Answer:
This demonstration is often done following a discussion of the ideal gas equation of state, PV=nRT.
We begin by weighing a balloon, then blowing it up and weighing it again. In the photo shown on right, the mass indication increased from 3.4 to 3.5 grams. At this point, it is important to note that the scale measures force, even though it reports a conclusion about mass based on the force measurement.
One assumption made in reaching the conclusion is that the buoyant force on the object being weighed is negligible. In the case of the balloon, this is incorrect. The buoyant force on this balloon is equal to the weight of the air displaced.
Since the volume of air inside the balloon is essentially the same as the volume of air displaced, we should expect that the buoyant force would support the weight of the air inside the balloon: The reported mass should not go up at all, because the force required of the scale should not change.
The increase in reported mass of .1 gram is attributed to the higher density of the air inside the balloon: The tension in the balloon compresses the air inside, as attested by the pressure required to blow the balloon up. Evidently, for this experiment, the pressure inside is greater than atmospheric by about 2%.
In the picture at right, the balloon is being pressed into a pan of liquid nitrogen. (The pan is the styrofoam lid of a small lunch box.) The balloon floats lightly on the liquid nitrogen unless pressed down. Pressing down places more surface area in contact with the cold nitrogen and speeds the demonstration. It is interesting to note the buoyant force by this liquified constituent of air.
The balloon shrinks dramatically, as indicated below. When left in contact with the liquid nitrogen long enough (perhaps 5 minutes) the oxygen inside the balloon liquifies, and then the nitrogen liquefies also. Close observation of the photo at the upper left corner of the pan shows some liquid nitrogen bubbles may forming above the dark spot in the center of the pan. One can also make out a faint line at the upper left corner of the pan which is the liquid nitrogen surface. The balloon still floats, riding rather high on that surface. Evidently, some of the balloon contents remain in the gas phase, making the mass of the balloon less than the mass of the displaced liquid nitrogen.
Next, we take the shrunken balloon and place it back on the scale, as above. In this instance, the reported mass is 8.7 grams, an increase of 5.2 grams.
A look at the figure on the right shows a faint line near the bottom of the cold balloon. Above that line, the balloon contains gas; below the liquid. That line represents the top surface of the liquid air inside the balloon. With this evidence, the easy thing to say would be, "Of course, liquids are heavier than gases," but that would be incorrect. We assert that the amount of air inside the balloon has not changed and that the mass of that air is not dependent on temperature.
If these assertions are true, then the force of gravity on the balloon has not changed. The scale reading is determined by the force which it must exert on the balloon in order to keep it stationary. Evidently, the required force is larger when the balloon is shrunken. The reason is that the buoyant force (upward) has decreased to practically zero, leaving the scale alone to balance the downward force by gravity.
From the data, we can say that the change in the buoyant force is equal to the weight associated with the apparent change in mass. The weight of 5.2 grams is about .052 newtons. The buoyant force is less now because the balloon displaces less air. If we could measure the change in volume of the balloon as DV, then the buoyant force would be (r g DV) upwards, where r is the density of air that was displaced by the balloon, and g is the gravitational field strength, 9.8 Newton/kg.
Note that the .052 newton force is not the weight of the air inside the balloon. Rather, it is the weight of the air that was displaced by the balloon. If we ignore the compression of air inside the balloon, the two numbers are the same. However, the two samples are completely different.
We can estimate the volume of the balloon by assuming that the hand in the photograph is about .1meters across. For purposes of estimation, we say that the volume shrank to almost zero when the balloon was cold so that the change in volume was nearly equal to the original volume. Plugging in numbers gives fair agreement with the book value of 1kg/cubic meter for the density of air.
The value for the density of air is secondary to two main features of this demonstration:
Large changes in temperature produce the large changes in volume that are indicated by the ideal gas equation.
The mass of air in a volume equal to the volume of a balloon can be determined provided that the buoyant force is understood.
A car goes from 60 m/s to 75 m/s in 10 seconds.calculate the car’s acceleration?
Explanation:
Uhhhh here is an example
a=vt. =27 m/s10 s. =2.7 m/s2.
