Answer:
atoms cannot go bad
Explanation:
Because they stay alive and get good nutriants
Let's assume raspberries are 10 wt% protein solids and the remainder water. When making jam, raspberries are crushed and mixed with sugar, in a 45:55 berry to sugar ratio, by mass. Afterward, the mixture is heated, boiling off water until the remaining mixture is 0.4 weight fraction water, resulting in the final product, jam. How much water, in kilograms, is boiled off per kilogram of raspberries processed
Answer:
The mass of water boiled off is [tex]0.0 \overline{185}[/tex] kg
Explanation:
The given percentage by weight of protein solids in raspberries = 10 weight%
The ratio of sugar to raspberries in ja-m = 45:55
The mass of the mixture after boiling = 0.4 weight fraction water
Let 's' represent the mass of sugar in the mixture, and let 'r' represent the mass of raspberry
The mass of raspberry, r = 1 kg
The percentage by weight of water in raspberry = 90 weight %
The mass of water in 1 kg of raspberry = 90/100 × 1 kg = 0.9 kg
The ratio of the mass of sugar to the mass of raspberry in jam = r/s = 45/55
∴ s = 1 kg × 55/45 = 11/9 kg
The mass of the mixture before boiling = 1 kg + 11/9 kg = 20/9 kg
The weight fraction of water in the remaining mixture after boiling = 0.4 weight fraction
Let 'w' represent the mass of water boiled off, we have;
(0.9 - w)/(20/9 - w) = 0.4
(0.9 - w) = 0.4 × (20/9 - w)
0.9 - w = 8/9 - 0.4·w
9/10 - 8/9 = w - 0.4·w = 0.6·w = (6/10)·w
(81 - 80)/(90) = (6/10)·w
1/90 = (6/10)·w
w = ((10/6) × 1/90) = 1/54
w = 1/54
The mass of water boiled off, w = (1/54) kg = [tex]0.0 \overline{185}[/tex] kg
Suppose you are standing in front of a flat mirror which is mounted to a vertical wall. For this problem you may suppose that your height is 1.70 m and your eyes are 12 cm below the top of your head. What is the smallest mirror that will still allow you to see the full length of your body
Answer:
The right approach is "0.85 m".
Explanation:
According to the question, the diagram will is provided below.
So that as per the diagram,
The values will be:
My height,
AO = 1.70 m
My eyes at,
AB = 12 cm
i.e.,
= 0.12 m
As we can see, the point of incidence lies between the feet as well as the eyes, then
BO = 1.58 m
Now,
⇒ [tex]O'D = \frac{1.58}{2}[/tex]
[tex]=0.79 \ m[/tex]
The point of incidence of the ray will be:
⇒ [tex]CO'=1.70-\frac{0.12}{2}[/tex]
[tex]=1.70-0.06[/tex]
[tex]=1.64 \ m[/tex]
Hence,
The smallest length of the mirror will be:
= [tex]CO'-O'D[/tex]
On substituting the values, we get
= [tex]1.64-0.79[/tex]
= [tex]0.85 \ m[/tex]
What are Heredity and Punnett Squares?
Answer: A Punnett square can be used to predict genotype and phenotypes of offspring from genetic crosses. ... In the P generation, one parent has a dominant yellow phenotype and the genotype YY, and the other parent has the recessive green phenotype and the genotype yy.
Explanation:
The acceleration used for the height is 9.81 m/s2 because it is the acceleration due to gravity. (true or false)
Answer:
true
Explanation:
Gravity (or the acceleration due to gravity) is 9.81 meters per second squared, on the surface of Earth, because of the size of Earth and the distance we are on its surface from its center.
)) What do these two changes have in common?
mixing chocolate syrup into milk
rain forming in a cloud
) Select all that apply.
Both involve chemical bonds breaking.
Both are changes of state.
Both are only physical changes.
Both are chemical changes.
Answer:
Both are only physical changes
Explanation:
A physical change is a change that does not involve or alter the chemical composition of the substances involved. Physical changes form no new substance and can be easily separated into individual constituents. Example of physical changes are change in state, boiling, melting etc.
