Answer:
All of then are true
I need brainliest so I can rank up
Explanation:
Answer:
I think all options are true is the right answer
Explanation:
Mark me the brainliest plzzz
Specify whether the boiling point, as determined in the miniscale boiling-point apparatus, is the temperature a.of the liquid at the timebubbles first emerge slowly from the liquid. b.at the vapor-liquid interface above the surface of the boiling liquid while a drop of liquid c.is suspended from the thermometer. d.of the liquid at the timebubbles emerge rapidly from the liquid. e.of the heating source at the timebubbles emerge rapidly from the liquid.
Answer:
a. of liquid at the time bubbles first emerge slowly from the liquid.
Explanation:
Boiling point of liquid happens due to heat energy. This is an exothermic reaction as heat is released in to the environment. The initial boiling vapors slowly move away from the liquid and as the temperature increases the vapors start moving quickly.
Tyler and Jim race each other up a mountain on their bicycles. Tyler rides a road bike on the switchbacks of the twisting and turning mountain road. Jim rides a mountain bike and follows a direct, but steeper, straight-line path up the mountain. They start at the same time and place at the bottom of the mountain and finish at the same time and place at the top of the mountain. From start to finish a. whose distance traveled was longer? b. whose displacement was longer? c. which rider had the faster average speed? d. which rider had the faster average velocity? e. who won the race?
Answer:
Explanation:
Displacement is minimum distance between initial and final point .
Distance is total length of path covered in a journey .
a )
Tyler covered a longer distance in the journey because total length of path covered by him is longer due to curved path .
b )
Both have same displacement , because minimum distance between initial and final point in both the case is same .
c )
average speed = distance / time
as time is same for both the case ,
average speed ∝ distance
As distance covered by Tyler is more , his average speed is more .
d )
average velocity = displacement / time
As both displacement and time are same in both the case , average velocity in both the case is same .
e )
They start at the same time and place at the bottom of the mountain and finish at the same time , both have tie and nobody won the race , in spite of speed of Tyler being greater .
A woman accidentally drops a flowerpot from a windowsill at a height d above the street towards a man of height h standing below. The woman calls out to the man in just enough time for the man to move out of the way. If the man needs a time interval of Δt to respond to the warning, at what height above the street will the flowerpot be when the woman calls out the warning? (Use the following as necessary: d, h, Δt, v for the speed of sound, and g for gravitational acceleration.)
Answer:
h^2 - ( 2t_o v_s + 2v_s^2 /g) h + v_s^2 \ t_o^2 =0
The correct result is that of a positive height
Explanation:
For this exercise we use the kinematic relations, let's start by finding the time it takes for the sound to reach the man
v_s = y / t
t = [tex]\frac{y}{ v_s}[/tex]
this height is y = h
t = \frac{h}{ v_s}
the man has a response time of t = t₀, therefore
time to move is
t' = t - t₀
the initial height of flower pot is
y = y₀ + v₀ t' - ½ g t'²
when it reaches the floor the height is zero y = 0 and as the pot is dropped its initial velocity is zero v₀ = 0
0 = y₀ +0 - ½ g (t -t₀)²
if the initial height is i = h,
h = ½ g ([tex]\frac{h}{v_s}[/tex] - t₀)²2
[tex]\frac{2}{g} h[/tex] = [tex]\frac{h^2}{v_s^2}[/tex] - [tex]\frac{2t_o }{v_s} h[/tex] + t₀²
[tex]\frac{h^2}{v_s^2} - ( \frac{2t_o}{v_s} + \frac{2}{g} ) h + t_o^2 = 0[/tex]h2 / vs2 - (2nd / vs + 2 / g) h + to2 - = 0
[tex]h^2 - ( 2t_o v_s + 2v_s^2 /g) h + v_s^2 \ t_o^2 =0[/tex]
To know the height, you must solve the second degree equation, it is much easier with numerical values.
The correct result is that of a positive height
How can you tell whether an object is neutral
or charged? What would you have to do to test
that object?
