When there is relative motion between the fluid layers or between the fluid and solid surface, the fluid's internal property known as viscosity is created. This relative motion is being resisted by viscosity.
Derivations of the equations of motion—the connections for the conservation of mass and momentum—in rectangular,
coordinates that are spherical and cylindrical. The entire body of experimental data suggests that the fundamental equations of fluid mechanics are in fact those listed below, and that they apply to any circumstance where a Newtonian fluid is flowing, in theory.
Viscosity, a characteristic of fluids, is produced when there is relative motion between the fluid layers or between the fluid and solid surface. Viscosity prevents this relative motion from happening.
Unfortunately, due to their all-encompassing nature, their analytical solution is challenging or impossible unless the circumstances are quite straightforward.
However,
These "Navier-Stokes equations" must be understood for the following reasons.
reasons:
1. They result in the analytical and precise resolution of some basic yet significant problems.
issues.
2. They serve as the foundation for additional chemical engineering research in various fields.
3. A few reasonable simplification assumptions can frequently result in
for many engineering problems, approximation solutions are eminently acceptable purposes. Typical examples can be found in the study of boundary layers, the use of films to cover substrates, waves, lubrication, and inviscid (irrotational) flow.
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two cars are diriving down the road. car a has a mass of 1,100 kg and is moving at 20 m/s. car b has a mass of 1000 kg and is moving at 30 m/s. which car has more kinetic energy and why
Answer:
Hope the pictures will help you
what diameter must a copper wire have if its resistance is to be the same as that of an equal length of aluminum wire with diameter 1.94 mm ?
The diameter of the copper wire if its resistance is to be the same as that of an equal length of aluminum wire with diameter 1.94 mm is 1.568mm
what is resistance?The obstruction to current flow in an electrical circuit is measured by resistance. The Greek letter omega Ω represents resistance, which is measured in ohms.
what is resistivity?The term "resistivity" refers to a quality that quantifies how much an object resists having current flow through it. It is a feature of the material itself, independent of the size or form of the sample, and is typically temperature dependent, though it could also be pressure-dependent.
by formula we have ;
R=ρL/A =ρL/πd²/4
where
R = resistance
ρ= resistivity
L = length ; and
A = area of cross section = πd²/4
we know that,
ρ₁ of aluminum = 2.63×10⁻⁸Ωm ; d₁ is diameter of aluminum
ρ₂ of copper=1.72×10⁻⁸Ωm ; d₂ is diameter of copper
given: L of copper= L of aluminum
also, R of copper= R of aluminum
on cancelling the equal terms we get,
ρ₁/d₁²=ρ₂/d₂²
here d₂=d₁[tex]\sqrt\frac{resistivity of copper}{resistivity of aluminum}[/tex]
d₂=1.94[tex]\sqrt \frac{1.72}{2.63}[/tex]
d₂=1.568mm
Hence the diameter of copper wire is 1.568mm
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A horse pulls a cart with force F. As a result of this force the cart accelerates with constant acceleration. The
magnitude of the force that the cart exerts on the horse
A) is zero newtons. B) greater than the magnitude of F. C) equal to the magnitude of F. D).
less than the magnitude of F.
Answer:
c. equal equal to the magnitude of f
Mr. Red Herring was found shot dead in his backyard. He was about a foot away from his back porch, lying next to his personal handgun. A bloody footprint was also found on the porch. Analysis of a bullet found nearby suggests that, based on the striation marks, the bullet that killed Herring came from a gun discarded in a nearby trash can. He apparently had a date that night with Mrs. Scarlet. Based on various evidence, the forensics team also created a digital rendering of the crime.
An example of demonstrative evidence is the digital rendering of the crime scene that shows where the gun was fired from. That is option B.
What is demonstrative evidence?A demonstrative evidence is the type of evidence that is being represented using visuals to help enhance the facts of a claim made against an opponent in the law court.
The components of demonstrative evidence include the following:
photos, x-rays, videotapes, movies, sound recordings, diagrams, forensic animation,maps and drawings.Based on the various evidence presented concerning the sudden death of Mr. Red Herring who was short dead in his backyard, the demonstrative evidence is when the forensics team also created a digital rendering of the crime showing where the gun was shot.