According to this question, two processes were given as follows:
1. mixing chocolate syrup into milk
2. rain forming in a cloud
These two processes are similar in the sense that they are both examples of physical changes.
The new roller coaster at Carowinds flies along at 80 m/s. How long does it take to
travel 16000 meters of track?
Which has the greater kinetic energy–a 1-ton car moving at 30 m/s or a half-ton car
moving at 60 m/s?
A) Both have the same kinetic energy
B) The half-ton car
C) The 1-ton car
D) It cannot be determined because the mass of the cars can't be found
Answer:
B). half-ton car
Explanation:
We can find Kinectic using below expresion.
m= mass of the object
v= velocity
mass (1) 1 ton= 1000kg
Mass(2)0.5= 500kg
K.E= 1/2 mv^2
For 1 ton car
Substitute the values
Kinectic energy= 1/2( 1000) 30^2
= 450,000J
For 0.5 ton car
Kinectic energy= 1/2(500)60^2
= 900,000J
Hence, half-ton car has greater Kinectic energy
he inductance of a tuning circuit of an AM radio is 4 mH. Find the capacitance of the circuit required for reception at 1200 kHz.
Answer:
4.4pF
Explanation:
the capacitance of the circuit required for reception is given:
wL = [tex]\frac{1}{wC}[/tex]
w = 2π [tex]f[/tex]
Using both equation
Capacitance is given
C = 1 - 4π2 f2 L
1- 4×9.8969×144×10 10 ×0.004
=4.4pF
A typical laboratory centrifuge rotates at 4000 rpm. Testtubes have to be placed into a centrifuge very carefully because ofthe very large accelerations.
Part A) What is the acceleration at the end of a test tubethat is 10 cm from the axis of rotation?
Part B) For comparison, what is the magnitude of theacceleration a test tube would experience if dropped from a heightof 1.0 m and stopped in a 1.0-ms-long encounter with a hardfloor?
Answer:
A) a_c = 1.75 10⁴ m / s², B) a = 4.43 10³ m / s²
Explanation:
Part A) The relation of the test tube is centripetal
a_c = v² / r
the angular and linear variables are related
v = w r
we substitute
a_c = w² r
let's reduce the magnitudes to the SI system
w = 4000 rpm (2pi rad / 1 rev) (1 min / 60s) = 418.88 rad / s
r = 1 cm (1 m / 100 cm) = 0.10 m
let's calculate
a_c = 418.88² 0.1
a_c = 1.75 10⁴ m / s²
part B) for this part let's use kinematics relations, let's start looking for the velocity just when we hit the floor
as part of rest the initial velocity is zero and on the floor the height is zero
v² = v₀² - 2g (y- y₀)
v² = 0 - 2 9.8 (0 + 1)
v =√19.6
v = -4.427 m / s
now let's look for the applied steel to stop the test tube
v_f = v + a t
0 = v + at
a = -v / t
a = 4.427 / 0.001
a = 4.43 10³ m / s²
1. Is it possible for the ball to move so quickly that the angle between the cable and vertical post stays at ninety degrees?
2. When the ball is moving in a horizontal circle, what vertical force (or component) balances gravity?
3. What happens to the centripetal force as the length of the cord increases?
Answer:
Tetherball is an interesting game in which two players tries to hit the ball hard so that it goes around the
pole.Each time the player hits the ball, it's orbit rises higher off the ground.Let's understand the physics
behind this.The motion of a tetherball is governed by two forces.These two forces combine to generate a
net force, i.e. centripetal force.If the ball is moving more quickly, it requires a greater centripetal force,
which in turn requires a greater tension force.Since the ball's weight hasn't changed, the angle of the
tension force changes until the ball is in vertical equilibrium.
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Ball Mass : This slider controls the mass of the ball. A ball with more mass will have more inertia, requiring
a greater net force to accelerate it. A ball with more mass will ALSO have a greater gravitational force
acting on it. Watch both of these effects occur when you manipulate this slider.
Cable Length : This slider controls the length of the cable. A longer cable is capable of allowing a greater
circular radius of motion for the ball. It is important to remember that the radius of the circular motion is
NOT equal to the length of the cable. Instead, if you want to understand the size of the circle of the ball's
motion, ignore the cable and just imagine the path of the ball.
