Answer:
I think it would be c, crashing into another car
the second law of thermodynamics imposes what limit on the efficiency of a heat engine?
A heat engine cannot have a thermal efficiency of 100% For all reversible processes, the second-law efficiency is 100%. The second-law efficiency of a heat engine cannot be greater than its thermal efficiency. The second-law efficiency of a process is 100% if no entropy is generated during that process
What is the potential energy of a 50kg car on top of a 600m hill?
Answer:
294,000 JExplanation:
The potential energy of a body can be found by using the formula
PE = mgh
where
m is the mass
h is the height
g is the acceleration due to gravity which is 9.8 m/s²
From the question we have
PE = 50 × 9.8 × 600 = 294,000
We have the final answer as
294,000 JHope this helps you
the following can increase the amount of friction in the surface, except ______.
a. glue
b. paper
c. lubricant
d. foil
Answer:
c. lubricant..........
SHORT ANSWER
Which of the balanced chemical equations below has a molar ratio
with the smallest difference between elemental sodium (Na) as a reac-
tant and the chemical formed as a product? Explain.
Use at least two sentences in your answer.
Word Count: (minimum - 10; maximum - 300)
a. 2Na + F2 = 2 NaF
b. 3Na + P = Na3P
c. Na + N2 2Na N
d. Na + 0 = 2 Na o
Answer:
A
Explanation:
the formula ,
From the given chemical equations,the one with molar ratio with smallest difference between elemental sodium (Na) as a reactant and the chemical formed as a product is equation 'a'.
What is chemical equation?Chemical equation is a symbolic representation of a chemical reaction which is written in the form of symbols and chemical formulas.The reactants are present on the left hand side while the products are present on the right hand side.
A plus sign is present between reactants and products if they are more than one in any case and an arrow is present pointing towards the product side which indicates the direction of the reaction .There are coefficients present next to the chemical symbols and formulas .
The first chemical equation was put forth by Jean Beguin in 1615.By making use of chemical equations the direction of reaction ,state of reactants and products can be stated. In the chemical equations even the temperature to be maintained and catalyst can be mentioned.
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Arthur (mass 79 kg) and Violet (mass 50 kg) are trying to play on a seesaw. If Violet sits 4 m from the fulcrum, at what distance from the fulcrum should Arthur sit
Hi there!
There are two torques acting on the system:
τ (Arthur) = RW = R(790N)
τ (Violet) = RW = 4(500N) = 2000N
∑τ = 0 = τ(Violet) - τ(Arthur)
Thus:
τ(Violet) = τ(Arthur)
2000 = 790R
R ≈ 2.53m
solve for the BMI weight 58kg Height 1.61
Answer:
Explanation:
BMI= weight/(height × height) ; weight in kilogram and height in metter
= 58kg / (1.61m × 1.61m )
= (58/ 2.5921) kg/[tex]m^{2}[/tex]
= 22.375 kg/[tex]m^{2}[/tex]
≈ 22.4 kg/[tex]m^{2}[/tex]
Nbel! You’re okay! What happened to you?
Answer:
Wanna be frie nds?
Explanation:
Im smort big brain
Answer:
hello wanna be fr iends and cha-t?
Explanation:
help please please please please
Answer:
lol can u tell what to help u with. cause I didn't understood
Find the gravitational potential energy of a body of mass 25kg,kept at a height of 4m
Answer:
Massm=2.5kg
Massm=2.5kgGravitational potential energy is the work done against force of gravity is stored in the body at a height h .
Massm=2.5kgGravitational potential energy is the work done against force of gravity is stored in the body at a height h .P.E.=U=mgh
Massm=2.5kgGravitational potential energy is the work done against force of gravity is stored in the body at a height h .P.E.=U=mghU=2.5×10×15
Massm=2.5kgGravitational potential energy is the work done against force of gravity is stored in the body at a height h .P.E.=U=mghU=2.5×10×15U=25/10×10×=375j
2. In a race, if a runner starts and stops at the same position, what is their
displacement? *
Answer:
It is the same
Explanation:
I Jsut know
How do you find the magnitude of the air inside a balloon?
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.