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Options of question:
In this story, which is an example of demonstrative evidence?
The fingerprint that showed that Mrs. Scarlet handled the gun
The digital rendering of the crime scene that shows where the gun was fired from
The report about bullet striations that prove the shot came from the discarded weapon
The DNA evidence gathered from the blood splatters
A force of -4.4 x 103 N exists between a positive charge of 8.0 x 10-4 C and a negative charge of -3.0 x 10-4 C. What is the distance that separates the charges?
The distance that separates the charges is 0.7 m.
What is the distance between the charges?The distance between the charges is determined by applying Coulomb's law of electrostatic force as shown below.
F = kq₁q₂/r²
where;
r is the distance between the chargesk is coulomb's constantF is the force between the chargesr² = kq₁q₂/F
r² = (9 x 10⁹ x 8 x 10⁻⁴ x 3 x 10⁻⁴) / (4.4 x 10³)
r² = 0.49
r = √0.49
r = 0.7 m
Thus, the distance that separates the charges is 0.7 m.
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A helicopter changes its velocity from 22.0 m/s [E] to 10.0 m/s [W] during a time interval of 8.0 s. What was the helicopter’s average acceleration?
The average acceleration of the helicopter, given that it changes its velocity from 22.0 m/s to 10.0 m/s is –1.5 m/s²
How do I determine the average acceleration ?We understood that acceleration is defined as the change in velocity with time i.e
a = (v – u) / t
a is the acceleration v is the final velocity u is the initial velocity t is the timeWith the above formula, we can determine the average acceleration of the helicopter. Details below
The following data were obtained from the question:
Initial velocity (u) = 22.0 m/sFinal velocity (v) = 10.0 m/sTime (t) = 8.0 s Average acceleration (a) =?a = (10 – 22) / 8
a = –12 / 8
a = –1.5 m/s²
Thus, we can conclude that the average acceleration is –1.5 m/s²
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Describe how well you think your modeled position matches the observed position for the man.
1 answer=10,543 points
Answer:
its upto your mind and magic
Explanation:
For questions 7 through 9, Albert and friends are stranded in a clearing on top of a hill at an
elevation of 350m above sea level. You are on a rescue plane flying in supplies to tide them over
until help arrives. Your plane is flying at a constant speed of 260 km/h from west to east at an
elevation of 810m.
PLEASE HELP WOTH THESE QUESTIONS!!!!!!
Answer:
Assume that the air resistance on the supplies is negligible, and that [tex]g = 9.8 \; {\rm m\cdot s^{-2}}[/tex].
The plane need to drop the supplies when it is horizontally approximately [tex]700\; {\rm m}[/tex] away from the hill.
The supplies will hit the tree.
Explanation:
Let [tex]u_{y}[/tex] and [tex]v_{y}[/tex] denote the initial and final vertical velocity of the supply; [tex]u_{y} = 0\; {\rm m\cdot s^{-1}}[/tex] since the plane was flying horizontally.
Let [tex]x_{y}[/tex] denote the vertical displacement of the supply; [tex]x_{y} = 350\; {\rm m} - 810\; {\rm m} = (-460)\; {\rm m}[/tex].
Let [tex]a_{y}[/tex] denote the vertical acceleration of the supply; [tex]a = (-g) = (-9.8)\; {\rm m\cdot s^{-2}}[/tex].
Make use of the SUVAT equation [tex]{v_{y}}^{2} - {u_{y}}^{2} = 2\, a_{y}\, x_{y}[/tex] to find [tex]v_{y}[/tex], the final vertical velocity of the supply:
[tex]\begin{aligned} {v_{y}}^{2} &= {u_{y}}^{2} + 2\, a_{y}\, x_{y} \end{aligned}[/tex].
[tex]\begin{aligned} v_{y} &= -\sqrt{{u_{y}}^{2} + 2\, a_{y}\, x_{y}} \\ &= -\sqrt{0^{2} + 2\, (-9.8)\, (-460)}\; {\rm m\cdot s^{-1}} \\ &\approx (-94.953)\; {\rm m\cdot s^{-1}} \end{aligned}[/tex].