Ball Speed : This slider controls the speed of the ball - imagine a kid just hit the ball and it sped up. A ball
moving more quickly is also accelerating more quickly because its velocity is changing as it moves in a
circle (remember that changes in DIRECTION of velocity 'count' as changes to velocity).
Force Diagram : This allows you to turn on or off the diagram of the forces acting on the ball. Look for the
ball to be in vertical force balance, which means the vertical component of tension is canceled by the
gravitational force. The ball should NOT be in horizontal force balance - it is accelerating towards the center
of the circle! It is important to note that this free body diagram should really be moving with the ball so that
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the tension force always points along the cord - we are just showing the forces at the moment the ball is at
the furthest-right on this screen.
Centripetal force vs Tetherball speed : This is a plot of centripetal force required to keep the ball in
circular orbit about the pole as a function of its speed. As expected, a more quickly-moving ball is changing
in velocity more often in a given amount of time, and so is accelerating more. This greater (centripetal, or
center-pointing) acceleration requires a greater net force.
Explanation:
A basketball player jumps straight up for a ball. To do this, he lowers his body 0.310 m and then accelerates through this distance by forcefully straightening his legs. This player leaves the floor with a vertical velocity sufficient to carry him 0.940 m above the floor. (a) Calculate his velocity (in m/s) when he leaves the floor. m/s (b) Calculate his acceleration (in m/s2) while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.310 m. m/s2 (c) Calculate the force (in N) he exerts on the floor to do this, given that his mass is 106 kg. N
Answer:A)u =4.295m/s , B)a = 29.746m/s² C) F=3,153N
Explanation:
Using the kinematic expression
v² = u² - 2as
where
u = initial velocity
v = final velocity
s = distance
g = acceleration due to gravity .
Given that he reaches a height of 0.940 m above the floor,
the final velocity = 0
Here, acceleration due to gravity is acting in opposite the initial direction of motion. So, a=-9.81 m/s.
v² = u² + 2as
0² - u² = 2 (- 9.81) × 0.940
- u² = 2 × - 9.81 × 0.920
- u² = -18.4428
cancelling the minus in both sides , we have that
u² = 18.4428
u = √18.4428
u =4.295m/s
(b) His acceleration (in m/s2) while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.310 m. m/s2
Using v² = u² + 2as
where u = initial speed of basketball player before lengthening = 0 m/s,
v = final speed of basketball player after lengthening = 4.295m/s,
a = acceleration while straightening his legs
s = distance moved during lengthening = 0.310m
v² = u² + 2as
a = (v² - u²)/2s
a = (4.29m/s)² - (0 m/s)²)/(2 × 0.310m)
a = (18.4428 m²/s² - 0 m²/s²)/(0.62 m)
a = (18.4428 m²/s²/(0.62 m)
a = 29.746m/s²
c) The force (in N) he exerts on the floor to do this, given that his mass is 106 kg. N
Force= mass x acceleration.
F = 106 kg X 29.746m/s²
F = 3,153.076 rounded to 3,153N
A researcher plans to release a weather balloon from ground level, to be used for high-altitude atmospheric measurements. The balloon is spherical, with a radius of 2.00 m, and filled with hydrogen. The total mass of the balloon (including the hydrogen within it) and the instruments it carries is 20.0 kg. The density of air at ground level is 1.29 kg/m3. (a) What is the magnitude of the buoyant force (in N) acting on the balloon, just after it is released from ground level
Answer:
B = 423.64 N
Explanation:
The thrust force in a fluid is equal to the weight of the displaced fluid,
B = ρ g V
where the density is that of air ρ=1.29 kg / m³ and the volume of the spherical balloon is
V = [tex]\frac{4}{3} \pi r^3[/tex]
v = 4/3 π 2³
V = 33.51 m³
let's calculate the thrust
B = 1.29 9.8 33.51
B = 423.64 N
What is one disadvantage of sending information over long distances
wirelessly using digital signals?
O A. The signals become weaker the farther the receiver is from the
source.
