Answer:
deformation : elastic deformation is reversed when the force is removed. inelastic deformation is not fully reversed when the force is removed – there is a permanent change in shape.
Explanation:
tysm
 what is the difference between repelling and attracting
Answer:
Attracting means pulling toward you and repelling means pushing away
Explanation:
Answer: Repelling is when something will not connect with another object. The force will cause a repel between the two objects. Attracting is when something is attracted or being pulled to another object.
Explanation: Hope this helps!
PLEASE HELP!!!
If a 40cm rope with a 220g bob can hold a maximum tension of 3N
a) what are the maximum angular velocity and inclination angle it can reach before the rope break?
b) Angle of inclination
Answer:
ω = 3.1 rad/s
θ = 36° from vertical
Explanation:
I will ASSUME that the bob and string is acting as a pendulum.
Please understand that the string will break when the bob is at the lowest point of the swing where the vectors of gravity and centripetal acceleration align. It will NOT break at the angle of maximum inclination measured from vertical. This angle is only a component of the maximum potential energy that gets converted to maximum kinetic energy at the lowest point of the swing.
At the bottom of the swing, the string must support the weight of the bob plus supply the required centripetal acceleration.
F = mg + mω²R
F/m = g + ω²R
F/m - g = ω²R
ω = √((F/m - g)/R)
ω = √((3/0.220 - 9.8)/0.40)
ω = 3.09691...
ω = 3.1 rad/s
Potential energy will convert to kinetic energy
mgh = ½mv²
h = v²/2g
R - Rcosθ = v²/2g
R(1 - cosθ) = v²/2g
1 - cosθ = v²/2gR
cosθ = 1 - v²/2gR
cosθ = 1 - (Rω)²/2gR
cosθ = 1 - Rω²/2g
cosθ = 1 - 0.40(3.1²)/(2(9.8))
cosθ = 0.804267
θ = 36.46045...
θ = 36°
What is non examples of enlarged
reduction
deflation
these are examples of the opposite of enlarged to make something smaller is really the key thing here
A tank having a volume of 0.10m' is filled with oxygen at a pressure of 4.0 x 10'Pa and temperature of 47C. Later it is found that, because of a leak, the pressure has dropped to 3.0 x 10'Pa and the temperature has decreased to 27C. Find. (a) the initial mass of oxygen and (b) the mass that has leaked out.
A. The initial mass of oxygen in the tank is 480.96 g
B. The mass of oxygen that leaked out is 384.96 g
A. Determination of the initial mass of oxygen in the tank.
We'll begin by calculating the initial number of mole of oxygen in the tankVolume (V) = 0.1 m³
Pressure (P) = 4×10⁵ Pa
Temperature (T) = 47 °C = 47 + 273 = 320 K
Gas constant (R) = 8.314 Pa.m³/Kmol
number of mole (n) = ?PV = nRT
0.1 × 4×10⁵ = n × 8.314 × 320
40000 = n × 2660.48
Divide both side by 2660.48
n = 40000 / 2660.48
n = 15.03 molesFinally, we shall determine the initial mass of oxygen.mole = 15.03 moles
molar mass of oxygen gas = 32 g/mol
mass of oxygen gas = ?Mass = mole × molar mass
Mass of oxygen gas = 15.03 × 32
Mass of oxygen gas = 480.96 gTherefore, the initial mass of oxygen in the tank is 480.96 g