(Negative since the supply would be travelling downwards.)
Let [tex]t[/tex] denote time it takes for the supply to land on the hill after being dropped from the plane. Make use of the SUVAT equation [tex]t = (v_{y} - u_{y}) / (a)[/tex] to find the value of [tex]t\![/tex]:
[tex]\begin{aligned} t &= \frac{v_{y} - u_{y}}{a} \\ &\approx \frac{(-94.953) - 0}{(-9.8)} \; {\rm s}\\ &\approx 9.6890 \; {\rm s} \end{aligned}[/tex].
Apply unit conversion and ensure that [tex]v_{x}[/tex], the horizontal speed of the plane is in the standard unit [tex]{\rm m\cdot s^{-1}}[/tex]:
[tex]\begin{aligned} v_{x} &= \frac{260\; {\rm km}}{1\; {\rm h}} \times \frac{1\; {\rm h}}{3600\; {\rm s}} \times \frac{1000\; {\rm m}}{1\; {\rm km}} \\ &\approx 72.222\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
Under the assumptions, the horizontal speed of the supply will be the same as that of the plane- [tex]v_{x} \approx 72.222\; {\rm m\cdot s^{-1}}[/tex]- until it lands.
While in the air, the supply will travel a horizontal distance of:
[tex]\begin{aligned}x_{x} &= v_{x}\, t \\ &\approx 72.222\; {\rm m\cdot s^{-1}} \times 9.6890\; {\rm s} \\ &\approx 699.76\; {\rm m}\end{aligned}[/tex].
Hence, for the supply to land exactly at the top of the hill, the plane need to drop the supply while at a horizontal distance of approximately [tex]700\; {\rm m}[/tex] away from the hill.
The horizontal distance between the trees and the location where the plane dropped the supply would be approximately [tex](700\; {\rm m} - 30\; {\rm m}) = 670\; {\rm m}[/tex]. The time required for the the supply to reach that horizontal position would be:
[tex]\begin{aligned} t &= \frac{x_{x}}{v_{x}} \approx \frac{669.76\; {\rm m}}{72.222\; {\rm m\cdot s^{-1}}} \approx 9.2736\; {\rm s}\end{aligned}[/tex].
Let [tex]h_{0}[/tex] denote the initial height of the supply (relative to the sea level.) In this question, [tex]h_{0} = 810\; {\rm m}[/tex].
Let [tex]h(t)[/tex] denote the height of the supply (relative to the sea level) after being dropped from the plane for time [tex]t[/tex].
The SUVAT equation [tex]h(t) = (1/2)\, a\, t^{2} + u_{y}\, t + h_{0}[/tex] gives an expression for [tex]h(t)[/tex]. Make use of this equation to find the height of the supply (relative to the sea level) when the supply reach the horizontal position of the trees at [tex]t \approx 9.2736\; {\rm s}[/tex]:
[tex]\begin{aligned} h(t) &= \frac{1}{2}\, a\, t^{2} + u_{y}\, t + h_{0} \\ &= \frac{1}{2}\times (-9.8)\, (9.2736)^{2} + 0\times 9.2736 + 810 \\ &\approx 388.60\; {\rm m} \end{aligned}[/tex].
Note that the altitude of the top of the trees is [tex]350\; {\rm m} + 40\; {\rm m} = 390\; {\rm m}[/tex] relative to the sea level. Since [tex]388.90\; {\rm m} < 390\; {\rm m}[/tex], the supplies will run into the trees.
what are the 4 types of graphs and when do we use each type?
a 0.11 kg tin can is resting on top of a 1.7 m high fence post. a 0.0020 kg bullet is fired horizontally at the can. it strikes the can with a speed of 900 m/s, passes through it, and emerges with a speed of 720 m/s. when the can hits the ground, how far is it from the fence post? disregard friction while the can is in contact with the post.
