitude= \sqrt{(-14)^2 } + 30^2}[/tex]
[tex]\longrightarrow \sf{Magnitude = 33.12}[/tex]
[tex]\longrightarrow \sf{The \: vector \: is \: (- 14, 30)}[/tex]
The angle between two vectors is given by the formula:
[tex]\sf{\longrightarrow \small \cos \emptyset = \dfrac{(a1b1 + a2b2)}{ \sqrt{(a1)^2 + (a2)^2√(b1)^2 + (b2)^2} } }[/tex]
In two dimensional, the x axis of vector form is:
[tex]\small\sf{\longrightarrow (b1, b2) = (1, 0) }[/tex]
[tex]\sf{\longrightarrow \small \cos \: \emptyset = \dfrac{(14 * 1 + 30 x 0)}{( \sqrt{(-14)^2 + (30)^2)(√(1)^2 + (0)^2)} } }[/tex]
[tex]\small\longrightarrow \sf{ \dfrac{14}{33.12} }[/tex]
[tex]\small\longrightarrow \sf{\emptyset \: = arcCos (\dfrac{ - 14}{33.12} )}[/tex]
[tex]\small\longrightarrow \sf{\emptyset= 115^\circ}[/tex]
[tex]\huge\underline{\underline{\boxed{\mathbb {ANSWER:}}}}[/tex]
[tex] \small\bm{The \: angle \: between \: the \: vector \: }[/tex]
[tex]\small\bm{and \: \: the \: \: positive \: \: x \: \: axis \: \: is \: \: \: 115^\circ .}[/tex]
Write a properly formatted hypothesis statement to answer this question: How does the amount of salt added to ice affect the rate at which the ice will melt?
Specify how you plan to change the independent variable by using terms such as increase or decrease. Also, specify how the dependent variable will change in response by using terms such as increase, decrease, or stays the same.
Criteria pts
Correct placement of IV 5
Correct placement of DV 5
If, then format 5
IV indicates either "increases" or "decreases" 5
DV indicates either "increases", "decreases", or "stays the same" 5
The hypothesis will be:
H₀ = The amount of salt added to ice will not affect the rate at which the ice will melt.
H₁ = The amount of salt added to ice will affect the rate at which the ice will melt.
The independent variable which is can be changed by increasing the rate of salt added to the equation.
The dependent variable which is ice will change or melt in response as it will decrease if the rate of the salt added increases.
What is the effect of salt on the melting temperature of ice?Salt does not really lower the temperature of an ice cubes, it is known to just lowers their freezing point, that is lowers their melting point.
Note that if salt is around, ice cubes are known to be colder to be solid, and they tend to melt at a temperature that is said to be lower than the freezing point of pure water.
If the ionic compound salt is known to be added, it tends to lowers the freezing point of the water, which implies that the ice on the ground is not able to freeze that layer of water at all.
Hence, the hypothesis will be:
H₀ = The amount of salt added to ice will not affect the rate at which the ice will melt.
H₁ = The amount of salt added to ice will affect the rate at which the ice will melt.
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An Airbus A380-800 passenger airplane is cruising at constant altitude on a straight line with a constant speed. The total surface area of the two wings is 395 m^2. The average speed of the air just below the wings is 259 m/s, and it is 288 m/s just above the surface of the wings.
What is the mass of the airplane? (The average density of the air around the airplane is ρair = 1.21 kg/m^3.)
The mass of an airplane with two wings that are 395m2 in size and average wing surface speeds of 259 and 288 m/s is 387x 10^3 kg.
We need to be aware of the Bernoulli principle in order to determine the solution.
How can I determine an airplane's mass?According to the Bernoulli's principle, the total amount of pressure energy, kinetic energy, and potential energy in a streamlined flow of an incompressible, non-viscous fluid is constant.It can be stated as follows:[tex]P+\frac{1}{2}dv^2+ dgh = constant.[/tex] We substitute d for to represent density.
We've done that,[tex]V_1=259m/s\\V_2=288m/s\\A=395m^2\\d=1.21kg/m^3[/tex]
We compare the governing idea for the wing's bottom and upper surfaces to:[tex]P_1+\frac{1}{2}dV_1^2+dgh=P_2+ \frac{1}{2}dV_2^2+dgh\\P_1-P_2=\frac{1}{2}d(V_2^2-V_1^2)\\\frac{F}{A}= \frac{1}{2}d(V_2^2-V_1^2)\\[/tex]
Consequently, using the aforementioned equation, the airplane's mass will be,[tex]m=\frac{\frac{1}{2}d(V_2^2-V_1^2)A\\}{g} \\m=387*10^3kg.[/tex]
Consequently, we can say that the mass of an airplane with two wings that are 395m2 in size and average wing surface speeds of 259 and 288 m/s is 387 x 10^3 kg.