Answer:
The number of electrons that surround the nucleus will determine whether or not it is electrically charged or electrically neutral
Explanation:
A vertical wire carries a current straight up in a region of the magnetic field directed north. What is the direction of the magnetic force on the current due to the magnetic field
Answer:
The direction of the force on the vertical wire is towards the East or right.
Explanation:
Using Fleming's right hand rule, the current is the middle finger pointing straight up, the magnetic field is the fore-finger pointing Northwards and then the thumb is the direction of the force on the vertical wire.
Following these conventions, the thumb points towards the East. So, the direction of the force on the vertical wire is towards the East or right.
Acceleration is the rate ot change of the velacity a -dejdt so it is the slope of the Velocity vs. Time graph Because it is dficult to drag the person in a consistent and reproducible way use the Expression Evakaator under the Special Features menu for this question lick Reset A and type in the hr on z t * t * t " t in the Expression Evaluator Click the Play button and let the simulation run roughly 5 sin ulation seconds before ressing the Pause but use the zoom buttons to a 쪄 the p s they the screen You should see 8 p at s ar l what you got in the previous question, but much smoother Look at the Postion vs Time. Velocity vs Time and Acceleration vs. Time piets h
a) the velocity is zero but the acceleration is negative
When the person is 8 to to the tight of the origin
b) the velocity is zero but the acceleration is positive
c) both the velocity and the acceleration are zero
d) both the velocity and the acceleraton are nonzero
Answer:
a) the body is changing direction,
b)the body must go to the left and the acceleration to the right
c) the movement has not started.
d) all points of the motion
Explanation:
In this exercise you are asked to find in which position you have the following characteristics of the movement
a) The velocity is zero and the acceleration is negative
This is when the body reaches the end of the travel and turns around, in this case the speed is zero and the acceleration has the opposite direction to the movement.
In this case the body moves to the right and the acceleration is to the left, therefore the speed decreases
b) The velocity is zero, but the acceleration is positive
This occurs at the points where the speed is changing direction, specifically for this case the body must go to the left and the acceleration to the right
c) Both are zero
This only occurs where the body is stopped and the movement has not started.
d) both the velocity and the relation are nonzero.
This is at all points of the motion since the velocity is constantly changing as long as there is an acceleration
Which part of the water cycle is where vapor from plants leaves the plants as they breath?
condensation
Transpiration
evaporation
Answer:
I think it is transpiration
Answer:
transpiration is the right answer
Canon launch is a 4.0 kg bowling ball with 50 J of kinetic energy what is the bowling ball speed
Answer:
5 m/s
Explanation:
50=1/2*4v^2
4*1/2=2
25*2=50
so...
square rood of 25 is 5
answer 5 m/s
sorry if that didn't make since
difine precision and accuracy
1. Clara stops for 10 minutes to catch up with a friend.
Answer:
Clara has speed of 80m/min
Explanation:
Clara was jogging at 600 m in 5 minutes. She stopped suddenly which reduced her velocity and then she waited for 10 minutes so that her friends comes near her. She stopped to catch her friend. During this 10 minutes the velocity of Clara is zero. She started to walk again at a slower speed of 80m/min.
Which statement is true of a glass lens that diverges light in air?
A.
It is thick near the center and thin at the edges.
B.
It is thin near the center and thick at the edges.
C.
It is uniformly thick.
D. It is uniformly thin.

Answer: it is thin near the center and thick at the edges
Explanation: took the test on Plato :)
Even though Alice visits the wishing well frequently and always tosses in a coin for good luck, none of her wishes have come true. As a result, she decides to change her strategy and make a more emphatic statement by throwing the coin downward into the well. If the water is 5.43 m below the point of release and she hears the splash 0.85 seconds later, determine the initial speed at which she threw the coin. (Take the speed of sound to be 343 m/s.)
Answer:
Explanation:
Total time taken = 0.85 s .