A mass of can 0.11 kg rests on a fence post 1.7 m high. a 0.0020 kg bullet is fired horizontally at a can. it hits the can at 900 m/s, passes through it, and emerges at 720 m/s.
when the can hits the ground, it is 1.93 m away from the fence post.
for vertical fall,
d = 0.5*g*t^2
1.7 = 0.5*9.8*t^2
t = 0.589 s
use conservation of momentum to calculate the initial horizontal velocity to tin
mb*vbi = mb*vbf + mt*vt
2*10^-3*900 = 2*10^-3*720 + 0.11*vt
vt = 3.273 m/s
horizontal distance = vt*time
= 3.273 m/s * 0.589 s
= 1.93 m
Answer= 1.93 m
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A car travels south at 30 m/s for 5 minutes. How many seconds does it travel
for?
A. 350 s
B. 250 s
C. 200 s
D. 300 s
Answer:
The Answer is D. 300 s because in 5 minutes there are 300 seconds
for the following trajectory, find the speed associated with the trajectory and then find the length of the trajectory on the given interval. r(t), for 0
The Fundamental Theorem of Calculus is used to evaluate the definite integral. A important tool for simulating the dynamics of moving objects and other physical phenomena is the parametric form.
For a parameter t, the dynamics or trajectory of a moving object is given in parametric form. We determine the speed of the object over the range of its parameter using derivatives and a definite integral. The Fundamental Theorem of Calculus is used to evaluate the definite integral. A important tool for simulating the dynamics of moving objects and other physical phenomena is the parametric form.
The velocity function or vector's absolute value or norm can be used to calculate an object's speed.
We obtain the velocity v(t) as the first derivative of the position trajectory given as follows since the velocity is the rate of change of the distance traveled:
v ( t ) = r ′ ( t ) = ⟨ ( 2 t 3 ) ′ , ( − t 3 ) ′ , ( 3 t 3 ) ′ ⟩ = ⟨ 6 t 2 , − 3 t 2 , 9 t 2 ⟩
The speed is given as either its norm or absolute value in the velocity function above as follows:
Speed(t)=∥v(t)∥=∥⟨6t2,−3t2,9t2⟩∥
=√(6t2)2+(−3t2)2(9t2)2=√36t4+9t4+81t4=√126t2(2)
The speed at t = 7 will be (2) from (2).
Speed(7)=√126×72=49√126.
Using (2) and the Calculus Fundamental Theorem, the trajectory's length L can be determined as follows:
L = ∫ 7 0 ∥ r ′ ( t )∥ d t = ∫ 7 0 ∥ v ( t ) ∥ d t = ∫ 7 0 √ 126 t 2 d t = √ 126 [ t 3 3 ] 7 0 = √ 126 [ 7 3 3 − 0 3 3 ] = 7 3 √ 126 3. \s.
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A bus and a Car are traveling along the EDSA having the same velocity. which if the two vehicles would have a greater momentum?a. the busb. the carc. both have the same momentumd. cant be easily determined
The momentum of an object can be calculated with the formula below:
[tex]p=mv[/tex]Where p is the momentum, m is the mass and v is the velocity.
The mass of the bus is greater than the mass of the car.
So, if they have the same velocity, the momentum of the bus will be greater.
Correct option: a.
A wire of a certain length (α = 0.0065 1/°C) has a resistance of 15 Ω at 20°C.
Calculate the temperature at which the resistance will be 22.8 Ω
The temperature at which the resistance will be 22.8 Ω is 100 °C
How to determine the temperatureThe following data were obtained from the question:
Coefficient of epansion (α) = 0.0065 °C¯¹ Original resistance (R₁) = 15 Ω Original temperature (T₁) = 20 °C New resistance (R₂) = 22.8 ΩNew temperature (T₂) =?The new temperature can be obtained as illustrated below:
α = R₂ – R₁ / R₁(T₂ – T₁)
0.0065 = 22.8 – 15 / 15(T₂ – 20)
0.0065 = 7.8 / 15(T₂ – 20)
Cross multiply
0.0065 × 15 (T₂ – 20) = 7.8
0.0975 (T₂ – 20) = 7.8
Divide both side by 0.0975
T₂ – 20 = 7.8 / 0.0975
T₂ – 20 = 80
Collect like terms
T₂ = 80 + 20
T₂ = 100 °C
Thus, the temperature is 100 °C
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An orange dropped from a tree and had a velocity of 8 m/s just before it hits the ground. How far is the ground from orange's starting position?
