Answer:
1.) Everything that moves, will eventually come to a stop. Rest is the “natural” state of all objects
Of all physics misconceptions, this is the most common. Even the great philosopher Aristotle, included it into his most important contribution to the field, his famous Laws of Motion. But now we know it is wrong because Newton’s First Law of Motion tells us that “everything at rest will stay at rest, and everything in motion will stay in motion, unless acted upon by an external force.”
The first statement seems reasonable enough, but the second part is a little bit murky. The reason this confusion persists boils down to the fact that we are unable to identify the force that stops all motion, which is friction. Friction is a force that acts between two objects that are in contact and are moving relative to each other. When we roll a ball, it stops because of the frictional force acting between it and the floor.
2.) A continuous force is needed for continuous motion
This misconception is a direct consequence of the first one. While this is true, if you are, for example, pushing a grocery cart in a supermarket, again this is only because there is friction involved. The force you apply to keep an object moving is only to counteract the frictional force. If you were to throw a rock on outer space, it would travel with a constant velocity forever, unless it hits something, of course. This is because space is mostly empty (it has trace elements of gas and dust throughout), and there would not be any frictional force acting on that rock.
3.) An object is hard to push because it is heavy
This is one of the most common misconceptions because it’s something we see and feel everyday. While a heavy object is really hard to push, it is not because of its weight, but because of its inertia or mass. Inertia is an objects resistance to change in motion. It is important to note that inertia is resistance to “change motion” rather than just motion itself. When, I was a kid, I imagined that it would be easy to carry and push massive objects when in outer space, but not surprisingly, my younger self was wrong.,
With that said… Since these objects still have mass despite being weightless, this mass represents the object’s inertia.
4.) Planets revolve around the sun because they are pushed by gravity
We have to remember that gravity — the weakest of the four fundamental forces — is an attractive force. The reason why planets revolve around the Sun can be chalked up to the fact that the planets were already spinning within the protoplanetary disk encircling a young Sun. Gravity merely keeps the planets in orbit around the Sun, but it isn’t necessarily the one thing pushing the planets along their orbital plane.
5.) Heavier objects fall faster than lighter ones
This misconception is already debunked long ago by Galileo on his experiment when he dropped two objects with different masses on the Leaning Tower of Pisa. He has shown on that experiment that objects move downward with the same acceleration.
Again, the problem comes from not being able to identify another force that is involved, which is air resistance. All objects moving through air, and hence, all falling objects, experience air resistance. This force is proportional to the area of the object in the direction of motion. Usually, this force is negligible, but for light objects — with weight comparable to the air resistance, like a feather — it will have a big effect. This is ultimately confirmed by the famous hammer and feather drop experiment on the moon.
6.) There is no gravity in outer space
There is gravity in outer space, it is just weaker than what we experience here on Earth. Astronauts that are orbiting the Earth don’t experience gravity because they are free-falling (yes, you read that right). All satellites, including the moon and the planets, are in a constant state of freefall.
They just also have a tangential velocity with their free fall, that is why they don’t crash to what they are orbiting. When something is in free fall, it becomes weightless. This is why Kate Upton can do a photo shoot in zero gravity here on Earth. The plane that they are riding in actually went into free fall to do that.
7.) Planets move in circular orbits around the Sun
Planets actually move in elliptical orbits around the sun (with the Sun being the focus of the ellipse). This is actually the first of Kepler’s Three Laws of Planetary Motion, which deals with precisely how planets orbit the Sun.
One misconception deals with our seasons. Some might wrongly come to the conclusion that Earth’s proximity to the Sun dictates the seasons (summer is when Earth is closest to the Sun and winter is when it’s farther away), but that’s not entirely true. In reality, our seasons are caused by the tilt of Earth’s axis.
What is the mass of an object that has a weight of 73.8 N
Answer:
[tex]we \: have \: \\ w = mg \\ 73.8 = m \times 9.8 \\ or \: m = \frac{73.8}{9.8} \\ or \: m = 7.5306[/tex]
According to O*NET, what are common work contexts for Foresters?
Answer: Foresters. Manage public and private forested lands for economic, recreational, and conservation purposes. May inventory the type, amount, and location of standing timber, appraise the timber's worth, negotiate the purchase, and draw up contracts for procurement.
Explanation:
which of these would most likely be a parts of a lab procedure?
A. write a hypothesis to answer a question
B. write a title at the top of a completed lab report
C. record the time to complete a chemical reaction
D. create a question on the cause of a chemical reaction
An object accelerates from 9 m/s to 26 m/s in 8 seconds. What is the acceleration? Round your answer to
the nearest tenth.
Explanation:
Acceleration is the change in velocity over time.
a = Δv / Δt
a = (26 m/s − 9 m/s) / 8 s
a ≈ 2.1 m/s²
A skyrocket travels 113m at an angle of 82.4 with respect to the ground and toward the south. What’s the rocket’a horizontal displacement?