[tex]▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪[/tex]
Acceleration of the car is ~
[tex] \boxed{1.5 \: \: m/s}[/tex]
[tex] \large \boxed{ \mathfrak{Step\:\: By\:\:Step\:\:Explanation}}[/tex]
We know that,
Acceleration = rate of change in velocity ~
that is ~
[tex]a = \dfrac{v - u}{t} [/tex]where,
v = final velocityu = initial velocity t = time taken ~now, let's solve ~
[tex]a = \dfrac{75 - 60}{10} [/tex][tex]a = \dfrac{15}{10} [/tex][tex]a = 1.5 \: \: m/s[/tex]if 20 boys can sweep the school compound in 8 minutes, how many boys can sweep it in 10minutes
Answer:
So this is just a ratio/proportion problem
so lets say its in boys:minutes
so in the first scenario
20:8
and we want to find
x:10
So proportion really
20/8=x/10
cross multiply
25
25 boys can sweep it in 10 minutes
A swimming pool has dimension of 30m×10m×3m. When it is filled with water , what is the thrust on the bottom and sides
Answer:
3,900m³
Explanation:
Two cars A and B are moving along a straight road in the OPPOSITE direction with velocities of 25 km/h and 40 km/h, respectively. Find the velocity of car B relative to car
Answer:
65km/h
Explanation:
Here,
Relative velocity of car moving in different directions with different velocities =Va + Vb
=25+40
=65 km/h
Please find attached photograph for your answer.
Hope it helps.
Do comment if you have any query.
What average force magnitude is required to stop a 15,000 kg train in 12.1 s if the train is traveling at 86 km/hr
Answer:
Explanation:
Givens
m = 15000 kg
t = 12.1 seconds
vi = 86 km / hr
vf =0
Formula
F = m*a
a = (vf - vi)/ t
vi has to be converted to m/s
86 km/hr [1000 m / 1 km] * [1 hour / 3600 seconds]
86 * 1000 / 3600 = 23.89 m/s
Solution
a = (vf - vi) / t
a = (0 - 23.89)/12.1
a = - 1.97 m/s^2 The minus means that the trains is slowing down.
F = 15000 kg * - 1.97 m/s^2
F = -29615.7 Newtons.
A ____ is believed to occur when energy, stored in a twist in the solar magnetic field above a sunspot, is suddenly released.
[tex]▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪[/tex]
A solar flare is believed to occur when energy, stored in a twist in the solar magnetic field above a sunspot, is suddenly released.
A man on the moon with a mass of 90 kilograms weighs 146 newtons. The radius of the moon is 1.74 x 10^6
meters. Find the mass of the moon
equation- F= G m1 m2/ d^2
Answer:
7.36 × 10^22 kg
Explanation:
Answer:
D
Explanation:
help with this question
Answer:
D and B.
Explanation:
The wavelength must be taken from the same corresponding point. D and B are both at the crests of a wave, with no other crests in between, which means the distance between them is equal to one wavelength.
A person who sits on the right-hand seat of the car that is making a left turn slides over to the right. What is the possible reason for this?
Answer:
zsjgbjidasngwhugbhwuabvhuvbwhuebewghvwev
Explanation:
11111111111111111111111111111111111111111111111111111
difference between 1 hectoliter and one kiloliter
Answer: 100 and 1000
Explanation:
When making a left turn, drive into the intersection, make the turn smoothly and without strain on the engine, braking to about ... to .........mph and stay on the brake until approximately half way into the intersection.
Answer:
whats the question
Explanation:
Is a football tackle an elastic or inelastic collision and why?
A: elastic, because momentum is conserved
B: inelastic, because kinetic energy is conserved
C: inelastic, because momentum is not conserved
D: inelastic, because kinetic energy is not conserved
If the kinetic energy of an object changes, then the collision is considered inelastic. This is regardless of whether the objects sticks together or not.
My best answer for this question would be D) inelastic, because kinetic energy is not conserved.
We can tell from the collision that it is not elastic.
Momentum is is conserved, it transfers to the other player.
Kinetic energy is not conserved, at it turns into internal friction.
I hope this helps! :)
A football tackle is an inelastic collision , because kinetic energy is not conserved.
What is collision?When two bodies or particles move towards enough to interact with each other it is called the particles or the bodies are colliding.