The orange was at a height of 3.26 m above the ground.
State third equation of motion.The third equation of motion is
v² - u² = 2aS
Given is an orange dropped from a tree and had a velocity of 8 m/s just before it hits the ground.
We can write -
[v] = 8 m/s
[a] = 9.8 m/s²
[u] = 0 m/s
Using third equation of motion, we get -
v² - u² = 2aS
64 - 0 = 2 x 9.8 x S
64 = 19.6 x S
S = 3.26 m
Therefore, the orange was at a height of 3.26 m above the ground.
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a child pushes a merry-go-round that has a diameter of 4.00 m and goes from rest to an angular speed of 17.0 rpm in a time of 45.0 s. 1) calculate the average angular acceleration of the merry-go-round. (express your answer to three significant figures.)
The total tangential speed = 3.56 m/s
, r =2 m ,
t =44sw
=17rpm = 17*2*3.14/60
= 1.7793 rad/swo =
01) from rotational kinematic equation
w =wo+t1.7793 =0 + *44 =0.0404 rad/s^2
(2) from rotationla kinematic equation
w^2 -wo^2 = 21.7793*1.7793
= (2*0.0404*)=39.18 rad
3) v =rw =2*1.7793
tangential speed = 3.56 m/s
What is average angular acceleration?
A spinning object's change in angular velocity per unit of time is expressed quantitatively as angular acceleration, also known as rotational acceleration. It is a vector quantity with either one of two predetermined directions or senses and a magnitude component.
The formula for average angular acceleration is: a = change in velocity divided by change in time. a=(w2−w1)/(t2−t1) a = ( w 2 − w 1 ) / ( t 2 − t 1 ) . w2 represents the final velocity measured in radians per second. w1 represents the initial velocity measured in radians per second.
Therefore, tangential speed = 3.56 m/s
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as the pressure of an enclosed gas decreases to half its original value, what happens to the volume of the gas if temperature is held constant?(1 point)
Answer:
See below
Explanation:
P1V1/T1 = P2V2/T2 IF T1 = T2
Then
P1V1 = P2V2 P2 = 1/2 P1
P1V1 = 1/2 P1 V2 DIVIDE BOTH SIDES BY P1
V1 = 1/2 V2 MULTIPLY BOTH SIDES BY 2
2V1 = V2 OR V2 = 2 V1 THE NEW VOLUME IS TWICE ORIGINAL
An object moves 60.0 m on a bearing (angle from North) of 60.0°. If the object then moves 30.0 m North, how far is it from the start point? You may use a scale diagram or trigonometry to answer this question.
Answer: x(t2)−x(t1) over the time interval [t1,t2]
Explanation: hope this helps
a heavy seesaw is out of balance. a lightweight girl sits on the end that is tilted downward, and a heavy boy sits on the other side so that the seesaw now balances. if the boy and girl both move forward so that they are one-half their original distance from the pivot point, what will happen to the seesaw?
The side the girl is sitting on will once tilt down.
The seesaw's left and right sides' opposing torques are balanced by the torque acting on the left side. The force will act on the seesaw itself at either the right or left side of the seesaw's center of gravity.
Formula: The expression for the amount of torque operating on a body is
torque =r F
where; r=position vector and F= force
if the force is applied at a right angle to the direction of r and the body is at a distance r from the pivot.
The seesaw's length is assumed to be l, and its left side's mass is assumed to be greater than its right side's mass.
[tex]m_{L}[/tex] > [tex]m_{R}[/tex]
Let the weight of the hefty body be [tex]M[/tex], and the mass of the girl be [tex]m[/tex].
The seesaw's left and right side rods' centers of gravity will be positioned [tex]\frac{l}{4}[/tex] of an inch apart from the center (or pivot). The girl and body are initially positioned [tex]\frac{l}{2}[/tex] apart from the pivot. as a result, initially
torque of left side= torque of right side
([tex]\frac{l}{4}[/tex])[tex]m_{L}[/tex][tex]g[/tex]+([tex]\frac{l}{2}[/tex])[tex]mg[/tex]=([tex]\frac{l}{4}[/tex])[tex]m_{R}[/tex][tex]g[/tex]+([tex]\frac{l}{2}[/tex])[tex]Mg[/tex]
we get,
[tex]\frac{m_{L} -m_{R} }{2}[/tex]= [tex]M-m[/tex]
Asking the body and the girl to move ([tex]\frac{l}{4}[/tex]) closer to the pivot will result in the following calculation of the net torque:
net torque= torque of left - torque of right
The seesaw is tilted to the left when we believe that the torque is greater on the left side.