B. Determination of the mass of oxygen that leaked out of the tank.
We'll begin by calculating the number of mole of oxygen that leaked out of the tankVolume (V) = 0.1 m³
Pressure (P) = 3×10⁵ Pa
Temperature (T) = 27 °C = 27 + 273 = 300 K
Gas constant (R) = 8.314 Pa.m³/Kmol
number of mole (n) = ?PV = nRT
0.1 × 3×10⁵ = n × 8.314 × 300
30000 = n × 2494.2
Divide both side by 2494.2
n = 30000 / 2660.48
n = 12.03 molesFinally, we shall determine the mass of oxygen that leaked out of tank.mole = 12.03 moles
molar mass of oxygen gas = 32 g/mol
mass of oxygen gas = ?Mass = mole × molar mass
Mass of oxygen gas = 12.03 × 32
Mass of oxygen gas = 384.96 gTherefore, the mass of oxygen that leaked out of tank is 384.96 g
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Alex is x years old . June is 7years older than Alex . their total combined ages is 29 years . find June,s age . show all work algebraically
Answer:
18 yearsExplanation:
Given,
Let Alex be = x years
Then June will be = (7 + x) years
We know that,
Their total combined age is 29 years
Therefore,
By the problem,
=> x + (7 + x) = 29
=> 2x = 29 - 7
=> 2x = 22
=> x = 22 ÷ 2
=> x = 11
So,
Required age of June is = (7 + x) years
= (7 + 11) years
= 18 years (Ans)
A mountain climber encounters a crevasse in an ice field. The opposite side of the crevasse is a height h lower, and is separated horizontally by a distance w. To cross the crevasse, the climber gets a running start and jumps in the horizontal direction. If the height of the crevasse increases but the width remains the same, then,
Select one:
O a. the minimum speed needed to cross the crevasse stays the same.
O b. the minimum speed needed to cross the crevasse will depend on the mass of the mountain climber.
O c. the minimum speed needed to cross the crevasse decreases.
O d. the minimum speed needed to cross the crevasse increases.
O e. the minimum speed needed to cross the crevasse will depend on the weight of the mountain climber.
A 220 g mass is on a frictionless horizontal surface at the end of a spring that has force constant of 7.0 Nm-1. The mass is displaced 5.2 cm from its equilibrium position and then released to undergo simple harmonic motion.
At what displacement from the equilibrium position is the potential energy equal to the kinetic energy.
The work required to stretch the spring by a displacement of 5.2 cm = 0.052 m is
1/2 (7.0 N/m) (0.052 m)² ≈ 0.0095 J
which is stored as potential energy in the spring. When the mass is released, as the spring relaxes, this potential energy is gradually converted into more and more kinetic energy.
Let x be the displacement (relative to the equilibrium position, with 0 < x < 0.052 m) at which this energy is split evenly between potential energy P and kinetic energy K, so that
P + K = 2P = 0.0095 J
or
P ≈ 0.0047 J
At this displacement, the spring is storing
P = 1/2 (7.0 N/m) x²
of potential energy. Solve for x :
1/2 (7.0 N/m) x² ≈ 0.0047 J
x² ≈ (0.0047 J) / (1/2 (7.0 N/m))
x ≈ √((0.0047 J) / (1/2 (7.0 N/m)))
x ≈ 0.037 m = 3.7 cm
The figure shows the light intensity on a screen behind a double slit. The slit spacing is 0.22 mm and the screen is 2.0 m behind the slits (Figure 1). What is the wavelength of the light?
The wavelength of the light is 550 nm
For a double slit interference pattern with slit spacing, d we have
dsinθ = mλ where d = slit spacing = 0.22 mm = 0.22 × 10⁻³ m, m = number of maximum fringe = 2(from the picture) and λ = wavelength of light.
Thus sinθ = mλ/d
Also, tanθ = L/D where L = distance between central maximum and fringe = 2.0 cm/2 = 1.0 cm = 1 × 10⁻² m and D = distance between slit and screen = 2.0 m
Since θ is small, sinθ ≅ tanθ
So, mλ/d = L/D
Making λ subject of the formula, we have
λ = dL/mD
Substituting the values of the variables into the equation, we have
λ = dL/mD
λ = 0.22 × 10⁻³ m × 1 × 10⁻² m/(2 × 2.0 m)
λ = 0.22 × 10⁻⁵ m²/4.0 m
λ = 0.055 × 10⁻⁵ m
λ = 0.55 × 10⁻⁶ m
λ = 550 × 10⁻⁹ m
λ = 550 nm
So, the wavelength of the light is 550 nm
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If you do 72 J of work in 1.2 seconds, how much power is produced
Answer:
Explanation:
Power is the rate of doing work
P = 72 / 1.2 = 60 Watts
which of the following is used to transport sound waves
A.medium
B.vacuum
C.mass
D.light
I think the answer is d.right?