B. The farther the signals travel, the more slowly they move.
C. The signals become stronger the farther the receiver is from the
source.
O D. The farther the signals travel, the easier they are to detect.
The disadvantage of sending information over long distances wirelessly using digital signals is "the signals become weaker the farther the receiver is from the source."
Since most of the signal which we use for communication are radio signal Radio signal are basically electromagnetic waves.As the wave moves forward it looses its amplitude.So basically radio signal becomes weaker for long distance.What are radio signals?Radio signals or radio waves are a form of electromagnetic wave. Although this may sound complicated, it is possibly sufficient to say that these waves have both electric and magnetic components. They are the same as light rays, ultra-violet and infra-red. The only difference is in the wavelength of the waves.Thus , The disadvantage of sending information over long distances wirelessly using digital signals is "the signals become weaker the farther the receiver is from the source."
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A truck with a mass of 1370 kg and moving with a speed of 12.0 m/s rear-ends a 593 kg car stopped at an intersection. The collision is approximately elastic since the car is in neutral, the brakes are off, the metal bumpers line up well and do not get damaged. Find the speed of both vehicles after the collision in meters per second.
Answer:
speed of car after collision, v2 =16.1 m/s and of the truck, v1 = 4.6 m/s
Explanation:
Given:
mass of truck M = 1370 kg
speed of truck = 12.0 m/s
mass of car m = 593 kg
collision is elastic therefore,
Applying law of momentum conservation we have
momentum before collision = momentum after collision
1370×12 + 0( initially car is at rest) = 1370×v1+ 593×v2 ....(i)
Also for a collision to be elastic,
velocity of approach = velocity of separation
12 -0 = v2-v1 ....(ii)
using (i) and (ii) we have
So speed of car after collision, v2 =16.1 m/s and of the truck, v1 = 4.6 m/s
Suppose you are an observer standing at a point along a three-lane roadway. All vehicles in lane 1 are traveling at 50 mi/hr with a constant 5-second time headway between them. All vehicles in lane 2 are traveling at 55 mi/hr with a constant 6-second time headway between them. All vehicles in lane 3 are traveling at 60 mi/hr with a constant 10-second time headway between them. You collected spot speed data for all vehicles as they crossed your observation point for 75 minutes. After 10 minutes, vehicles in lane 1 stopped arriving, and after 30 minutes, vehicles in lane 2 stopped arriving. What is the flow and density of the observed traffic stream
Answer:
lane 3 Ф = 450 vehicles, ρ = 0.1 vehicle / s
lane 2 Ф_{average} = 300 vehicles, ρ _{average} = 6.66 10⁻² vehicles/s
lane 1 Ф_{average} = 300 vehicles, ρ_{average} = 2.66 10⁻² vehicle/s
Explanation:
Before solving this exercise we must clarify the concepts the flow is defined as the occurrence of an event in a time interval, in this case the passage of a car through time
Flux Density is the flux between unit area or unit time
Let's start by calculating the calculation for lane 3
the flow.
Let's use a direct rule of proportions (rule three) if the number of vehicles per unit of time (t₀ = 10s), for the observation time how many vehicles passed in the observation time (t_total = 75 * 60 = 4500 s)
Ф = 4500 s (1 vehicle / 10 s)
Ф = 450 vehicles
The flux density is the flux per unit area, in this case the area is not indicated, so we can define the flux density as the flux per unit of time.
