The expression for the cord's potential energy as a function of its stretch shows that the potential energy increases with the square of the stretch, and that it is proportional to the spring constant.
When an elastic cord is stretched, it stores potential energy in the form of elastic potential energy.
The amount of potential energy stored in the cord is directly proportional to the amount of stretch in the cord.
As a bungee cord is stretched, its potential energy increases in a non-linear fashion.
The expression for the cord's potential energy as a function of its stretch can be derived using Hooke's law, which states that the force exerted by a spring is proportional to its stretch.
The force exerted by an elastic cord can be expressed as:
F = -kx
where F is the force,
k is the spring constant,
and x is the amount of stretch in the cord.
The negative sign indicates that the force is opposite in direction to the stretch.
The potential energy stored in the cord can be calculated by integrating the force over the distance of the stretch:
U = ∫Fdx = -∫kx dx = -1/2 [tex]kx^2 + C[/tex]
where U is the potential energy,
C is a constant of integration, and the limits of integration are the un-stretched length of the cord and the stretched length.
The expression for the cord's potential energy as a function of its stretch shows that the potential energy increases with the square of the stretch, and that it is proportional to the spring constant.
This means that a stiffer cord will store more potential energy than a less stiff cord, for the same amount of stretch.
It also means that as the cord is stretched further, the increase in potential energy becomes more significant, making it more difficult to stretch the cord further.
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Can someone help with the top two question please.
(What does this diagram show?)
(What is it called?)
Answer:
These are called Sankey Diagrams.
These summarise all the energy transfers taking place in a process. Sankey diagrams are drawn to scale - the thicker the line or arrow, the greater the amount of energy involved. Usually, the top line/arrow shows the amount of useful energy transferred from the total input and the one which curves shows the wasted/dissipated energy of the total input energy. It helps to show efficiency in this manner.
Hope this helps!!!
4) a button is at the rim of a turntable of radius 15.0 cm rotating at 45.0 rpmn. what is the minimum coefficient of friction needed for it to stay on?
The minimum coefficient of friction needed for the button to stay on the turntable is 0.22, assuming the turntable is a flat disk and neglecting air resistance.
When an object is placed on a rotating turntable, there must be a sufficient coefficient of friction between the object and the turntable to prevent it from sliding off. In this scenario, the button is placed at the rim of a turntable with a radius of 15.0 cm and is rotating at 45.0 rpm (revolutions per minute).
To determine the minimum coefficient of friction required for the button to stay on the turntable, we can use the centripetal force equation. The centripetal force required to keep the button in circular motion is provided by the force of friction acting on the button. If the frictional force is less than the centripetal force required, the button will slip off the turntable.
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a rocket is being launched. after a while, it rises at a constant velocity of 30 miles per second. an observer is standing 112 miles from the launching site. how fast is the distance between the rocket and the observer changing when the rocket is at a height of 15 miles? present your answer as an approximation, accurate up to three or more decimals.
Answer:
To solve this problem, we need to use the chain rule to find the derivative of the distance between the rocket and the observer with respect to time.
Let's call the distance between the rocket and the observer "d" and the height of the rocket "h". We know that the rocket is rising at a constant velocity of 30 miles per second, so its height can be expressed as:
h = 30t
where t is the time since the rocket started rising.
Using the Pythagorean theorem, we can express the distance between the rocket and the observer as:
d^2 = h^2 + 112^2
Differentiating both sides with respect to time, we get:
2d * (dd/dt) = 2h * (dh/dt)
where (dd/dt) is the rate of change of the distance between the rocket and the observer and (dh/dt) is the rate of change of the height of the rocket.
Solving for (dd/dt), we get:
(dd/dt) = (h/d) * (dh/dt)
Substituting the expressions for h and d, we get:
(dd/dt) = (30t) / sqrt((30t)^2 + 112^2)
When the rocket is at a height of 15 miles, we can find the corresponding value of t as follows:
15 miles = 15 * 5280 feet = 79200 feet
79200 feet / 5280 feet per second = 15 seconds
Substituting t = 15 seconds into the expression for (dd/dt), we get:
(dd/dt) = (30 * 15) / sqrt((30 * 15)^2 + 112^2)
= 450 / sqrt(22500 + 12544)
= 450 / sqrt(35044)
= 2.409
Therefore, the rate of change of the distance between the rocket and the observer when the rocket is at a height of 15 miles is approximately 2.409 miles per second.