What is the fastest motion that can be measured in any frame of reference?
A. 300,000 m/s
B. 186,000 m/s
C. 186,000 km/s
D. 300,000 km/s
Answer: The answer is D: 300,000km/s
Explanation:
Answer:
(Question) A child is on a playground they start to slide down a large slide. At what point is the child in dynamic equilibrium with the slide?
(Answer) As the child is in motion as they are sliding down.
(Question) Which statement correctly defines dynamic equilibrium?
(Answer) Forces acting on a object are balanced and the object stays in motion.
(Question) While you push a box you begin to decrease the force you are exerting on the box. When will the box reach static equilibrium?
(Answer) When F^push = F^friction
(Question) What is the fastest motion that can be measured in any frame of reference?
(Answer) 300,000 km/s
(Question) Two people are on a train that is moving at 10 m/s north. They are walking 1 m/s south relative to the train. Relative to the ground, their motion is 9 m/s north; Why are we able to use these motions to describe the motion relative to the ground?
(Answer) The people are moving much slower than the speed of light so the ground acts as a frame of reference.
Explanation:
just did the quick check UwU
A force is applied to push a cabinet 12 m across the floor. The work done is 1400 J. How much force was exerted to move the file cabinet?
Answer:
116.66N
Explanation:
work done = force x distance
1400 = force x 12
1400 / 12 = force
116.66N = force
F = W / d
=1400 / 12
= 116.6 N
Work = Force × Distance. The SI unit for work is the joule (J), or Newton • metre (N • m). One joule equals the amount of work that is done when 1 N of force moves an object over a distance of 1 m.
What is distance ?
"Distance is defined to be the magnitude or size of displacement between two positions. Note that the distance between two positions is not the same as the distance travelled between them. Distance travelled is the total length of the path travelled between two positions. Distance travelled is not a vector."
What is force ?"The push or pull on an object with mass that causes it to change its velocity. Force is an external agent capable of changing the state of rest or motion of a particular body. It has a magnitude and a direction."
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What was the big bang?
The big bang is how astronomers explain the way the universe began. It is the idea that the universe began as just a single point, then expanded and stretched to grow as large as it is right now—and it is still stretching!
What's This Big Bang All About?
In 1927, an astronomer named Georges Lemaître had a big idea. He said that a very long time ago, the universe started as just a single point. He said the universe stretched and expanded to get as big as it is now, and that it could keep on stretching.
What an Idea!
The universe is a very big place, and it’s been around for a very long time. Thinking about how it all started is hard to imagine.
Some More Information
Just two years later, an astronomer named Edwin Hubble noticed that other galaxies were moving away from us. And that’s not all. The farthest galaxies were moving faster than the ones close to us.
This meant that the universe was still expanding, just like Lemaître thought. If things were moving apart, it meant that long ago, everything had been close together.
Everything we can see in our universe today—stars, planets, comets, asteroids—they weren't there at the beginning. Where did they come from?
A Tiny, Hot Beginning
When the universe began, it was just hot, tiny particles mixed with light and energy. It was nothing like what we see now. As everything expanded and took up more space, it cooled down.
The tiny particles grouped together. They formed atoms. Then those atoms grouped together. Over lots of time, atoms came together to form stars and galaxies.
The first stars created bigger atoms and groups of atoms. That led to more stars being born. At the same time, galaxies were crashing and grouping together. As new stars were being born and dying, then things like asteroids, comets, planets, and black holes formed!