Time taken by sound to travel 5.43 m + time taken by coin to fall by 5.43 m = .85
5.43 / 343 + time taken by coin to fall by 5.43 m = .85
time taken by coin to fall by 5.43 m = .85 - 5.43/343 = .834 s
Let the initial velocity of throw of coin = u
displacement of coin s = 5.43 m
time take to fall t = .834 s
s = ut + 1/2 gt²
5.43 = u x .834 + .5 x 9.8 x .834²
5.43 = u x .834 + 3.41
u x .834 = 2.02
u = 2.42 m /s .
Any change in the cross section of the vocal tract shifts the individual formant frequencies, the direction of the shift depending on just where the change in area falls along the standing wave. Constriction of the vocal tract at a place where the standing wave of a formant exhibits minimum-amplitude pressure oscillations generally causes the formant to drop in frequency; expansion of the tract at those same places raises the frequency. Three other major tools for changing the shape of the tract in such a way that the frequency of a particular formant is shifted in a particular direction are the jaw, the body of the tongue and the tip of the tongue. Moving the various articulatory organs in different ways changes the frequencies of the two lowest formants over a considerable range [18].
One way to increase formant frequency is to ________ the vocal tract at a place where the standing wave of a formant frequency exhibits minimum-amplitude pressure oscillations.
a. Stretch
b. Vibrate
c. Contract
d. Expand
Answer:
The correct answer is option D.
Explanation:
It is stated in the question that constriction of the vocal tract at a place where the standing wave of a formant exhibits minimum-amplitude pressure oscillations generally causes the formant to drop in frequency so to increase formant frequency, the vocal should expand where the standing wave of a formant exhibits minimum-amplitude pressure oscillations. The answer is D.
I hope this helps.
An ordinary ruler is used to measure the area and its error of a rectangle. It is found that their sides are 5.0 cm long and 2.0 cm width. The error in area (in cm) is
Answer:
You need to know the accuracy to which you can read the ruler:
Suppose that you can read the read the ruler to the nearest milimeter
A = L * W your calculated area of the rectangle
A + ΔA = (L + ΔL) * (W + ΔW) = L W + L ΔW + W * ΔL + ΔL ΔA
Or ΔA = L ΔW + W ΔL
Where we have subtracted A = L * W and the term ΔL * ΔA is very small
So (5 + .1) * (2 + .1) - 5 * 2 = .1 * 2 + .1 * 5 = .7 cm^2
Then you report A = 10 cm^2 +- .7 cm^2 including the - sign for completeness
Magnus has reached the finals of a strength competition. In the first round, he has to pull a city bus as far as he can. One end of a rope is attached to the bus and the other is tied around Magnus's waist. If a force gauge placed halfway down the rope reads out a constant 2100 Newtons while Magnus pulls the bus a distance of 1.30 meters, how much work does the tension force do on Magnus
Answer:
Workdone = -2730 J
Explanation:
Formula for workdone is;
W = Force × Displacement
Now, according to Newton's 3rd law of motion, to every action, there is an equal and opposite reaction.
In the question given, we are told that a force gauge placed halfway down the rope reads out a constant 2100 Newtons while Magnus pulls the bus. This means that the force exerted by the rope on Magnus acts in an opposite direction to that which Magnus does to the rope.
Therefore, the force will be in the negative direction.
So;
Workdone = -2100 N × 1.3 m
Workdone = -2730 J
In a physics lab experiment for the determination of moment of inertia, a team weighs an object and finds a mass of 2.15 kg. They then hang the object on a pivot located 0.163 m from the object's center of mass and set it swinging at a small amplitude. As two of the team members carefully count 113 cycles of oscillation, the third member measures a duration of 241 s. What is the moment of inertia of the object with respect to its center of mass about an axis parallel to the pivot axis
Answer:
0.339 kgm²
Explanation:
We know the period of this pendulum, T = 2π√(I/mgh) where I = moment of inertia of the object about the pivot axis, m = mass of object = 2.15 kg, g = acceleration due to gravity = 9.8 m/s² and h = distance of center of mass of object from pivot point = 0.163 m.
Since T = 2π√(I/mgh), making I subject of the formula, we have
I = mghT²/4π²
Now since it takes 241 s to complete 113 cycles, then it takes 241 s/113 cycles to complete one cycle.