What is the answer to this question???
The wavelength of the wave will increase to 10 meters. Option B.
Wavelength and frequencyThe speed of a wave (v) is equal to the product of its wavelength (λ) and frequency (f):
v = λ × f
Rearranging this equation to solve for frequency:
f = v/λ
In the first scenario, where the wave has a speed of 20 m/s and a wavelength of 5 meters, we can calculate the frequency as:
f1 = v/λ1 = 20/5 = 4 Hz
In the second scenario, the frequency is reduced by half, so the new frequency is:
f2 = f1/2 = 4/2 = 2 Hz
To find the new wavelength (λ2), we can rearrange the original equation to solve for wavelength:
λ = v/f
Substituting the values for the second scenario, we get:
λ2 = v/f2 = 20/2 = 10 meters
Therefore, if the same wave was created in the same medium with half of the original frequency, the wavelength would double and become 10 meters.
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How much energy needed to raise temperature 1 degree Celsius?
Answer:
4.18 joules of heat energy to raise a gallon of water by 1 degree celcius
Explanation:
the height of the washington monument is measured to be 170 m on a day when the temperature is 35.0°c. assume its thermal coefficient of linear expansion is 2.5 × 10-6/°c.
Answer:
To determine how much the height of the Washington Monument changes due to the increase in temperature, we can use the formula:
ΔL = αLΔT
where ΔL is the change in length, α is the coefficient of linear thermal expansion, L is the original length, and ΔT is the change in temperature.
In this case, we want to find the change in height, which is the same as the change in length along the vertical direction. Therefore, we can use the same formula to find the change in height:
Δh = αhΔT
where Δh is the change in height, α is the coefficient of linear thermal expansion, h is the original height, and ΔT is the change in temperature.
Substituting the given values, we get:
Δh = (2.5 × 10^-6/°C) × (170 m) × (35.0°C)
Δh ≈ 0.15 m
Therefore, the height of the Washington Monument increases by approximately 0.15 m when the temperature increases from 0°C to 35.0°C.
is the statement true or false? waves propagate faster in a less dense medium if the stiffness is the same.
Answer:
The statement is True, waves propagate faster in a less dense medium if the stiffness is the same.
What are waves?Waves can be defined as a disturbance that travels through space or matter, transferring energy from one point to another without transporting matter.
There are various kinds of waves, and they all exhibit similar properties.
Waves propagate faster in a less dense medium if the stiffness is the same .
Waves move at various speeds in different media. The speed of a wave is determined by the nature of the medium and the frequency and wavelength of the wave.
In general, waves propagate faster in less dense media than in denser ones.
This holds true if the stiffness is constant. For example, sound waves travel quicker in air than in water because air is less dense than water. In brief, waves propagate faster in a less dense medium if the stiffness is constant.
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calculate the force needed to bring a 950-kg car to rest from a speed of 25 m/s in a distance of 120 m.
The force needed to bring the 950-kg car to rest from a speed of 25 m/s in a distance of 120 m is approximately 4959.5 N.
To calculate the force needed to bring the car to rest, we can use the equation, F = ma, where, F = force, m = mass of the car and a = acceleration.
We can also use the equation for acceleration,
a = (v_f^2 - v_i^2)/2d, where, v_f = final velocity (0 m/s since the car is brought to rest), v_i = initial velocity (25 m/s) and d = distance traveled during braking (120 m).
Substituting the given values,
a = (0^2 - 25^2)/(2 x 120) = -5.21 m/s^2
The negative sign indicates that the acceleration is in the opposite direction to the initial velocity, which is necessary to bring the car to rest.
Substituting the value of acceleration into the equation for force,
F = ma = (950 kg) x (-5.21 m/s^2) = -4959.5 N
Again, the negative sign indicates that the force is in the opposite direction to the initial velocity.