τnet=[tex]\frac{l}{4} m_{L}[/tex][tex]g[/tex]+[tex]\frac{l}{4}[/tex][tex]mg[/tex]−[tex]\frac{l}{4} m_{R}[/tex][tex]g[/tex]−[tex]\frac{l}{4}[/tex][tex]Mg[/tex]
We obtain the following values by replacing the value of [tex]m_{L} -m_{R}[/tex] in terms of [tex]m[/tex] and [tex]M[/tex]
therefore,
τnet=[tex]lg(\frac{M-m}{4} )[/tex]
Our presumption that the left side will tilt because M is heavy and m is tiny is valid since the torque on the left side will predominate.
The girl will again be sitting on the side that is tilting downward because she was previously sat on the left.
REMARK: We would have discovered negative net talk if we had thought that the right hand side was tilting or had more torque. As a result, in such scenario, we would have discovered that the left side tilts, contrary to what we had assumed. Both strategies are thus appropriate. But it is important to avoid guessing at all costs while answering these questions because doing so reduces the likelihood of getting the correct response.
what is torque?The force that can cause an object to rotate along an axis is measured as torque. An object acquires angular acceleration due to torque.
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A projectile is fired into the air from the edge of a 100 m high cliff at an angle of 37° above the horizontal. The projectile hits a target 400 m away from the base of the cliff. What is the initial velocity of the projectile, vo? (Neglect air friction and assume x-axis to be horizontal and y-axis to be vertical). r 100m 400 m m
The initial velocity of the projectile, Vo is 25√5 m/s.
What is initial velocity?
initial velocity is the velocity at time interval t=0 and it's representative as u.
It is velocity which the motion start.
Sol-as per the question
Sx=400 Vx=Vo cos 37°=4/5Vo
Sy= -100 Vy=Vo sin 37° 3/5Vo ay=-10 m/s^2
Let the time of flight be t
Sx=u^xt +1/2axt^2
=>400= 4/5Vot----------(eq1)
Sy=uyt +1/2 ay t^2
=>-100=3/5 Vot -5t^2-----(eq2)
By solving this we are get-
Vo=25√5 m/s
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How long does it take a car to travels 40 m/s to 80 m/s with an acceleration of 20 m/s2 on a highway?
What is the distance from the moon to the sun?
Answer:
About 150 million kilometers
Explanation:
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What is the instantaneous velocity of a freely falling object 15 s after it is released from a position of rest? Express your answer to two significant figures and include the appropriate units. v = nothing nothing Request Answer Part B What is its average velocity during this 15- s interval? Express your answer to two significant figures and include the appropriate units. v = nothing nothing Request Answer Part C How far will it fall during this time? Express your answer to two significant figures and include the appropriate units. d = nothing nothing
Using concepts of acceleration, velocity and position, it is found that:
A. The instantaneous velocity in 15 seconds is of -147 m/s².
B. The average velocity during this interval is of -9.8 m/s.
Cc. The distance fallen during this time is of 1102.5 meters.
Acceleration, velocity and positionThe velocity after t seconds is given by the following equation:
v(t) = v(0) + at.
In which:
v(0) is the initial velocity.a is the acceleration.In the context of this problem, the values of these parameters are given as follows:
v(0) = 0, as the object was released from rest.a = -9.8, as the object is freely falling in the air, hence the acceleration is the gravity.Then the velocity equation is given as follows:
v(t) = -9.8t.
In 15 seconds, the instantaneous velocity is given as follows:
v(15) = -9.8(15) = -147 m/s².
The average velocity during this interval is of given by the change in velocity divided by the change in time, hence:
v(0) = 0.v(15) = -147.Average velocity = (-147 - 0)/(15 - 0) = -9.8 m/s.