What is the vertical component of a ball thrown at a 27 degree angle at 16 m/s?
y = y 0 + v 0 y t − 1 2 g t 2 . If we take the initial position y 0 to be zero, then the final position is y = 10 m. The initial vertical velocity is the vertical component of the initial velocity: v 0 y = v 0 sin θ 0 = ( 30.0 m / s ) sin 45 ° = 21.2 m / s .
Find the gravitational potential energy of a body of mass 25kg,kept at a height of 4m If g=10m/s'.
Answer:
Explanation:E=mgh=25×4×10=1000J
v=VyV67.2 -8.2 m/s
Part B
Now let's see if your prediction agrees with the experimental results. There isn't any data at y=-3.5 m, so you'll need to interpolate by zooming in
on the relationship between y velocity (W) and y displacement. You'll use Tracker's Data tool to help with this analysis. Follow these steps:
• Change the vy versus t graph to a vy versus y graph. (Click the t label on the horizontal axis and select y. position y-component from the
menu.)
• It's a little hard to see what's going on with this graph at y = -3.5, so zoom in. Double-click on the graph to open it up in the Data tool.
. Put your cursor on the point where the graph line intersects y = -3.5. Click and hold to read the vy at that point shown in a yellow box in the
bottom left comer).
What's the interpolated graph value of v, at y = -3.57 Comment on how well this value agrees with the value of v, that you calculated in part A of
this activity. Make a hypothesis about any notable difference.
Answer:
-3.5
Explanation:
b/c the distance is moving in to two A And B A +B or S1 +S2 that is it
Answer:
The interpolated velocity value from the graph is -8.1 meters/second. The calculated value based on the average acceleration is -8.2 meters/second. This is very good agreement.
Explanation:
Menciona tres diferencias entre un planeta y un satélite natural.
Answer:
1. Un planeta gira alrededor del Sol mientras que un satélite gira alrededor de un planeta.
2. Tamaño
3. Aspectos Físicos
4. Creación
4. According to Newton's First Law of Motion, an object with a net force acting on it will
Answer:
a object with net force on it will stay in motion unless acted on by an equal opposite force.
Question 5: A heavy rope is flicked upwards, creating a single pulse in the rope. Make a drawing of the rope and indicate the following in your drawing
a. the direction of motion of the pulse
b. Amplitude
c. Pulse length
d. Position of the rest
Answer:
the direction of motion of the pulse
How would the gravity between two objects change if one object got bigger?
The correct answer is the 3rd answer, gravity would increase its pull on the small object
hope this helps! :)
if there is a gravitational force between all objects, why do we not feel or observe it?
Answer:
We do. It's just way too small compared to the force between the objects we're observing and the earth. It's like looking inside a room with an elephant and a grain of rice. The rice is there, it's just too small compared to the room and the elephant inside for you to notice it. Or, if you ever traveled on a plane, you can easily see the towns, or the roads, but not the single people walking the street.
A student has fallen behind on her homework and suddenly realizes that a big research paper is due the next day and she has not even started working on it. She panics, staying up all night to write her paper, but still runs out of time. At the last minute, she finds a research paper online that covers the same topic as her paper, and she copies and pastes large portions of text from the website into her paper. The student’s behavior is an example of what? a violation of intellectual property rights an online invasion of privacy inappropriate use of Internet sites an incident of cyberbullying
Answer:
Violation of intellectual property rights
Explanation:
Im sorry I am late but for the future people its A! :)) Have a great day :)
unlike other kinds or muscle cells, ____muscle cells are able to contract by themselves
A. epithelial
B. Liver
C. Intestinal
D. cardiac
Answer:
cardiac
Explanation:
cardiac muscle can contact themselves
it is not 132m and plzz explain how the answer is 66m
Distance = 1/2 x velocity x time
Distance = 1/2 x 0.40 x 330
Distance = 66 meters
What is the force that counteracts the thrust force for flight?
The force which counteracts the thrust force for the flight is known as the drag force, as it opposes the flow.
What is drag force?Drag is a force that opposes an object's relative motion to a fluid environment in the field of fluid dynamics. It may be among two liquid film (or surfaces) or in between a liquid and a flat wall. The drag force is influenced by velocity, as opposed to other resistive forces like dry contact, which are essentially independent of it.
When a flow is moving at low or high speed, the drag force is equal to the speed for low pressure and to the square of the velocity for high-speed flow. Although viscous friction is what ultimately causes drag, turbulent drag is unaffected by viscosity.