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A ship is 2.40km from a large rock cliff when it sounds its foghorn at a frequency of 178Hz. How long will it take for the sounds to return to the ship? (Sound travels 343m/sec)
A ship sounds its foghorn at a frequency of 178Hz when it is 2.40km away from a sizable rock face. (Audio travels at 343 m/s.) Reconstruction succeeded in resolving the federalism controversy that had dogged the country since the 1790s.
By most other standards, reconstruction was a failure: radical Republican legislation eventually failed to shield former slaves from white persecution and to bring about significant alterations to the South's socioeconomic structure. The federalism question, which had been a topic since the 1790s practically immediately, was at hand when President Rutherford B. Former Confederate officials and slaves returned to the South after Hayes withdrew federal soldiers from the region in 1877. These newly powerful white southern legislators established anti-progressive laws like voter ID requirements and black codes with the help of a conservative Supreme Court in an effort to roll back the rights that blacks had won during Radical Reconstruction. With its rulings in the Slaughterhouse Cases, the Civil Rights Cases, and United States v., the U.S. Supreme Court strengthened this anti-progressive movement federalism.
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An ultracentrifuge is spinning at a speed of 80,000 rpm. The rotor that spins with
the sample can be roughly approximated as a uniform cylinder of 10 cm radius
and 8 kg mass, spinning about its symmetry axis). In order to stop the rotor in
under 30 s from when the motor is turned off, find the minimum braking torque
that must be applied.
O-19.2 Nm
-17.2 Nm
O -15.2 Nm
O-11.2 Nm
O None of the above
D. The minimum braking torque that must be applied is -11.2 Nm.
Moment of inertia of the uniform cylinderThe moment of inertia of the uniform cylinder is calculated as follows;
I = ¹/₂MR²
where;
M is mass of the cylinderR is radius of the cylinderI = (0.5)(8)(0.1²)
I = 0.04 kgm²
Minimum braking torqueτ = -Iα
where;
α is angular accelerationα = ω/t
α = (80,000 x 2π/rev x 1 min/60s) / (30 s)
α = (80,000 x 2π)/(60 x 30)
α = 279.25 rad/s²
τ = - ( 0.04 kgm²) x (279.25)
τ = -11.2 Nm
Thus, the minimum braking torque that must be applied is -11.2 Nm.
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How far would you push a car if you did 28,000J of work, exerting a force of 825N?
We push a car to the distance of 33.939m if we do 28000J work , exerting a force of 825N.
What is work done and force ?work done: The amount of energy transferred to a body .Force : force is an influence that can change the velocity of an object .How to calculate distance moved from work done and force ?we know ; work done =Force ×distance moved in the direction of force Mathematically, W=F.Swhere W= work doneF = applied force
S = distance moved
So S=W/F=28000J/825N
=33.939meter
Thus, we can conclude that the distance moved by the car is 33.939m.
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What is an example of total internal reflection at work?
A.
A ray of light has the same intensity both entering and exiting a fiber optic cable.
B.
A ray of light entering a glass cube gets refracted.
C.
A ray of light in air hits a shiny surface and bounces off.
D.
A ray of light entering a ruby gets refracted.
Answer:
A i think...
Explanation:
Sorry if its wrong
Two uncharged spheres are separated by 1.70 m. If 2.40 ✕ 10¹² electrons are removed from one sphere and placed on the other, determine the magnitude of the Coulomb force (in N) on one of the spheres, treating the spheres as point charges.
_______N
**Hint** Find the net charge on each sphere and substitute values into Coulomb's law.
The magnitude of the Coulomb force (in N) on one of the spheres, given the data is 4.59×10⁻⁴ N
How to determine the charge on each spheresSphere 1 losses 2.40×10¹² electrons
But
1 electron = 1.6x10¯¹⁹ C
Thus,
Charge on sphere 1 = +1.6x10¯¹⁹ × 2.40×10¹² = +3.84×10¯⁷ C
Sphere 2 gains 2.40×10¹² electrons
But
1 electron = 1.6x10¯¹⁹ C
Thus,
Charge on sphere 2 = -1.6x10¯¹⁹ × 2.40×10¹² = -3.84×10¯⁷ C
How to determine the coulomb forceCharge on sphere 1 (q₁) = +3.84×10¯⁷ CCharge on sphere 2 (q₂) = 3.60 mC = -3.84×10¯⁷ CElectric constant (K) = 9×10⁹ Nm²/C²Distance apart (r) = 1.7 mForce (F) =?