What are the types of collision?Elastic collisioninelastic collisionWhat is an elastic collision?In this type of collision , there is no loss of kinetic energy.The momentum and kinetic energy both are conserved.What is an inelastic collision?in this type of collision , there is a loss of kinetic energy.The momentum of the system is conserved but the kinetic energy is not conserved .So, when the football tackles it's kinetic energy changes.
Thus, A football tackle is an inelastic collision , because kinetic energy is not conserved.
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To minimize signal distortion, at each end of the J-1939 CAN-bus there is a(n)_____________resistor.
Noah runs 200 m to the east and then 400 m north. What is Noah's distance and displacement?
Answer:
Noah has run 600 metres north east
60W light bulb is left on for 5mins. How much energy does it use in KWH
Answer:
Energy = Power × Time
time must be hoursPower must be in kilowatts[tex]{ \tt{E = ( \frac{60}{1000} ) \times ( \frac{5}{60}) }} \\ \\ { \tt{E = 0.005 \: kwh {}^{ } }}[/tex]
Suppose measure the mass of an orange. Then you peel the orange and separate out the slices. If you measure the orange peel and all the slices, will that mass equal the mass of the whole orange?
yes
no
The mass of the orange after it has been cut into pieces will still remain the same as the mass of the whole orange.
Let us recall that the law of conservation of mass states that, mass can neither be created nor destroyed but is converted from one form to another. In accordance with this law of conservation of mass. The mass of the orange must remain the same after it has been cut in pieces.
Hence, If the mass of an orange is measured after it has been cut into pieces, the mass of the orange after it has been cut into pieces will still remain the same as the mass of the whole orange.
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which acts as a transverse wave with particle motion perpendicular to wave motion?
Answer:
transverse wave, motion in which all points on a wave oscillate along paths at right angles to the direction of the wave's advance. Surface ripples on water, seismic S (secondary) waves, and electromagnetic (e.g., radio and light) waves are examples of transverse waves.
Which will be different on the moon than it is on Earth?
weight
mass
Answer:
Below
Explanation:
On the moon, an objects weight will be different than it is on earth. This is because we cannot change the mass of an object, the mass of an object is the measure of matter an object has. However the weight is depended on the gravitational pull of whatever planet you are on. In this case, weight will be lighter on the moon than it is on earth because the moon's gravitational pull is 1.62 m/s^2 while earths is around 9.8 m/s^2.
Hope that helps!
A 1,000 kg truck is traveling at 3 m/s. Suddenly, the driver sees a herd of cows on the road ahead and applies the brakes. The truck's tires could fail after doing 5,000 J of work to slow the vehicle. Can the truck stop before the tires fail?
A. Yes, the total KE the tires need to transfer out of the system is less than 5,000 J.
B. Yes, the tires do not do any work, it is only the brakes that do work.
C. No, the truck had to stop suddenly and the quick change in KE will cause the tires to fail.
D. No, the total KE the tires need to transfer out of the system is more than 5,000 J.
This question involves the concepts of the law of conservation of energy and kinetic energy.
The correct option is "A. Yes, the total KE the tires need to transfer out of the system is less than 5,000 J".
According to the law of conservation of energy:
Loss in Kinetic Energy = Work done by the tires
[tex]\frac{1}{2}mv^2=W[/tex]
where,
W = work done by tires = ?
m = mass of the truck = 1000 kg
v = speed of the truck = 3 m/s
Therefore,
[tex]W=\frac{1}{2}(1000\ kg)(3\ m/s)^2[/tex]
W = 4500 J
Since the failure limit of work done by the tire is 5000 J, which is greater than the actual work done by the tire in this scenario. Hence, the tire will not fail in this case.
Learn more about the law of conservation of energy here:
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The attached picture explains the law of conservation of energy.
What is the correct definition of rarefaction
Answer:
Explanation:
A decrease in the density of something is rarefaction. ... Most of the time, rarefaction refers to air or other gases becoming less dense. When rarefaction occurs, the particles in a gas become more spread out. You may come across this word in the context of sound waves.
You are walking back toward the back of a bus that is moving forward with a constant velocity. Describe your motion relative to the bus and relative to a point on the ground.
Answer: Relative to the bus, you are stationary. Relative to a point on the ground, you are walking forward with velocity equal to the velocity of the bus.