So, T = 241 s/113 = 2.133 s
So, Substituting the values of the variables into I, we have
I = mghT²/4π²
I = 2.15 kg × 9.8 m/s² × 0.163 m × (2.133 s)²/4π²
I = 15.63/4π² kgm²
I = 0.396 kgm²
Now from the parallel axis theorem, I = I' + mh² where I' = moment of inertia of object with respect to its center of mass about an axis parallel to the pivot axis
I' = I - mh²
I' = 0.396 kgm² - 2.15 kg × (0.163 m)²
I' = 0.396 kgm² - 0.057 kgm²
I' = 0.339 kgm²
An astronaut named Sandra Bullock has drifted too far away from her spaceshuttle while attempting to repair the Hubble Space telescope. She realizes that theshuttle is moving away from her at 3 m/s. On her back is a 10 kg jetpack which consistsof an 8 kg holding tank filled with 2 kg of pressurized gas. Without the jetpack, sheand her space suit have a mass of 80 kg.
Required:
a. She is able to use the gas to propel herself in the same direction as the shuttle. The gas exits the tank at a uniform rate with a constant velocity of 100 m/s, relative to the tank (and her). After the gas in the tank has been released, what is her velocity?
b. After this, she throws her empty tank into space and relies on the conservation of momentum to increase her speed to match that of the shuttle. With what velocity (in her frame of reference!) will she have to throw the tank?
Answer:
a) v_f = 0.898 m / s, b) v₂ = -6.286 m / s
Explanation:
a) For this exercise we use the conservation of momentum, we define a system formed by the astronaut, her equipment and the expelled gases. We must also define a stationary frame of reference, let's place the system on the platform, so the speed of the subject is v = -3 m / s
Initial instant. Before you start to pass gas
p₀ = (M + Δm) v
M is the mass of the astronaut M = 80Kg and Δm the masses of the gases
Final moment. When you expel the gases
p_f = M (v + Δv) + Δm (v-v_e)
where v_e is the gas velocity v_e = 100 m / s
momentum is conserved
p₀ = p_f
M v + Δm v = Mv + M Δv + Δm v -Δm ve
0 = M Δv - Δm v_e
if we make the very small quantities Δv → dv and Δm → dm, furthermore the quantity of output gas is equal to the decrease in the total mass dm = -dM
M dv = -v_e dM
∫ dv = - v_e ∫ dM / M
We solve, between the lower limits v₀ = v with M = M₀ and the upper limit v = v_f for M = M_f
v_f - v₀ = - v_e (ln M_f - Ln M₀)
v_f - v₀ = v_e ln ([tex]\frac{M_o}{M_f}[/tex])
v_f = v₀ + v_e ln (\frac{M_o}{M_f})
let's calculate
v_f = -1.3 + 100 ln (80 + 10 + 2/80 + 10)
v_f = -1.3 +2.20
v_f = 0.898 m / s
b) launch the jetpack to increase its speed up to the speed of the platform
initial instant. Before launching the tanks
p₀ = (M + m') v_f
final instnte. After launching the tanks
p_f = M v₁ + m' v₂
indicate that the final velocity of the astronaut is the platform velocity v₁=0 m / s, since the reference system is fixed on it
p₀ = p_f
(M+ m) v_f = M v₁ + m v₂2
v₂ = [tex]\frac{ M ( v_f - v_o) + m' v_f}{m'}[/tex]
v₂ = [tex]\frac{M}{m}[/tex] (v_f -v₁) + v_f
let's calculate
v₂ = 80/10 (0.898 - 0) + 0.898
v₂ = -7.1874 + 0.898
v₂ = -6.286 m / s
point.
4.
i. Explain why a particle moving with a constant speed along a circular
path has a radial acceleration.
ii. Show that the acceleration of a body moving in a circular path of
radius r with uniform speed v is → and draw a diagram to show the
direction of the acceleration.
2
iii. Show that the expression à = † is dimensionally correct.
Explanation:
Explanation: When a particle moves along a straight path, then the radius of curvature is infinitely great. This means that v2/r is zero. Explanation: When a particle moves with a uniform velocity, then dv/dt will be zero.