Using the Coulomb's law equation, the force can be obtained as illustrated below:
F = Kq₁q₂ / r²
F = (9×10⁹ × 3.84×10¯⁷ × 3.84×10¯⁷) / (1.7)²
F = 4.59×10⁻⁴ N
Thus, the magnitude of the Coulomb's force is 4.59×10⁻⁴ N
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The gravitational force exerted by the Sun on the Earth holds the Earth in an orbit around the Sun. Let us assume that the orbit is perfectly circular. The work done by this gravitational force during a short time interval in which the Earth moves through a displacement in its orbital path is (a) zero (b) positive (c) negative (d) impossible to determine
The work done by this gravitational force during a fast time interval in which the Earth drives through a displacement in its orbital path is (a) zero.
What is the unit of gravitational force?Gravitational Force
The kilogram force (kgf) is a gravitational unit of force. The force exercised by the earth on a body of mass 1 kg is regarded as 1 kgf. Force = Mass × Acceleration. 1 kgf = 9.8 N.
The gravitation force and perpendicular to the path of movement
hence angle between gravitational force [ F ]and displacement (d)=90°
work done (W)= F d cos 90°
= 0
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A 0.550 kg air-track glider is attached to each end of the track by two coil springs. It takes a horizontal force of 0.500 N to displace the glider to a new equilibrium position, x= 0.070 m.
1. Find the effective spring constant of the system.
2. The glider is now released from rest at x= 0.070 m. Find the maximum x-acceleration of the glider.
3. Find the x-coordinate of the glider at time t= 0.650T, where T is the period of the oscillation.
4. Find the kinetic energy of the glider at x=0.00 m.
(1) The effective spring constant of the system is 7.14 N/m.
(2) The maximum x-acceleration of the glider is 0.9 m/s².
(3) The x-coordinate of the glider at time t= 0.650T is 0.28 m.
(4) The kinetic energy of the glider at x=0.00 m is zero.
The effective spring constant of the system
The effective spring constant of the system is calculated as follows;
F = kx
where;
k is spring constantk = F/x
k = 0.5/0.07
k = 7.14 N/m
Maximum acceleration of the glidera = ω²x
where;
ω is angular speedω = √k/m
ω = √(7.14/0.55)
ω = 3.6 rad/s
a = (3.6)² x 0.07
a = 0.9 m/s²
Period of the oscillationT = 2πx/v
T = 2πx/(ωx)
T = 2π/ω
T = 2π/(3.6)
T = 1.75 seconds
t = 0.65T
t = 0.65 x 1.75
t = 1.14 seconds
x = vt
x = (ωx)t
x = (3.6 x 0.07) x 1.14
x = 0.28 m
kinetic energy of the gliderAt position x = 0, the glider is at rest, the velocity is zero and the kinetic energy will be zero.
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A sine bar is used to determine the angle of a part feature. The length of the sine
bar=8.000in. The rolls have a diameter of 1.000 in. All inspection is performed on a surface plate. In
order for the sine bar to match the angle of the part, the following gage blocks must be stacked: 2.0000,
0.5000, 0.2500, and 0.0050. Determine the angle of the part feature.
A sine bar is used to get the angular measurement of a part feature. The angle of the part feature is 18.24°
From the question
Given that,
Length of the sine bar, L = 8.000 in
Diameter of the rolls = 1.000 in
Height under the roll, H = 2.0000 + 0.5000 + 0.0050
= 2.505 in
From sine bar formula,
we know that,
H = sin A x L
sin A = H ÷ L
where,
A ⇒ angle of part feature
H ⇒ height under the roll
L ⇒ length of the sine bar
Substituting values in the above equation,
sin A = H / L
A = sin⁻¹ ( 2.505 ÷ 8 )
A = sin⁻¹ (0.3131)
A = 18.24⁰
Hence the angle of the part feature = 18.24°
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A mass M is suspended from a spring and oscillates with a period of 0.840 s. Each complete oscillation results in an amplitude reduction of a factor of 0.96 due to a small velocity dependent frictional effect. Calculate the time it takes for the total energy of the oscillator to decrease to 0.50 of its initial value.
The energy becomes 0.50 times in 6.72 s.
Let E represent the oscillator's initial energy, Et be the energy's final value at time t, where A is its beginning amplitude, At amplitude at time t, be. as the oscillator's energy increases to 0.50 times its initial value. We can replace the oscillator's total energy for the energy at time t to obtain the amplitude as shown below.
Et=0.50E
1
k(4₂)² = (0.5) - kA²
(4₂)² = (0.5) A²
At = 0.71A
So, the amplitude of the oscillator becomes 0.71 times its initial ar
0.71A = = A(0.96)¹2
log(0.71)
log(0.96)
8.4
n=
So, the time taken for n oscillation is obtained as,
t = n (0.800 s)
= (8.4) (0.800)
= 6.72 s
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