The position function is the integral of the velocity function, hence:
s(t) = -4.9t². (integral of -9.8t = -9.8t²/2 = -4.9t²).
The distance fallen in 15 seconds is:
s(t) = -4.9(15)² = -1102.5 (1102.5 meters fallen).
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a star of apparent magnitude 1 appears a star of apparent magnitude 1 appears either brighter or fainter than a star of apparent magnitude 2, depending on the distance to the stars. brighter than a star of apparent magnitude 2. fainter than a star of apparent magnitude 2. farther away than a star of apparent magnitude 2.
The likelihood that this star is not "venerable," or old, is low; but, we would need more details, particularly the star's spectral class, to be certain.
The relationship between the absolute magnitude M and the apparent magnitude m and luminosity distance DL is as follows:
M = m 5log10DL10 pc
The distance of 100 pc in this instance provides us a fairly straightforward relationship:
M=m−5=−1.0
Let's first calculate the star's luminosity in relation to the Sun's luminosity to understand why. This star is 5.7 magnitudes brighter than the Sun, which has an absolute magnitude of +4.7. Using the traditional scale where 5 magnitude equal a brightness factor of 100: LL=100(5.7)/5=190.5 where the subscript denotes the Sun.
Our star therefore has 200 times the brightness of the Sun. Depending on the spectral class, we might determine whether this star was a hot main sequence star or a low-temperature red giant. It's likely to be the latter if we were to attempt a guess. Smaller stars have not had the time to develop into red giants during the existence of our galaxy.
Larger star than the Sun will go through the red giant phase of their evolution much more quickly. In just 40 million years, even our Sun will increase from 100 to 2,000 times its current brilliance.
Therefore, if a red giant is visible to us at only 200 times the solar luminosity, we would have to observe it very quickly in cosmic time.
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PLEASE SOLVE NO I NEED IT DONE IM ON A DEADLINE!!! a bird flying 3.45 m / s directly north feels a wind directly East that accelerates at 0.558 m / S^2. What is its velocity 5.25 s later
The velocity of the bird can be obtained as 6.38 m/s.
What is the velocity?Let us recall that the acceleration is the rate at which the velocity is changed. In this case, we are told that; a bird flying 3.45 m / s directly north feels a wind directly East that accelerates at 0.558 m / S^2. Having seen the question we can pick the following parameters out from the question;
Initial velocity u = 3.45 m / s
Acceleration a = 0.558 m / S^2
Time taken = 5.25 s
Given that;
v = u + at
v = final velocity
u = initial velocity
a = acceleration
t = time taken
v = 3.45 + (0.558 * 5.25)
v = 6.38 m/s
We can see that as the bird tends to be flying as we can see in the question that have been shown above that the final velocity that it can have after 5.25 seconds is 6.38 m/s.
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3. Hakeem was participating in a psychology experiment. For the first ten minutes Hakeem would hear a click then a puff of air would blow in Hakeem's eye. His eyes would blink and tear up. For the second ten minutes Hakeem would hear a click but the puff of air would only blow in his eye every third time. Hakeem knows this but his eyes would blink and tear up after every click.
UCS: UCR:
CS: CR:
NS:
In the psychology experiment in which Hakeem participates, UCS is click, UCR is blink and tear up, CS is blowing a puff of air, CR is thought of the puff of air and NS is the click before puff of air was introduced.
In psychology the abbreviations of the given terms are as follows:
UCS - Unconditioned Stimulus UCR - Unconditioned ResponseCS - Conditioned Stimulus CR - Conditioned ResponseNS - Neutral StimulusHere the Unconditioned Response is the blinking and tearing up because he does that even though there is no puff of air. Since clicking provokes the Unconditioned Response, clicking is the Unconditioned Stimulus. The clicking is paired with puff of air for multiple times, that is why Unconditioned Response is stimulated. So the puff of air is the Conditioned Stimulus.