Q5. Use Superposition to V. in the circuit below? (5 points)
4 mA
12V
2 ΚΩ
2 mA
1 ΚΩ
2 ΚΩ
Answer:
4va
12va
2jk
1jk
2jk
Concept Simulation 4.1 reviews the central idea in this problem. A boat has a mass of 4490 kg. Its engines generate a drive force of 4520 N due west, while the wind exerts a force of 890 N due east and the water exerts a resistive force of 1210 N due east. Take west to be the positive direction. What is the boat's acceleration, with correct sign
Answer:
-0.54m/s²
Explanation:
According to Newton's second law of motion
F = ma
Force = mass * acceleration
Given
Mass m = 4490kg
Take the sum of forces
Sum of force along the east = 890+1210 = 2100N
Sum of forces along the west = -4520N
Net force = -4520+2100
Net force = -2420N
Acceleration = Net force/Mass
Acceleration = -2420/4490
Acceleration = -0.54m/s²
Hence the boat acceleration is -0.54m/s²
A shuttle bus slows down with an average acceleration of -2.4 m/s2. How long does it
take the bus to slow from 9.0 m/s to rest?
Answer:
[tex]\boxed {\boxed {\sf 3.75 \ seconds }}[/tex]
Explanation:
Average acceleration is found by dividing the change in acceleration by the time.
[tex]a=\frac{ v_f-v_i}{t}[/tex]
The shuttle bus has an acceleration of -2.4 meters per square second. It slows from 9.0 meters per second to rest, or 0 meters per second. Therefore:
[tex]a= -2.4 \ m/s^2 \\v_f= 0 \ m/s \\v_i= 9 \ m/s[/tex]
Substitute the values into the formula.
[tex]-2.4 \ m/s^2=\frac{0 \ m/s - 9 \ m/s}{t }[/tex]
Solve the numerator.
[tex]-2.4 \ m/s^2 = \frac{-9 \ m/s}{t}[/tex]
We want to solve for t, the time. We have to isolate the variable. Let's cross multiply.
[tex]\frac{-2.4 \ m/s^2}{1} = \frac{-9 \ m/s}{t}[/tex]
[tex]-9 \ m/s *1= -2.4 \ m/s^2 *t[/tex]
[tex]-9 \ m/s=-2.4 m/s^2*t[/tex]
t is being multiplied by -2.4. The inverse of multiplication is division, so divide both sides by -2.4
[tex]\frac{-9 \ m/s }{-2.4 \ m/s^2} =\frac{ -2.4 \ m/s^2*t}{-2.4 \ m/s^2}[/tex]
[tex]\frac{-9 \ m/s }{-2.4 \ m/s^2} =t[/tex]
[tex]3.75 \ s=t[/tex]
It takes 3.75 seconds.
A mole of a monatomic ideal gas at point 1 (101 kPa, 5 L) is expanded adiabatically until the volume is doubled at point 2. Then it is cooled isochorically until the pressure is 20 kPa at point 3. The gas is now compressed isothermally until its volume is back to 5 L (point 4). Finally, the gas is heated isochorically to return to point 1.
a. Draw the four processes and label the points in the pV plane.
b. Calculate the work done going from 1 to 2.
c. Calculate the pressure and temperature at point 2.
d. Calculate the temperature at point 3.
e. Calculate the temperature and pressure and point 4.
f. Calculate the work done going from from 3 to 4.
g. Calculate the heat flow into the gas going from 3 to 4. g
Answer:
(a). Check attachment.
(b). 280.305 J.
(c). 31.81 kpa; 38.26K.
(d). 24.05K.
(e). 24.05k; 40kpa.
(f). -138.6J.
Explanation:
(a). Kindly check the attached picture for the diagram showing the four process.
1 - 2 = adiabatic expansion process.
2 - 3 = Isochoric process.
3 - 4 = isothermal process.
4 - 1 = isochoric process.
(b). Recall that the process from 1 to is an adiabatic expansion process.
NB: b = 5/3 for a monoatomic gas.