ρ = 450/4500
ρ = 0.1 vehicle / s
Lane 2
we look for the flow
we can have separates the interval into two parts
* for the first t₁1 = 30 * 60 = 1800 s
Ф₁ = 1800 s (1 vehicle / 6s)
Ф₁ = 300 vehicles during t₁
* for the rest of the time t₂ = 4500-1800 = 2700 s
Ф₂ = 0
the average density is the total number of vehicles between the total time
#_ {vehicle} = 300 +0
Ф_{average) = # _vehicle
Ф_{average} = 300 vehicles in all time
The density is
ρ 1 = fi1 / t1
ρ1 = 300/1800
ρ1 = 1.66 10-1 vehicles / s
the average density is
ρ_{average} = [tex]\frac{\phi_1 + \phi_2}{ t_{total}}[/tex]
ρ _{average} = (300 +0) / 4500
ρ _{average} = 6.66 10⁻² vehicles / s
Lane 1
flow
* first time interval t₁ = 10 * 60 = 600 s
Ф₁ = 600 s (1 vehicle / 5s)
Ф₁ = 120 vehicles in interval t₁
* second interval t₂ = 4500-600 = 3900 s
Ф2 = 0
average flow
Ф = Ф1 + Ф2
Ф = 120 vehicles at all time
Density
* first interval
ρ₁ = 120/600
ρ₁ = 0.2 vehicles / s
* second interval
ρ₂ = 0
average density
ρ+{average} = 120/4500
ρ_{average} = 2.66 10⁻² vehicle/s
HELP PLEASE DUE IN 3 MINUTES
Answer:
Tectonic Plate Movement
Explanation:
Each continent and ocean sits on its own tectonic plate which floats on the Earths upper mantle. They move very little over time.
Answer:
tectonic plates movement
Under the influence of its drive force, a snowmobile is moving at a constant velocity along a horizontal patch of snow. When the drive force is shut off, the snowmobile coasts to a halt. The snowmobile and its rider have a mass of 128 kg. Under the influence of a drive force of 195 N, it is moving at a constant velocity whose magnitude is 5.90 m/s. The drive force is then shut off. Find (a) the distance in which the snowmobile coasts to a halt and (b) the time required to do so.
Answer:
a) Δx = 11.6 m
b) t = 3.9 s
Explanation:
a)
Since the snowmobile is moving at constant speed, and the drive force is 195 N, this means that thereis another force equal and opposite acting on it, according to Newton's 2nd Law, due to there is no acceleration present in the horizontal direction .This force is just the force of kinetic friction, and is equal to -195 N (assuming the positive direction as the direction of the movement).Once the drive force is shut off, the only force acting on the snowmobile remains the friction force.According Newton's 2nd Law, this force is causing a negative acceleration (actually slowing down the snowmobile) that can be found as follows:[tex]a = \frac{F_{fr} }{m} = \frac{-195N}{128kg} = -1.5 m/s2 (1)[/tex]
Assuming the friction force keeps constant, we can use the following kinematic equation in order to find the distance traveled under this acceleration before coming to an stop, as follows:[tex]v_{f} ^{2} -v_{o} ^{2} = 2* a* \Delta x (2)[/tex]
Taking into account that vf=0, replacing by the given (v₀) and a from (1), we can solve for Δx, as follows:[tex]\Delta x =- \frac{v_{o}^{2}}{2*a} =- \frac{(5.90m/s)^{2}}{2*(-1.5m/s2)} = 11.6 m (3)[/tex]
b)
We can find the time needed to come to an stop, applying the definition of acceleration, as follows:[tex]v_{f} = v_{o} + a*\Delta t (4)[/tex]
Since we have already said that the snowmobile comes to an stop, this means that vf = 0.Replacing a and v₀ as we did in (3), we can solve for Δt as follows:[tex]\Delta t = \frac{-v_{o} }{a} = \frac{-5.9m/s}{-1.5m/s2} = 3.9 s (5)[/tex]
A cat dozes on a stationary merry-go-round, at a radius of 7.0 m from the center of the ride. The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 6.9 s. What is the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding
Answer:
0.6
Explanation:
Given that :
Radius, R = 7m
Period, T = 6.9s
The Coefficient of static friction, μs can be obtained using the relation :
μs = v² / 2gR
Recall, v = 2πR/T
μs becomes ;
μs = (2πR/T)² / 2gR
μs = (4π²R² / T²) ÷ 2gR
μs = (4π²R² / T²) * 1/ 2gR
μs = 4π²R / T²g
μs = 4π²*7 / 6.9^2 * 9.8
μs = 28π² / 466.578
μs = 276.34892 / 466.578
μs = 0.5922887
μs = 0.6
why do the stars rotate
Answer:
Angular momentum
Explanation:
Stars are formed as a result of a collapse of a low-temperature cloud of gas and dust. During the colapse conservation of angular momentum causes any small net rotation of the cloud to increase thus forcing the material into rotating
A block starting from rest slides down the length of an 18 plank with an acceleration of 4.0 meters per second. How long does the block take to reach the bottom?