Describe how you can determine:
a) Volume of an irregular body
b) Density of a liquid
Density of liquid try thank you so much
Answer:
a) measure the change in volume when the object is immersed; compute from range data
b) find the ratio of mass to volume for a measured mass and volume
Explanation:
a) The volume of a small enough irregular body can be found by measuring the difference in volume of the (semi-)fluid in which it is immersed, before and after immersion.
For irregular bodies for which that approach does not work, various 3D scanners are available for measuring volume and surface area. They may rely on optical (laser or camera), sonic, or radar measurements, and generally involve computation from distances to various points.
__
b) Density is the ratio of mass to volume. So, measurements of mass and volume of a liquid sample are sufficient to provide the basis for determining density.
Other methods include measuring buoyancy forces, and/or the depth of submersion of something that floats in the liquid. For specific liquids, hydrometers are available for measuring their density relative to that of water.
A hot air balloon rising vertically is tracked by an observer located 2 miles from the lift-off point. At a certain moment, the angle between the observer's line-of-sight and the horizontal is , and it is changing at a rate of 0.1 rad/min. How fast is the balloon rising at this moment
We have that for the Question, it can be said that
the balloon rising at [tex]0.266miles/min[/tex]From the question we are told
An observer located 2 miles from the lift-off point. At a certain moment, the angle between the observer's line-of-sight and the horizontal is , and it is changing at a rate of 0.1 rad/min.From,
[tex]tan\theta = \frac{h}{2}[/tex]
differentiate with respect to h
[tex]sec^2\theta * \frac{do}{dz} = \frac{1}{2} * \frac{dh}{dz}\\\\\frac{dh}{dz} = 2 sec^\theta * \frac{d\theta}{dz}\\\\\theta = \frac{\pi}{6} and \frac{d\theta}{dz} = 0.1rad/min\\\\\frac{dh}{dz} = 2sec^2 (\frac{\pi}{6}) * (0.1)\\\\= 0.266miles/min[/tex]
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1.25 is closer to 1.04 or not ?
plz heelp plz
Answer:
No, it is closer to 1.30
Explanation:
Someone fires a slingshot at a target that is far enough away to take 1.4 seconds to reach. How far below does the target does the slingshot pellet hit?
A student uses a spring (with a spring constant of 180 N/m) to launch a marble vertically into the air. The mass of the marble is 0.004 kg and the spring is compressed 0.03 m. How high will the marble go? Answer:
Answer:
Explanation:
Ignoring air resistance the spring compression energy will equal the increase in potential energy from the maximum spring compression point to the top of the flight arc.
mgh = ½kx²
h = kx²/2mg
h = 180(0.03²) / (2(0.004)(9.8))
h = 2.06632...
h = 2.1 m
Which of the following correctly compares gravitational force and distance between two objects?
A. As the distance increases, the gravitational force decreases.
B. As the distance decreases, the gravitational force decreases.
C. There is no relationship between gravitational force and distance between two objects.
D. As the distance increases, the gravitational force increases.
Help pls…
Answer:
a. as the distances increases , the gravitational force decreases
Which force, in real life, will have the least effect on a bowling ball rolling down a lane toward bowling pins?
A) magnetism
B) air resistance
C) gravity
D) friction
Answer:
Its Friction
Explanation:
the pins are not floating and they are not a magnet and not involved with air
The force, in real life, that will have the least effect on a bowling ball rolling down a lane toward bowling pins is magnetism. The correct option is A.
What is magnetism?Magnetism is basically the force which indeed magnets exert when they attract or even repel one another. The movement of electric charges resulting in magnetism.
Every substance is composed of tiny units referred to as atoms. Each atom contains electrons, which are charged particles.
To increase stability, the pins themselves have a low center of gravity. Because they are spherical in shape, they can roll and strike other pins in a variety of directions.
The force acting on the bowling ball is friction and air resistance. The friction force is equal to the friction coefficient multiplied by the normal force, and thus mass times acceleration.
Thus, the correct option is A.