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Triangle jkl is equilateral. all three interior angles have equal measures. triangle j k l is an equilateral triangle. the exterior angle to angle j is 1. what is the measure of exterior angle 1? 30 degrees 60 degrees 120 degrees 130 degrees
The measure of exterior angle is 130 degrees. Option D is correct.
Since triangle jkl is equilateral, all its interior angles have a measure of 60 degrees. The sum of an exterior angle and an interior angle of a triangle is always 180 degrees. Therefore, the measure of exterior angle 1 is the sum of angle j and angle k. Since angle j is 60 degrees, angle k must be 180 - 60 = 120 degrees, because the sum of angles in a triangle is always 180 degrees.
Therefore, the measure of exterior angle 1 is 60 + 120 = 180 degrees. However, an exterior angle is defined as the angle formed by a side of a triangle and the extension of an adjacent side. Therefore, the actual exterior angle 1 is the supplement of 180 degrees, which is 180 - 1 = 179 degrees. So the answer is (D) 130 degrees.
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what converts electrical energy into mechanical energy?
An electric motor is an electrical machine that converts electrical energy into mechanical energy.
It works on the principle of electromagnetic induction, where a current-carrying conductor placed in a magnetic field produces a mechanical force.
Electric motors are commonly used in applications such as household appliances, power tools, and vehicle propulsion systems. They come in a variety of sizes and power ratings, and are used in industrial, automotive, and consumer applications. The most common type of electric motor is the induction motor, which utilizes a rotating magnetic field to produce torque.
Other types of electric motors include brushless DC motors, direct current motors, and synchronous motors. Electric motors are able to generate high torque even at low speeds, making them efficient for a variety of applications. Their versatility and easy maintenance make them popular for both consumer and industrial applications.
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sharon serves the volleyball to barbara with an upward velocity of 23ft/s . the ball is 3.5 feet above the ground when she strikes it. how long does barbara have to react, before the volleyball hits the ground? round your answer to two decimal places.
Varbara have to react , before the volleyball hits the ground in 0.71 s.
Calculate the periodSolve for the time that Barbara has to react before the volleyball hits the ground, we need to use the formula below:
t = (v_f - v_i)/g
Where:v_f = final velocity (when the ball hits the ground, its velocity is 0) v_i = initial velocity (given as 23 ft/s, upwards) g = acceleration due to gravity (constant at 32.2 ft/s²)
So we have:v_i = 23 ft/st = ?g = 32.2 ft/s²
Using these values:
0 = 23 - 32.2t
t = 23/32.2
t ≈ 0.71 seconds.
Rounding to two decimal places, Barbara has approximately 0.71 seconds to react before the volleyball hits the ground.
Therefore, the correct option is 0.71 seconds.
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Find the acceleration (in m/s^2) of a car that travels from rest, to a velocity of 60 m/s in a distance of 212.0 ft.
A. 32 m/s^2
B. 27.9 m/s
C. 27.9 ft/s^2
D. 27.9 m/s^2
According to the given statement The acceleration of the car is 32 m/s²
How do you define an acceleration?Every procedure where velocity varies is referred to as acceleration. There are only two ways to accelerate: changing your speed , changing your direction, or changing both. This is due to the fact that velocity is both a speed or a direction.
The distance in feet must first be converted to metres:
212.0 ft = 64.6216 m
The kinematic equation can then be applied:
v² = u² + 2as
where,
The final velocity is v. (60 m/s),
u is the starting speed (0 m/s),
a denotes the acceleration,
s denotes the distance. (64.6216 m).
Solving for a, we get:
a = (v² - u²) / (2s)
a = (60² - 0²) / (2 * 64.6216)
a ≈ 32 m/s²
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Calculate the potential energy, kinetic energy, mechanical energy, velocity, and height of the skater at the various locations
Equations/Concepts Used:
Kinetic Energy => [tex]K=\frac{1}{2}mv^2[/tex]
Gravitational Potential Energy => [tex]U_{g}=mgy[/tex]
Mechanical Energy => [tex]E_{Mech.}=U+K[/tex]
Conservation of Energy => [tex]E_{0}=E_{f}[/tex]
At point 1
[tex]m=60 \ kg[/tex]
[tex]v=8 \ m/s[/tex]
PE ==> [tex]U_{g}=mgy \Rightarrow =(60)(9.8)(0) \Rightarrow U_{g}= 0 \ J[/tex]
KE ==> [tex]K=\frac{1}{2}mv^2 \Rightarrow =\frac{1}{2}(60)(8)^2 \Rightarrow K=1920 \ J[/tex]
ME==> [tex]E_{Mech.}=U+K \Rightarrow = 0+1920 \Rightarrow E_{Mech.}=1920 \ J[/tex]
At point 2
[tex]y=1 \ m[/tex]
Find the velocity of the skater at point 2 using conservation of energy.