Answer:
[tex]\boxed{\text{\sf \Large 3.0 s}}[/tex]
Explanation:
Use distance formula
[tex]\displaystyle d=ut+\frac{1}{2} at^2[/tex]
[tex]u= \text{\sf initial velocity}\\d= \text{\sf distance}\\a= \text{\sf acceleration}\\t= \text{\sf time taken}[/tex]
[tex]\displaystyle 18=0 \times t+\frac{1}{2} \times 4 \times t^2[/tex]
[tex]t=3[/tex]
A block starting from rest slides down the length of an 18 m plank with an acceleration of 4.0 meters per second square. Time taken by the block to reach at the bottom is 3 sec.
What is acceleration?The rate at which an object changes its velocity is known as acceleration. Acceleration is a vector quantity. If an object's velocity is changing then it is accelerating and an object with a constant velocity is not accelerating.
The speed at which something moves in a specific direction is known as its velocity. As an illustration, think of the speed of a car travelling north on a highway or the speed at which a rocket takes off.
Given that in the question block slides down an 18 m plank length with an acceleration of 4.0 meters per second square when it begins at rest.
Using equation of motion,
S = ut + (1/2)at²
s is distance, s = 18 m
u is initial velocity, u = 0
a is acceleration, a = 4 m/sec²
t is time
18 = 0*t + (1/2)*4*t
solving we get t = 3 sec.
Time taken by the block to reach at the bottom is 3 sec.
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A spring in a toy gun has a spring constant of 10 N/m and can be compressed 4 cm.
It is then used to shoot a 1 g ball out of the gun. Find the velocity of the ball as it
leaves the gun
A metal pot feels hot to the touch after a short time on the shove. what type of material is the metal pot
how were the outer planets formed?
Answer:
All planets including the outer larger planets were formed at the same time somewhere around 4.5 Billion years ago.
Explanation:
the young sun drove away most of the gas from the inner solar system, leaving behind the rocky cores also known as the terrestrial planets.
A toy projectile is fired vertically from the ground upward with a velocity of +29 meters per second. It arrives at its maximum altitude in 3.0 seconds. How high does the projectile go?
Answer:
[tex]\boxed{\text{\sf \Large 42 m}}[/tex]
Explanation:
Use height formula
[tex]\displaystyle \sf H=\frac{u^2 sin(\theta)^2}{2g}[/tex]
u is initial velocity
θ = 90° (fired vertically upward)
g is acceleration of gravity
[tex]\displaystyle \sf H=\frac{29^2 \times sin(90 )^2}{2 \times 10}=42.05[/tex]
PLEASE ANSWER THIS WILL MARK AS BRAINLIEST PLEASE USE TRUTHFUL ANSWERS PLEASE
Which best explains why species living in Australia are found nowhere else on Earth? This is an example of Geologic Evolution.
A.
Australia has an ecosystem different from any other area on Earth.
B.
Humans have genetically altered many Australian species in laboratories.
C.
Australian species were genetically altered after a comet hit the landmass.
D.
Australia separated from other continents and species there evolved independently.
10. A person is about to kick a soccer ball. Consider the leg and foot to be a single rigid body and that it rotates about a fixed axis through the knee joint center. Immediately prior to impact the leg and foot are in the vertical direction and the distal end of the foot has an acceleration of 10 g in the horizontal direction. The muscle force vector makes an angle of 15 degrees with the vertical and has a moment arm of 5 cm from the knee joint center. Assume the person is 1.7 m tall and has a mass of 75 kg. Find the force in the muscle
Answer:
i don't know but i hope you get it right
Explanation
PLEASE HELP!
what would the answer be?
Answer:
Tie aluminium foil on each end of the battery using rubber band, then use copper wire to attach the led to the aluminium foil...... probably would work
Fat Albert (the TV show character) runs up the stairs on Monday. On Tuesday, he walks up the same set of stairs. Which day did he do more work?