Therefore,
UCS - Clicking UCR - Blinking and tearing upCS - Blowing of puff of air in the eyeCR - Thought of puff of air being blown into the eye NS - Clicking before any puff of air was blownTo know more about stimulus and response
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how would i find acceleration? im really confused rn. i can solve it, i just cant remember the right equation. help
Answer:
See below
Explanation:
acceleration = change in velocity / change in time
Or F = m * a
So the first one Accel = 40 m/s / 20 s = 2 m/s^2
Or 100 N = 50 * a A = 2 m/s^2 Same answer two different ways...
The others are similar ......
if 10 light rays shined on a mirror what would happen to the 10 light rays?
If 10 light rays shined on a mirror than the 10 light rays light will converge or diverge.
What is mirror?A mirror is a wave reflector. Light consists of waves, and light waves reflect from the flat surface of a mirror.
There are two types of mirror concave and convex.
When parallel light beams strike a concave mirror, all of the rays that are reflected meet at the same location (focus). A concave mirror is also known as a converging mirror as a result.
Parallel light beams that strike a convex mirror's surface and are reflected appear to meet (diverge) at a single location, although they do not. so convex mirror is defined as a diverging mirror as a result.
So, rays of light will get converged or diverged.
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Light travels at a speed of about 3.0 108 m/s.(a) How many miles does a pulse of light travel in a time interval of 0.1 s, which is about the blink of an eye?Δx = mi(b) Compare this distance to the diameter of Earth. (Use 6.38 106 m for the radius of the Earth.)ΔxDE =
Given:
Speed of light = 3 x 10⁸ m/s
Let's solve for the following:
• (a). How many miles does a pulse of light travel in a time interval of 0.1 s, which is about the blink of an eye?
Apply the formula:
[tex]\Delta x=v*t[/tex]Where:
v is the speed of light
t is the time.
Thus, we have:
[tex]\begin{gathered} \Delta x=3.0\times10^8*0.1 \\ \\ \Delta x=3.0\operatorname{\times}10^7\text{ m} \end{gathered}[/tex]Now let's convert the answer from meters to miles.
Where:
1 mile = 1609.34 meters
[tex]\begin{gathered} 3.0\times10^7=\frac{3.0\times10^7}{1609.34} \\ \\ =18641.14\text{ mi} \end{gathered}[/tex]Δx = 18641.14 mi
• (b). Compare this distance to the diameter of Earth.
Apply the formula:
[tex]\frac{\Delta x}{D_E}=\frac{\Delta x}{2*r}[/tex]Where:
r = 6.38 x 10⁶ m.
Thus, we have:
[tex]\frac{\Delta x}{D_E}=\frac{3.0\times10^7}{2*6.38\times10^6}=2.35[/tex]ANSWER:
• (a). 18641.14 mi
,• (b). 2.35
on a two lane highway, a car is following a pickup truck. at one instant, the car has a speed of 32 m/s and is 184 m behind the truck. at the same time, the truck has a speed of 28 m/s. if neither vehicle accelerates, how long will it take the car to catch up to the truck?
The vehicle traveling in the opposing direction travels the following distance in the same amount of time: 200/9 (2.5) equals 99.38 meters. The safe separation between the two vehicles is therefore: 129.38 + 99.38 = 228.76 meters.
To obtain the velocity function, we must calculate the integral of the acceleration function:
V ( t ) = ∫ a ( t ) d t V ( t ) = ∫ 3 d t = 3 t + c 1
To find the integral's constant, we use our initial conditions as a guide. As of now
t = 0
The speed is 80 km per hour.
80 km/h is equal to (80)(1000)/3600, or 200/9 meters per second.
V ( 0 ) = 3 ( 0 ) + c 1
200/ 9 = 0 + c 1
c 1 = 200/ 9
The distance the vehicle travels in
the number of seconds required to travel the same distance as the truck plus 30 meters (15 meters in front of and 15 meters behind the truck).
The distance covered by the vehicle is
In the same number of seconds, 200/9 t.
This is the amount of time it takes for the car to overtake and pass the truck.
15 metres.
To do this, the automobile travels a distance of:
S ( t ) = 3 /2 ( 20 ) + 200/ 9 ( 2 5 ) = 129.38 meters
The vehicle traveling in the opposing direction travels the following distance in the same amount of time:
99.38 meters is equal to 200/9 ( 2 5).
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