As a person walks toward the back of a bus, they are walking towards a stationary object according to their own perspective. However, from an outside perspective, the bus is moving forward with a constant velocity and so relative to that point in motion, you are actually walking backwards with respect to it. This difference in motion means you would not be able have the same experience as the person on the bus when getting off because when you get off you would stop while they continue going forward. The last sentence could be considered a footnote because it only applies if this was an actual situation rather than just an analogy question in a physics test.
Relative to the bus, you move at walking speed. The motion relative to the ground is in the direction same as the bus and equal to the difference between the speed of the bus and walking speed.
What is relative motion?When an object has a certain velocity, then this velocity is w.r.t. some frame that is called the reference frame. When we measure the velocity of an object, the ground or the earth is taken as the reference frame.
The motion of an object observed by the observer depends on the frame of the observer and this type of motion is known as relative motion.
For example, if you are traveling in a train and the train is moving at a speed of 80 km/hr, then your speed according to another passenger sitting on the same train is equal to zero.
Similarly, if you are walking toward the back of a bus, then your motion relative to the bus is walking speed. Relative to the ground, it is the difference between the speed of the bus and walking speed.
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The maximum allowable potential difference across a 250 mH inductor is 360 V . You need to raise the current through the inductor from 1.5 A to 2.5 A . What is the minimum time you should allow for changing the current?
This question involves the concepts of potential difference, inductance, and current.
The minimum time that should be allowed for the current to change is "0.694 ms".
The inductance of an inductor is given by the following formula:
[tex]E=L\frac{\Delta I}{\Delta t}\\[/tex]
where,
E = potential difference across the inductor = 360 volts
L = inductance of the inductor = 250 mH = 0.25 H
ΔI = change in current = 2.5 A - 1.5 A = 1 A
Δt = time required = ?
Therefore,
[tex]\Delta t = \frac{(0.25\ H)(1\ A)}{360\ volts}[/tex]
Δt = 6.94 x 10⁻⁴ s = 0.694 ms
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In your response be very specific and break down each item on the list, then state if they would be classified the same or differently.
Consider the definition for each as you respond.
• atoms
• elements
• compounds
• molecules
• matter
A bowling ball and a ping-pong ball are each tied to a string and hung from the ceiling. The distance from the ceiling to the CM of each object is the same. Which object would have a longer period of motion if they were set swinging
what noncontact forces act on an object that are balanced
Answer:
I can give you a list of noncontact forces off of my head, if that is what you need.
Explanation:
First, gravity. It acts everywhere and pulls down every object that is in the air towards the ground, unless they have an equal and opposite force that counters it.
Second, wind. I don't think this will count because wind is a physical thing, but it can be used in the same way as we did with gravity. If the force counteracting it is larger, equal, and opposite of the force it is countering, then it will move forward, through the wind.
Do you think humans will ever walk on the sun?
[tex] \: \: \: \: \: [/tex]
yesbecause sun have a super nova and if someone get it there that person definitely get burnhope it helps
[tex] \: \: \: \: \: [/tex]
A ball is released freely from the top of a building 80m high. Find its speed.
At the top of the building, the speed of the ball is zero.
The speed increases steadily as the ball falls.
When it hits the street, its speed is 39.6 m/s.
The acceleration of an object is inversely proportional to its mass, so if you decrease its mass while keeping the net force the same, the acceleration will increase.
true or false
Answer:
True
Explanation:
Since acceleration is inversely proportional to mass, decreasing mass will make the object lighter, and thus easier to speed up. So acceleration increases as mass decreases and vice versa
Answer: TRUE Explanation:Newton's second law of motion can be formally stated as follows: The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.
HELP PLZZZZZZ AHHHHH
Answer: I think it's larger
Explanation: man im just trying to help you
a box takes 350 N to start moving the coefficient of static friction is 0.35. what is the weight of the box?
Answer:
101.937 kg
Explanation:
The force needed to get the box moving must just cancel the static friction force:
F = µsmg = 0.35•m•9.81 = 350 ---> m = 350 / (0.35•9.81) = 101.937 kg
Again, with units shown, and using 1 N = 1 kg•m/s2:
0.35•m•9.81(m/s2) = 350 N
solving for m:
m = 350 N / (0.35•9.81 m/s2) = (350 kg•m/s2) / (3.434 m/s2 ) = 101.937 kg
______________________
(Hope this helps can I pls have brainlist (crown)☺️