Then, the workdone = (1/ 1 - 1.66) [ (p1 × v1^b)/ v2^b × v2 - (p1 × v1)].
= ( 1/ 1 - 5/3) [ (101 × 5^5/3) × 10^1 -5/3] - 101 × 5.
Thus, the workdone = 280.305 J.
(c). P2 = P1 × V1^b/ V2^b = 101 × 5^5/3/ 10^5/3 = 31.81 kpa.
T2 = P2 × V2/ R × 1 = 31.81 × 10/ 8.324 = 38.36k.
(d). The process 2 - 3 is an Isochoric process, then;
T3 = T2/P2 × P3 = 38.26/ 31.82 × 20 = 24.05K.
(e). The process 3 - 4 Is an isothermal process. Then, the temperature at 4 will be the same temperature at 3. Tus, we have the temperature; point 3 = point 4 = 24.05k.
The pressure can be determine as below;
P4 = P3 × V3/ V4 = 20 × 10/ 5 = 200/ 5 = 40 kpa.
(f) workdone = xRT ln( v4/v3) = 1 × 8.314 × 24.05 × ln (5/10) = - 138.6 J
Which statement best compares coal and ores?
Both are burned for energy.
Both take millions of years to form.
Both require oxygen to form.
Both are used to make coins.
Answer:
Option 2 both take millions of years to form
Explanation:
Both coal and ores take millions of years to form.
What are ores?Ore is a naturally occurring rock or silt that has precious minerals in it that may be extracted, processed, and sold for a profit. These minerals are usually metals. Mining is the process of removing ore from the soil. The valuable metals or minerals are then removed by treating or refining the ore, frequently through smelting.
The concentration of the desired ingredient in an ore is referred to as its grade. To decide if a rock has a high enough grade to be worth mining and is therefore regarded as an ore, the value of the metals or minerals it contains must be evaluated against the expense of extraction.
Typically, oxides, sulphides, silicates, or native metals like copper or gold are the minerals of interest. To separate the valuable components from the waste rock, ore must be treated. Numerous geological processes collectively known as ore genesis are responsible for the formation of ore deposits.
Learn more about ore here:
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A child holds a sled at rest on frictionless snow covered hill. if the sled weighs 77N,find the force T exerted by the rope on the sled and the force n exerted by the hill on the sled
Answer:62
Explanation:
The weight of the sled is 77 N. The force by the hill on the sled is equal to its weight that is 77 N. Then the tension force exerted by the rope on the sled is being 77N sin θ, where θ be the angle of inclination.
What is force?Force is an external agent acting on an object to change its motion or to deform it. There are various kinds of force like magnetic force, tension force, frictional force, gravitational force etc.
The weight that an object experience on earth is due to the gravitational force. The force that is exerted by a rope on an object is tension force since it is pulling from a side.
The normal force by the hill on the sled is equal to its weight that is 77 N. The tension force on the sled by the rope is dependent on the angle of inclination θ. If know the angle we can find T by the equation:
T = 77 sin θ.
Find more on tension force:
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The electric field from two charges in the plane of the paper is represented by the dashed lines and arrows below.
Select a response for each statement below. (Use 'North' towards top of page, and 'East' to the right)
The magnitude of the E-field at Ris .... than at M.
The force on a (+) test charge at P is zero.
The magnitude of the charge on the left is .... that on the right.
The force on a (+) test charge at L is directed ....
The force on a (-) test charge at J is directed
The force on a (-) test charge at N is directed ....
The sign of the charge on the right is negative.
Answer:
a) electric field at point P must be zero
b) harged must be positive
c) force ais in the direction of the electric field
d) force is in the opposite direction to the electric field
e) force is in the opposite direction to the field
Explanation:
After reading your exercise, it is unfortunate that the diagram did not come out, but we are going to answer the questions in general.
a) force on a charge (+) is zero
this implies that the electric field at point P must be zero
F = q E
b) the magnitude of the charge on the left is on the right
this indicates that the charged must be positive since the lines must exit the charge
c) force on load directed towards (direction not indicated)
since the charge is positive the force at point L is in the direction of the electric field at this point
d) force on test load (-) does not indicate direction
The force on a negative charge is in the opposite direction to the electric field at point J
e) Force on a test load (-) at point N
the force is in the opposite direction to the field at point N
If a wave has a speed of 1000 m/s and frequency of 500 Hz, what is the wavelength?