Answer:
∆x = 14.9 m, south
Explanation:
∆x = d(cos q) = (113 m)(cos 82.4°)
If a skyrocket travels 113 meters at an angle of 82.4 with respect to the ground and toward the south, then the horizontal displacement of the rocket would be 14.945 meters southwards.
What is displacement?An object's position changes if it moves in relation to a reference frame, such as when a passenger moves to the back of an airplane or a professor moves to the right in relation to a whiteboard. Displacement describes this shift in location.
As given in the problem if a skyrocket travels 113 meters at an angle of 82.4 with respect to the ground and toward the south.
The horizontal displacement = 113 Cos82.4
= 14.945 meters
Thus, the horizontal displacement of the rocket would be 14.945 meters southwards.
To learn more about displacement here, refer to the link;
https://brainly.com/question/10919017
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An object of irregular shape has a characteristic length of L = 0.5 m and is maintained at a uniform surface temperature of Ts = 400 K. When placed in atmospheric air at a temperature of T[infinity] = 300 K and moving with a velocity of V = 25 m/s, the average heat flux from the surface to the air is 10,000 W/m2 . If a second object of the same shape, but with a characteristic length of L = 2.5 m, is maintained at a surface temperature of Ts = 400 K and is placed in atmospheric air at T[infinity] = 300 K, what will the value of the average convection coefficient be if the air velocity is V = 5 m/s?.
Answer:
The value of the average convection coefficient is 20 W/Km².
Explanation:
Given that,
For first object,
Characteristic length = 0.5 m
Surface temperature = 400 K
Atmospheric temperature = 300 K
Velocity = 25 m/s
Air velocity = 5 m/s
Characteristic length of second object = 2.5 m
We have same shape and density of both objects so the reynold number will be same,
We need to calculate the value of the average convection coefficient
Using formula of reynold number for both objects
[tex]R_{1}=R_{2}[/tex]
[tex]\dfrac{u_{1}L_{1}}{\eta_{1}}=\dfrac{u_{2}L_{2}}{\eta_{2}}[/tex]
[tex]\dfrac{h_{1}L_{1}}{k_{1}}=\dfrac{h_{2}L_{2}}{k_{2}}[/tex]
Here, [tex]k_{1}=k_{2}[/tex]
[tex]h_{2}=h_{1}\times\dfrac{L_{1}}{L_{2}}[/tex]
[tex]h_{2}=\dfrac{q}{T_{2}-T_{1}}\times\dfrac{L_{1}}{L_{2}}[/tex]
Put the value into the formula
[tex]h_{2}=\dfrac{10000}{400-300}\times\dfrac{0.5}{2.5}[/tex]
[tex]h_{2}=20\ W/Km^2[/tex]
Hence, The value of the average convection coefficient is 20 W/Km².
What are spooky rays? How were they discovered? How are they related to visible light?
Which is the average velocity of a 35 kg kid sliding for 3.66 m on ice?
Answer:
Calculate the displacement of the car during the above acceleration. { ⃑ = –130 m} c. ... A 2.0 kg brick has a sliding coefficient of friction of 0.38.
Explanation:
What equal amounts of positive charge would have to be placed on the Earth and on the Moon to neutralize their gravitational attraction? Do you need to know the Moon’s distance to solve this problem? Why or why not? (b) How many metric tons of hydrogen would be needed to provide the positive charge calculated in part (a)? The molar mass of hydrogen is 1.008 g/mol.
Answer:
Fg = G M m / R^2 gravitational attraction
Fe = k Q^2 / R^2 electric repulsion
G M m = k Q^2
Q = (G M m / k)^1/2
Since m = .0123 M
Q = (.0123 G / k)^1/2 * M
Q = (.0123 * 6.67 * 10E-11 / 9 * 10E9)^1/2 * 5.98 * 10E24
Q = 5.71 * 10E13 C charge required on each body
n = 5.71 * 10E13 / 1.6 * 10E-19 = 3.57 * 10E32 atoms
N = 3.57 * 10E32 / 6.02 * 10E23 = 5.93 * 10E8 g-mol = 5.93 * 10E*5 kg-mol
Since 1 metric ton = 1000 kg
One would need 593 metric tons of hydrogen
The mass of an astronaut on Earth is 70 kg and her weight is 686N. What will the weight of the astronaut be on the moon? The moon has a gravitational pull of 1.62m/s².
A. 6.86kg
B. 70kg
C. 162.2N
D. 113.4N
Answer: C is correct
Explanation:
its a fill in the blank can someone plz help me
Answer:
That is Melon Collision and the second blank is Lone Collision.
Explanation:
5.16 An insulated container, filled with 10 kg of liquid water at 20 C, is fitted with a stirrer. The stirrer is made to turn by lowering a 25-kg object outside the container a distance of 10 m using a frictionless pulley system. The local acceleration of gravity is 9.7 m/s2. Assume that all work done by the object is transferred to the water and that the water is incompressible. A. Determine the work transfer (kJ) to the water. B. Determine the increase in internal energy (kJ) of the water. C. Determine the final temperature ( C) of the water. HINT: Assume that the temperature change is small enough that a constant value of the specific heat is a good approximation. D. Determine the heat transfer (kJ) from the water required to return the water to its initial temperature. Turns, Stephen R.. Thermodynamics (p. 294). Cambridge University Press. Kindle Edition.