A force in physics is an input that has the power to change an object's motion. A mass-containing object's velocity can vary, or accelerate, as a result of a force. Intuitively, a push or a pull can also be used to describe forces.
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The drag force, which resists the flow, is the force that balances the propulsion force for flight.
What is Drag force?In the study of fluid dynamics, drag is a force that opposes an object's relative motion to a fluid environment. It could be situated between two liquid surfaces (or films) or between a liquid and a flat wall.
Unlike other resistive forces like dry contact, which are largely independent of velocity, the drag force is affected by it.
For low pressure and high speed flows, respectively, the drag force is equal to the speed for low pressure and the square of the velocity. Although drag is ultimately caused by viscous friction, turbulent.
Thus, The drag force, which resists the flow, is the force that balances the propulsion force for flight.
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Question 2 (1 point)
Potential energy depends on...
Speed and velocity
Position or shape
Length and color
Density and volume
Answer:
potential energy depends on position or shape
The part of the shore that represents the water's edge and migrates back and forth with the tide is known as the ________.
Answer :It is know as the shore line
Explanation:
1 =
Q=240c
t=300s
Please explain Im so confused
Answer:
Answer:
0.8 A
Explanation:
Given: Q = 240 C
t = 300 s
Required: I
Equation: I = Q / t
Solution: I = 240 C / 300 s
Answer: I = 0.8 A
If a person is pushing a cart with a force of 40 Newtons and it
accelerates at 0.5 m/s², what is the mass of the cart?
Answer:
80kg
Explanation:
A = Fm
0.5 = 40m
m = 40/0.5
m = 80kg
If the velocity is 50 and the time is 5 seconds what is the acceleration?
Just divide the both, you will get the answer!
does it sound rude?
im sorry for that!
Answer:
10
Explanation:
the ground wire on a household circuit is typically what color
Answer:
Hey mate....
Explanation:
This is ur answer.....
Green & YellowHope it helps!
Brainliest pls!
Follow me! :)
The ground wire on a household circuit is green in color in a typical household
What is an electric charge?Charged material experiences a force when it is subjected to an electromagnetic field due to the physical property of electric charge. You can have positive or negative electric charges (commonly carried by protons and electrons respectively). opposing charges attract one another whereas similar charges repel one another. We refer to an object as neutral if it has no net charge.
The charge on one electron is -1.6 ×10⁻¹⁹ coulomb.
The ground wire is either bare or has green insulation in most residential branch circuit cables in the US.
The norm in Europe and the UK is
live, brown
neutral is blue.
stripes with green and yellow: earth.
Thus,the ground wire on a household circuit is green in color in a typical household
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A source produces 25 waves every second. The wavelength of the waves is
50 cm. How fast will the waves be travelling?
Answer:
1 s
Explanation:
The speed of the sound, distance travelled and the time taken relationship is given as follows.
Speedofsound,v=Distancetravelled×Timetaken ----eqn 1
With the given values of wavelength = 50 cm = 0.5 m and frequency =1000 Hz, the speed of the wave is calculated as follows.
Speedofsound,v=Frequency(ν)×Wavelength(λ) -- eqn 2
That is, v=ν×λ=1000Hz×0.5m=500m/s
Substituting the value of the speed of sound in eqn 1 gives the time taken for the sound wave to travel a distance of 500 m.
That is, Timetaken=
Distancetravelled
Speedofsound
=
500m
500m/s
=1second.
Hence, the time taken for the sound wave to travel is 1 second.
The speed with which the waves will be travelling is 12.5m/s.
Given the data in the question;
Frequency; [tex]f = 25\ waves\ per\ second = 25s^{-1}[/tex]Wavelength; [tex]\lambda = 50cm = 0.5m[/tex]Speed of the wave; [tex]v =\ ?[/tex]
To determine how fast the waves will be travelling, we use the expression for the relations between wavelength, frequency and speed of wave.
[tex]\lambda = \frac{v}{f}[/tex]
Where [tex]\lambda[/tex] is wavelength, f is frequency and v is speed of wave
We substitute our values into the equation
[tex]0.5m = \frac{v}{25s^{-1}} \\\\v = 0.5m * 25s^{-1}\\\\v = 12.5ms^{-1}[/tex]
Therefore, the speed with which the waves will be travelling is 12.5m/s.
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