• 1500 Hz
• 2 m
• 0.05 m
Answer:
2 m
Explanation:
speed=frequency×wavelength
wavelength=speed/frequency
wavelength=1000/500
=2 m
How far can a bus carrying small children, travel at a rate of 60 km per hour travel in 2 1/2 hours?
Explanation:
speed = 60km/hr.time = 2¹/2 hr = 5/2 hrdistance = speed × time = 60 ×5/2 = 150kmMARK ME AS BRAINLISTFor the questions below, include units if applicable. If necessary, use a separate sheet of paper for 1, 6c and 7c. Tire pressure is in part a function of the temperature of the tire.
1. Based on everyday experience, state (in words) the relationship between tire pressure and temperature. Look at the data below and see if the numbers support your statement.
2. Prepare a hand-drawn plot of the two variables on the reverse side of this worksheet. Include a title, axis labels (with units), and a trendline. Estimate the tire pressure when the temperature is 18.6°C: Estimate the temperature of the air in the tire when the pressure is 37.0 psi: 3.
a. Prepare a plot using graphing software. Include a title, axis labels (with units), the equation of the best-fit Line and the R? value on the graph.
b. Re-write the equation of the best-fit line substituting "Temperature" for x and "Pressure" for y directly on the graph.
c. Attach the fully labeled graph to this worksheet.
4. What is the value of the slope for the relationship between temperature and pressure?
5. Determine the percent error using the definition of percent error: Use 0.145 psi/" for the "Actual" value of the slope. %error = Actual-Experimental % Error Actual
6. Based on your computer-generated graph,
a. visually estimate the tire pressure when the temperature is 18.6°C:
b. calculate the tire pressure at this temperature using the equation of the best fit line: the graph to ensure that this value is reasonable.
c. compare the calculated pressure to the two visually interpolated values (Steps 2 and 6a). Comment on any discrepancies.
7. Based on your computer-generated graph,
a. visually estimate the temperature of the air in the tire when the pressure is 37.0 psi:
b. calculate the temperature of the air in the tire at this pressure: Use the graph to ensure that this value is reasonable.
c. compare the calculated temperature to the two visually interpolated values (Steps 2 and 7a). Comment on any discrepancies.
Data:
Temperature (x) Tire Pressure, psi (y)
12.9°C 3.39 x 10
15.4C 34.25
-2.10 F 2.68 x 10
19.5 °C 3.50 x 10
29.6 'F 36.53
Answer:
All answer are explained below in the explanation section.
Explanation:
1. The pressure varies proportionally with the change in temperature. It can also be observed in our daily lives.
As for example, a pressure cooker uses the same principal to cook food faster. With the increasing temperature, the pressure inside the cooker increases.
Thus after a while, the excess pressure inside is released through the top nozzle. The data shown below supports that pressure and temperature varies linearly.
2. Hand drawn plot is attached in the attachment please refer to the attachment for the hand drawn plot.
Tire pressure at temperature 18.6 degree C is ~ 35 psi.
Temperature at air pressure of 37 psi is ~26.1 degree C
3. a.) Necessary values are included in the stat box. It is attached in the attachment please refer to the attachment.
3. b) The equation becomes: Pressure = 0.176 x temperature + 32.32
3. c) It is already done in part a of this question.
4. The value of the slope estimated from the linear fit is 0.176 +/- 0.094.
5. % Error = [tex]\frac{Actual - Experiment}{Actual} x 100[/tex]
Plugging in the values, we get:
Actual = 0.145, Experimental = 0.176. Thus, percentage error is given by:
% Error = 21.33%
6. a.) Visual estimation of tire pressure at t = 18.6 degree C is ~ 35 psi
6. b.) Estimation of pressure from the best fit line is given by 35.6 psi, which is consistent with the eye estimation value.