Answer:
a) W=2.425kJ
b) [tex]\Delta E=2.425kJ[/tex]
c) [tex]T_f=20.06^{o}C[/tex]
d) Q=-2.425kJ
Explanation:
a)
First of all, we need to do a drawing of what the system looks like, this will help us visualize the problem better and take the best possible approach. (see attached picture)
The problem states that this will be an ideal system. This is, there will be no friction loss and all the work done by the object is transferred to the water. Therefore, we need to calculate the work done by the object when falling those 10m. Work done is calculated by using the following formula:
[tex]W=Fd[/tex]
Where:
W=work done [J]
F= force applied [N]
d= distance [m]
In this case since it will be a vertical movement, the force is calculated like this:
F=mg
and the distance will be the height
d=h
so the formula gets the following shape:
[tex]W=mgh[/tex]
so now e can substitute:
[tex]W=(25kg)(9.7 m/s^{2})(10m)[/tex]
which yields:
W=2.425kJ
b) Since all the work is tansferred to the water, then the increase in internal energy will be the same as the work done by the object, so:
[tex]\Delta E=2.425kJ[/tex]
c) In order to find the final temperature of the water after all the energy has been transferred we can make use of the following formula:
[tex]\Delta Q=mC_{p}(T_{f}-T_{0})[/tex]
Where:
Q= heat transferred
m=mass
[tex]C_{p}[/tex]=specific heat
[tex]T_{f}[/tex]= Final temperature.
[tex]T_{0}[/tex]= initial temperature.
So we can solve the forula for the final temperature so we get:
[tex]T_{f}=\frac{\Delta Q}{mC_{p}}+T_{0}[/tex]
So now we can substitute the data we know:
[tex]T_{f}=\frac{2 425J}{(10000g)(4.1813\frac{J}{g-C})}+20^{o}C[/tex]
Which yields:
[tex]T_{f}=20.06^{o}C[/tex]
d)
For part d, we know that the amount of heat to be removed for the water to reach its original temperature is the same amount of energy you inputed with the difference that since the energy is being removed this means that it will be negative.
[tex]\Delta Q=-2.425kJ[/tex]
A racecar goes around a 400m track in 58 seconds, what is the speed and velocity?
Answer:
v = 6.89 [m/s]
velocity = 0
Explanation:
To solve this problem we must use the definition of speed which is defined by the following expression of physics, such as the relationship between distance and time.
v = x/t
where:
v = speed [m/s]
x = distance = 400 [m]
t = time = 58 [s]
v = 400/58
v = 6.89 [m/s]
The velocity is a vector, since the racing car moves in circles IE its movement started where it ended, its velocity is zero.
A graduated cylinder has 20mL of water. A rock is placed in the graduated cylinder and the volume rises to 30mL. How can you calculate the volume of the rock?
Answer:
B. Subtract the new volume from the original volume
Explanation:
You have just moved into a new apartment and are trying to arrange your bedroom. You would like to move your dresser of weight 3,500 N across the carpet to a spot 5 m away on the opposite wall. Hoping to just slide your dresser easily across the floor, you do not empty your clothes out of the drawers before trying to move it. You push with all your might but cannot move the dresser before becoming completely exhausted. How much work do you do on the dresser? You have just moved into a new apartment and are trying to arrange your bedroom. You would like to move your dresser of weight 3,500 across the carpet to a spot 5 away on the opposite wall. Hoping to just slide your dresser easily across the floor, you do not empty your clothes out of the drawers before trying to move it. You push with all your might but cannot move the dresser before becoming completely exhausted. How much work do you do on the dresser? W>1.75×104J W=1.75×104J 1.75×104J>W>0J W=0J
Answer:
W = 0J
Explanation:
The work done by the dresser is described as
W = f d (cos θ)
F has been given as the weight of this dresser. And it is 3500 N
d = 0 m
When you put these values into the equation
W = 3500 x 0 x cosθ
W = 0 J
This value tells us that the work done on this dresser is zero. No work has been done. Therefore the last option answers the question.
Equation of Motion Problems
Displacement. Distance. Speed and Velocity
1. Donovan Bailey ran the 100 m dash at the Atlanta Olympics in 9.84 s. Michael Johnson ran the 200 m in 19.32
s and the 400 m in 43.49 s. Find their average speed in each case.
Wom
kilometrom
Answer:
Below in bold.
Explanation:
1. Speed = distance / time
= 100 / 9.84
= 10.16 m/s.
Speed = 200/19.32 = 10.35 m/s.
Speed = 400/43.49 = 9.20 m/s.
Property of matter which remains at rest, or in motion, unless acted upon
by an external force. *
Inference
Theory
Hypothesis
Inertia