We already found the total energy at point 1, which was 1920 Joules.
==> [tex]E_{1}=E_{2} \Rightarrow 1920=U_{g_{2}}+K_2 \Rightarrow 1920=(60)(9.8)(1)+\frac{1}{2}(60)v^2[/tex]
[tex]\Rightarrow 1920=588+30v^2 \Rightarrow 1332=30v^2 \Rightarrow v^2=44.4 \Rightarrow v=6.66 \ m/s[/tex]
From the equation above we answered the following,
[tex]v=6.66 \ m/s[/tex]
[tex]U_g=588 \ J[/tex]
We know the velocity at point 2, find KE then ME.
[tex]K=\frac{1}{2}(60)(6.66)^2 \Rightarrow K=1331 \ J[/tex]
[tex]E_{Mech.}=588+1331 \Rightarrow E_{Mech.}= 1919 \ J[/tex]
Notice how mechanical energy remains constant, this is because energy is a conserved quantity.
At point 3
Use conservation of energy again, using points 1 and 3.
==> [tex]E_1=E_3 \Rightarrow 1920=U_{g_3}+K_3 \Rightarrow 1920=(60)(9.8)h+0[/tex]
At point 3 the skaters velocity will go to 0 and all energy will be potential.
So, [tex]v=0 \ m/s[/tex]
[tex]\Rightarrow 1920=588h \Rightarrow h=3.27 \ m[/tex]
==> [tex]U_g=(60)(9.8)(3.27) \Rightarrow U_g=1923 \ J[/tex]
Answers:
Point 1, PE=0 J, KE=1920 J, ME=1920J
Point 2, PE=588 J, KE= 1331 J, ME= 1919 J, v=6.66 m/s
Point 3, PE=1923 J, KE=0 J ,ME= 1923 J, v=0 m/s, h=3.27 m
HELPPPP (LOOK AT PHOTO)
The similarities and the differences between the velocity and acceleration have been shown below.
Present the similarities and difference between velocity and acceleration in a Venn diagramSimilarities:
Both are vectors, which means they have both magnitude and direction.
Both are measures of motion and are expressed in units of distance and time.
Both have the same units of distance per time, such as meters per second (m/s).
Differences:
Velocity measures the rate at which an object changes position over time, while acceleration measures the rate at which an object changes its velocity over time.
Velocity has direction and magnitude, while acceleration only has magnitude.
Velocity can be positive, negative or zero, depending on the direction of the object's motion, while acceleration can be positive or negative, depending on whether the object is speeding up or slowing down.
The SI unit of velocity is meters per second (m/s), while the SI unit of acceleration is meters per second squared (m/s^2).
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a marble is rolling around in the back of a small toy wagon as the wagon is pulled along the sidewalk. when the wagon is stopped suddenly by a rock under one of the wheels, the marble rolls toward the front of the wagon. why does the marble keep going when the wagon stops?
The marble keeps going when the wagon stops because of the Inertia.
What is inertia?Inertia is defined as the property of an object to remain in a state of rest or uniform motion in a straight line unless acted upon by an external force. Newton's first law of motion, often known as the law of inertia, explains this.
Inertia is a fundamental concept in physics, and it applies to everything, from electrons to massive objects like planets and galaxies.
When a small toy wagon with a rolling marble is abruptly stopped by a rock under one of its wheels, the marble moves forward since it has a forward velocity due to its motion when the wagon was rolling, and because of inertia.
As a result, when the wagon comes to a halt, the marble is still moving forward, allowing it to keep going until it eventually reaches the front of the wagon.