6. c.) The eye estimation and the estimation from the line fit are quite comparable. The discrepancy of +/-0.5 psi is within the percentage error calculated in 5.
7. a.) Visual estimation of temperature of the air for a tire-pressure of 37 psi is ~ 26 degree C.
7. b.) Estimation of temperature from best fit value of line is = 26.64 degree C
7 c) The values from eye estimation and evaluated from the fit are quite consistent within a random fluctuation of +/- 0.64 degree C.
A block of mass m1 = 19.5 kg slides along a horizontal surface (with friction, μk = 0.35) a distance d = 2.6 m before striking a second block of mass m2 = 8.25 kg. The first block has an initial velocity of v = 6.5 m/s.
(a) Assuming that block one stops after it collides with block two, what is block two's velocity after impact in m/s?
(b) How far does block two travel, d2 in meters, before coming to rest after the collision?
Answer:
19.5 m/s
87.8 m
Explanation:
The acceleration of block one is:
∑F = ma
-m₁gμ = m₁a
a = -gμ
a = -(9.8 m/s²) (0.22)
a = -2.16 m/s²
The velocity of block one just before the collision is:
v² = v₀² + 2aΔx
v² = (8.25 m/s)² + 2 (-2.16 m/s²) (2.3 m)
v = 7.63 m/s
Momentum is conserved, so the velocity of block two just after the collision is:
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
m₁u₁ = m₂v₂
(18.5 kg) (7.63 m/s) = (7.25 kg) v
v = 19.5 m/s
The acceleration of block two is also -2.16 m/s², so the distance is:
v² = v₀² + 2aΔx
(0 m/s)² = (19.5 m/s)² + 2 (-2.16 m/s²) Δx
Δx = 87.8 m
Explanation:
By using conservation of linear momentum and also by equating work done to kinetic energy, [tex]V_{2}[/tex] = 15.36 m/s and [tex]d_{2}[/tex] = 4.32 meters
Parameters given are :
[tex]m_{1}[/tex] = 19.5 kg
friction, μk = 0.35
distance d = 2.6 m
mass [tex]m_{2}[/tex] = 8.25 kg.
initial velocity of [tex]U_{1}[/tex] = 6.5 m/s.
a.) Since we assumed that the block one stops after it collides with block two, the final velocity for block one will be zero. That is, [tex]V_{1}[/tex] = 0 so its final momentum = 0
Let us also assume that block two was initially at rest. Therefore, it initial velocity and its momentum will be equal to zero.
The formula to use will be :
[tex]m_{1}U_{1} = m_{2}V_{2}[/tex]
Substitute all the parameters into the formula above
19.5 x 6.5 = 8.25[tex]V_{2}[/tex]
Make [tex]V_{2}[/tex] the subject of formula
[tex]V_{2}[/tex] = 126.75/8.25
[tex]V_{2}[/tex] = 15.36 m/s
b.) Let us first calculate the work done in by block one.
The K.E = [tex]1/2mU^{2}[/tex]
substitute its mass and velocity into the formula
K.E = 1/2 x 19.5 x [tex]6.5^{2}[/tex]
K.E = 411.94 Joule
The work done = Kinetic energy
But the resultant Force F = force f - friction
where Frictional force = 0.35 x 19.5 x 9.8
Frictional force = 66.89N
Work done will be the product of resultant Force F and the distance travelled
(F - 66.89) x 2.6 = 411.94
F - 66.89 = 411.94/2.6
F - 66.89 = 158.44
F = 225.3 N
The second block will experience the same force which is equal to 225.3N
Find the kinetic energy of the second block.
K.E = [tex]1/2mV^{2}[/tex]
K.E = 0.5 x 8.25 x 15.36^2
K.E = 973.2
Using The work done = Kinetic energy
225.3[tex]d_{2}[/tex] = 973.2
[tex]d_{2}[/tex] = 973.2/225.3
[tex]d_{2}[/tex] = 4.32 meters
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help me pls it’s a usa test prep pretty easy
Answer:
Im 99.99999% sure its c
Explanation:
i cant see the pictures too well