Answer:
Tuesday bc instead of running he/she was walking bc he/she might not have as much energy
Explanation:
Attempt 2 You have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 15151515 lb and was traveling eastward. Car B weighs 11251125 lb and was traveling westward at 42.042.0 mph. The cars locked bumpers and slid eastward with their wheels locked for 17.517.5 ft before stopping. You have measured the coefficient of kinetic friction between the tires and the pavement to be 0.7500.750 . How fast (in miles per hour) was car A traveling just before the collision
Answer:
v = 28.98 ft / s
Explanation:
For this problem we must solve it in parts, let's start by looking for the speed of the two cars after the collision
In the exercise they indicate the weight of each car
Wₐ = 1500 lb
W_b = 1125 lb
Car B's velocity from v_b = 42.0 mph westward, car A travels east
let's find the mass of the vehicles
W = mg
m = W / g
mₐ = Wₐ / g
m_b = W_b / g
mₐ = 1500/32 = 46.875 slug
m_b = 125/32 = 35,156 slug
Let's reduce to the english system
v_b = 42.0 mph (5280 foot / 1 mile) (1h / 3600s) = 61.6 ft / s
We define a system formed by the two vehicles, so that the forces during the crash have been internal and the moment is preserved
we assume the direction to the east (right) positive
initial instant. Before the crash
p₀ = mₐ v₀ₐ - m_b v_{ob}
final instant. Right after the crash
p_f = (mₐ + m_b) v
the moment is preserved
p₀ = p_f
mₐ v₀ₐ - m_b v_{ob} = (mₐ + m_b) v
v = [tex]\frac{ m_a \ v_{oa} - m_b \ v_{ob} }{ m_a +m_b}[/tex]
we substitute the values
v = [tex]\frac{ 46.875}{82.03} \ v_{oa} - \frac{35.156}{82.03} \ 61.6[/tex]
v = 0.559 v₀ₐ - 26.40 (1)
Now as the two vehicles united we can use the relationship between work and kinetic energy
the total mass is
M = mₐ + m_b
M = 46,875 + 35,156 = 82,031 slug
starting point. Jsto after the crash
K₀ = ½ M v²
final point. When they stop
K_f = 0
The work is
W = - fr x
the negative sign is because the friction forces are always opposite to the displacement
Let's write Newton's second law
Axis y
N-W = 0
N = W
the friction force has the expression
fr = μ N
we substitute
-μ W x = Kf - Ko
-μ W x = 0 - ½ (W / g) v²
v² = 2 μ g x
v = [tex]\sqrt{ 2 \ 0.750 \ 32 \ 17.5}[/tex]Ra (2 0.750 32 17.5
v = 28.98 ft / s
A trough is 10 meters long, 1 meters wide, and 2 meters deep. The vertical cross-section of the trough parallel to an end is shaped like an isoceles triangle (with height 2 meters, and base, on top, of length 1 meters). The trough is full of water (density 1000kg/m3 ). Find the amount of work in joules required to empty the trough by pumping the water over the top. (Note: Use g
Answer:
The amount of work required to empty the trough by pumping the water over the top is approximately 98,000 J
Explanation:
The length of the trough = 10 meters
The width of the through = 1 meter
The depth of the trough = 2 meters
The vertical cross section of the through = An isosceles triangle
The density of water in the through = 1000 kg/m³
Let 'x' represent the width of the water at a depth
x/y = 1/2
∴x = y/2
The volume of a layer of water, dV, is given as follows;
dV = 10 × y/2 × dy = 5·y·dy
The mass of the layer of water, m = ρ × dV
∴ m = 1000 kg/m³ × 5·y·dy m³ = 5,000·y·dy kg
The work done, W = m·g·h
Where;
h = The the depth of the trough from which water is pumped
g = The acceleration due to gravity ≈ 9.8 m/s²
[tex]\therefore \, W \approx \int\limits^2_0 {5,000 \times y \times 9.8 \, dy} = \left[24,500\cdot y^2 \right]^2_0 = 98,000[/tex]
The work done by the pump to pump all the water in the trough, over the top W ≈ 98,000 J