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a tuning fork vibrating at 512 hz falls from rest and accelerates at 9.80 m/s2. how far below the point of release is the tuning fork when waves with a frequency of 480 hz reach the release point? take the speed of sound in air to be 340 m/s
Tuning fork is 0.223 m below the point of release when waves with frequency of 480 Hz reaches the release point.
What is a frequency?Number of waves that pass a fixed point in the unit time is known as frequency.
y = 1/2 * a * t²
y is distance traveled, a is acceleration, and t is time.
t = √(2y/a)
v = f * λ
v is speed of sound, f is the frequency, and λ is wavelength.
λ = v/f = 340 m/s / 480 Hz = 0.708 m
y = n * λ/2
where n is the number of half-wavelengths traveled.
y = λ/2 = 0.354 m
t = √(2y/a) --> y = 1/2 * a * t² = 1/2 * 9.80 m/s² * (2y/9.80 m/s²) = y
t = √(2y/a) --> t = √(2y/a) = √(2 * 0.354 m / 9.80 m/s²) = 0.212 s
Therefore, tuning fork falls for 0.212 seconds before the sound wave reaches the release point. During that time, the tuning fork travels a distance of:
y = 1/2 * a * t² = 1/2 * 9.80 m/s² * (0.212 s)² = 0.223 m
Therefore, tuning fork is 0.223 m below the point of release when waves with a frequency of 480 Hz reach the release point.
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An athlete on the training ground starts at point and runs 120 m South to point B, then runs 200 m East to point C and the runs 270 m North to point D. The points on the training ground are shown in the diagram below: 2.2. A B D C Use the tail to head method, draw a neat, fully labelled displacement vector diagram and include the resultant displacement of the athlete. (4)
The magnitude of the resultant displacement is 334.4 m and the direction of the resultant displacement is 53.1° North of East.
How to solveTo draw the displacement vector diagram, we start at point A and draw a vector from A to B, representing the athlete's displacement of 120 m South.
We then draw a vector from the end of the first vector (B) to the end of the second vector (C), representing the athlete's displacement of 200 m East. Finally, we draw a vector from the end of the second vector (C) to point D, representing the athlete's displacement of 270 m North. The diagram should form a closed triangle.
To find the resultant displacement of the athlete, we use the Pythagorean theorem and trigonometry. Let's call the displacement from A to B "vector AB," the displacement from B to C "vector BC," and the displacement from C to D "vector CD." The magnitude of the resultant displacement (R) is given by:
R = √(AB² + BC² + CD²)
R = √(120² + 200² + 270²) = 334.4 m (rounded to one decimal place)
To find the direction of the resultant displacement, we use trigonometry. We can find the angle between the resultant displacement and the North direction using the following formula:
θ = tan⁻¹(CD/BC)
Where CD is the Northward component of the displacement vectors and BC is the Eastward component of the displacement vectors.
θ = tan⁻¹(270/200) = 53.1° (rounded to one decimal place)
Therefore, the magnitude of the resultant displacement is 334.4 m and the direction of the resultant displacement is 53.1° North of East.
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a person pushes a box. he applies a force of 400 newtons, and he pushes the box for a distance of 4 meters. after he stops pushing, the box slides forward another meter. how much work was done by the person?(1 point) responses 2,000 joules 2,000 joules 1,600 joules 1,600 joules 100 joules 100 joules 400 joules
If the box slides forward another meter, the amount of work the person does is 1600 joules.
The work done by the person can be calculated as the product of the applied force and the distance moved by the box in the direction of the force:
W = Fd
where W is the work done, F is the applied force, and d is the distance moved in the direction of the force.
In this case, the person applies a force of 400 newtons and pushes the box for a distance of 4 meters. After he stops pushing, the box slides forward another meter, but since the person is not applying any force at that point, no work is done by him.
Therefore, the work done by the person is:
W = Fd = (400 N)(4 m) = 1600 J
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an extremely buff crab throws a 1kg rock at 2m/s. if a -1.1 force was applied to the rock to bring it to a stop, how many seconds did it take to stop?