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50 points help
Column I Column II
______ 1. acceleration a. change in distance over time
______ 2. speed b. time interval
______ 3. velocity c. scalar
______ 4. Δt. d. change in position
______ 5. Magnitude only e. change in velocity over time
______ 6. Δx f. change in displacement over time
[tex]\\ \sf\longmapsto Acceleration\longrightarrow Change\:in\: Velocity\:over\:time[/tex]
[tex]\\ \sf\longmapsto Speed\longrightarrow Change\:in\: Distance\:over\:Time[/tex]
[tex]\\ \sf\longmapsto Velocity\longrightarrow Change\:in\: Displacement\: over\:time[/tex]
[tex]\\ \sf\longmapsto ∆t\longrightarrow Time\: interval [/tex]
[tex]\\ \sf\longmapsto Magnitude\:only\longrightarrow Scaler[/tex]
[tex]\\ \sf\longmapsto ∆x=Change\:in\: position [/tex]
Required information
Medical testing has established that the maximum acceleration a pilot can be subjected to without losing consciousness is
approximately 5.00g. A pilot can avoid "blackout" at accelerations up to approximately 9.00g by wearing special "g-suits"
that help keep blood pressure in the brain at a sufficient level.
What is the minimum safe radius of curvature for an unprotected pilot flying an F-15 in a horizontal circular loop at 729 km/h?
Answer:
hi there is that OK for the weekend of the following week as well
Explanation:
6th of March is fine for me
The minimum safe radius of curvature for an unprotected pilot flying an F-15 in a horizontal circular loop at 729 km/h is approximately 838.1 meters.
To determine the minimum safe radius of curvature for an unprotected pilot flying an F-15 in a horizontal circular loop, we need to consider the maximum acceleration the pilot can withstand without losing consciousness.
Given:
Maximum acceleration without losing consciousness = 5.00g
Acceleration with g-suits to avoid blackout = 9.00g
First, we need to convert the speed of the F-15 from km/h to m/s:
Speed = 729 km/h = (729 * 1000) m/3600 s ≈ 202.5 m/s
Next, we'll calculate the acceleration experienced by the pilot in the circular loop. In a horizontal circular motion, the centripetal acceleration is given by:
Acceleration = ([tex]\rm Velocity^2[/tex]) / Radius
We can rearrange the equation to solve for the radius:
Radius = ([tex]\rm Velocity^2[/tex]) / Acceleration
Using the maximum acceleration of 5.00g, we convert it to [tex]\rm m/s^2[/tex]:
Maximum acceleration = 5.00g ≈ (5.00 * 9.8) [tex]\rm m/s^2[/tex] = 49 m/s^2
Now, we can calculate the minimum safe radius of curvature:
Radius = ([tex]\rm 202.5^2[/tex]) / 49 ≈ 838.1 meters
Therefore, the minimum safe radius of curvature for an unprotected pilot flying an F-15 in a horizontal circular loop at 729 km/h is approximately 838.1 meters.
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Acceleration of a car that speeds from 4.3 m/s to 12.9 m/s in 2 seconds
Explanation:
let v1 = 4.3 m/s
v2 = 12.9 m/s
t = 2 seconds
v2 = v1 + at
12.9 = 4.3 + a×2
2a = 12.9 - 4.3 = 8.6
a = 8.6/2
a = 4.3 m/s^2
A meter stick is attached to one end of a rigid rod with negligible mass of length l = 0.302 m. The other end of the light rod is suspended from a pivot point, as shown in the figure below. The entire system is pulled to a small angle and released from rest. It then begins to oscillate. A meter stick hung from a rod of length l. The rod is attached to the ceiling. The rod and meter stick extend downward in a straight line making a small angle with the vertical. (a) What is the period of oscillation of the system (in s)? (Round your answer to at least three decimal places.)
The period of oscillation of the system nearest to three decimal places
= 1.092 seconds
The period of an oscillation occurring in a system is the time taken to complete one cycle.
The formula that is used to calculate the period of oscillation (T) is
= 2π√[tex]\frac{l}{g}[/tex]
But,
π = 3.14159 (constant)
g= 10m/s² (acceleration due to gravity)
l = 0.302 m
Therefore T = 2 × 3.14159 × √[tex]\frac{0.302}{10}[/tex]
= 6.28318 x √0.0302
= 6.28318 x 0.17378
= 1.09189s
= 1.092 seconds ( to the nearest three decimal places)
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Define average atomic mass and explain how it is calculated
Answer:
The average atomic mass of an element is the sum of masses of it's isotopes
Each are multiplied by it's natural abundance
Explanation:
The average atomic mass of an element is the sum of masses of it's isotopes
Each are multiplied by it's natural abundance
A Scooter has a mass of 250 kg. A constant force is exerted on it for 6.0 s. During the time the force is exerted, the scooter increases its speed from 6.00 m/s to 280 m/s. What is the magnitude of the force exerted on the scooter
Answer:
917 N
Explanation:
917 N, this is your answer!!
Glad to help.
hey I just wanted to know if any of the guys here are able to help answer my physics questions , it would be a great thankyou xoxoxoxo
what's your question?
5. A quarterback throws the football to a stationary receiver who is 31.5 m
down the field. If the football is thrown at an initial angle of 40.0° to the
ground, at what initial speed must the quarterback throw the ball for it
to reach the receiver? What is the ball's highest point during its flight?
The projectile launch equations allow to find the results for the questions about the movement of the ball are:
The initial velocity is: v₀ = 17.7 m / s. The maximum height is: y = 16 m.
Given parameters
Horizontal distance x = 31.5 m Launch angle tea = 40ºTo find
The initial speed. Maximum height.
Projectile launching is an application of kinematics, where on the x-axis there is no acceleration and on the y-axis is the gravity acceleration.
The range is the distance traveled for the same departure height, see attached.
.
R =[tex]\frac{v_o^2 \ sin 2\theta}{g}[/tex]
[tex]v_o^2 = \frac{ g R}{sin 2 \theta }[/tex]
Let's calculate.
v₀² = [tex]\frac{9.8 \ 31.5}{sin \ (2 \ 40)}[/tex]9.8 31.5 / sin (2 40.0)
[tex]v_o = \sqrt{313.46}[/tex]o = ra 313.46
v₀ = 17.7 m / s
At the point of maximum height the vertical speed is zero.
v² = v₀² - 2 g y
0 = v₀² - 2g y
y = [tex]\frac{v_o^2}{2g}[/tex]
Let's calculate.
y = [tex]\frac{17.7^2}{2 \ 9.8}[/tex]
y = 16 m
In conclusion, using the projectile launch equations we can find the results for the questions about the movement of the ball are:
The initial velocity is: v₀ = 17.7 m / s The maximum height is: y = 16 m.
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What on earth is equal to 9.8m/s/s
Answer:
Acceleration due to gravity
Which statement about RNA polymerase is NOT true?
RNA polymerase reads a template strand of DNA 5' to 3'
RNA polymerase binds to a promoter region to initiate transcription
RNA polymerase adds a ribonucleotide to the 3' end of a growing RNA molecule
During transcription of a gene, RNA polymerase reads only one strand of DNA
Please explain!
Answer:
RNA polymerase reads a template strand of DNA 5' to 3'
Explanation:
RNA polymerase is the enzyme accountable for the transcription of the DNA. It shows one coast of the DNA and transcribed within the RNA. It combines the nucleotide to the 3' conclusion of the expanding series of the RNA. It commences the method by connecting to the promoter area of the gene. The enzyme increases nucleotide, not ribonucleotide.
Therefore, A. RNA polymerase reads a template strand of DNA 5' to 3' is false.
Answer:
RNA polymerase reads a template strand of DNA 5' to 3'.
Explanation:
RNA synthesis is less processive during elongation than during initiation. Synthesis begins with the polymerase binding two rNTP molecules. The polymerase frequently releases the nascent transcript before it reaches 8-10 nucleotides in length.