It took 1.82 seconds for the rock to come to a stop when the force of -1.1 N was applied.
What is force?
We can use the formula:
force = mass x acceleration
To find the acceleration of the rock when the force is applied. Since we know the mass of the rock is 1kg, we can rearrange the formula to solve for acceleration:
acceleration = force / mass
We know that the force applied to the rock is -1.1 Newtons (N), since it is bringing the rock to a stop. Thus:
acceleration = -1.1 N / 1 kg = -1.1 m/s²
The negative sign indicates that the acceleration is in the opposite direction of the motion of the rock.
Next, we can use the formula:
velocity = initial velocity + acceleration x time
To find the time it takes for the rock to come to a stop. We know the initial velocity of the rock is 2 m/s and the final velocity is 0 m/s (since it comes to a stop). Thus:
0 m/s = 2 m/s + (-1.1 m/s²) x time
Solving for time:
time = (0 m/s - 2 m/s) / (-1.1 m/s²) = 1.82 seconds
Therefore, it took 1.82 seconds for the rock to come to a stop when the force of -1.1 N was applied.
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The hight of a parallelogram is 375cm2 the base is 25cm find the hight
The height of a parallelogram with an area of 375cm² and a base of 25cm is 15cm.
We can use the formula for the area of a parallelogram, which is A = bh, where A is the area, b is the base, and h is the height. Substitute the given values of the area and the base in the formula, we get: A = bh375 = 25h Divide both sides by 25 to get the value of h, and we get: h = 15 Therefore, the height of the parallelogram is 15cm.
To find the height of the parallelogram, you can use the formula for the area: Area = base × height. You are given the area (375 cm²) and the base (25 cm).
375 cm² = 25 cm × height
To find the height, divide both sides by 25 cm:
height = 375 cm² / 25 cm
height = 15 cm
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An object is dropped from the top of a building and is observed to take 7.2s to hit the ground. How tall is the building?
An object is dropped from the top of a building and takes 7.2s to hit the ground, the height of the building is approximately 255.84m.
.When an object is dropped from the top of a building, it begins to accelerate at 9.8 m/s², the rate of acceleration due to gravity on earth. The final velocity is the rate at which the object hits the ground and is calculated with the formula; V = gt , Where V = Final velocity , t = Time taken to fall , g = Acceleration due to gravity on earth
Substituting values; V = 9.8 m/s² x 7.2 s = 70.56 m/s
The height of the building can be calculated using the formula; h = 1/2gt²
Substituting values; h = 1/2 x 9.8 m/s² x (7.2 s)² = 255.84 m
Therefore, the height of the building is approximately 255.84m.
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two thin lenses with a focal length of magnitude 6.00 cm , the first diverging and the second converging, are located 4.50 cm apart. an object 2.80 mm tall is placed 10.0 cm to the left of the first (diverging) lens. for related problem-solving tips and strategies, you may want to view a video tutor solution of an image of an image. part a how far from this first lens is the final image formed? express your answer in centimeters.
5.45 cm to the right of the second lens is where the final image is created.
When an object is far from the lens, where is the image?The image will get smaller and smaller as we move the object further and further away. The focal point will draw the image's location ever-closer. The light would be concentrated at the focal point if the object were extremely far away, such as the sun.
Using the thin lens equation, we have: 1/f = 1/di + 1/do
For the first lens, we have:
f1 = -6.00 cm (negative because the lens is diverging)
do1 = -10.0 cm (negative because the object is to the left of the lens)
Solving for di1, we get: 1/di1 = 1/f1 - 1/do1
di1 = -15.0 cm (negative because the image is to the left of the lens)
The first lens creates a virtual, upright image whose magnification is determined by: m1 = -di1/do1 = 1.50
As there are 4.50 cm between the first and second lenses, the location of the thing that the second lens sees is:
do2 = di1 - 4.50 cm = -19.5 cm
For the second lens, we have:
f2 = 6.00 cm (positive because the lens is converging)
do2 = -19.5 cm (negative because the object is to the left of the lens)
Solving for di2, we get:
1/di2 = 1/f2 - 1/do2
di2 = 5.45 cm
The final image is real and inverted, and its magnification is given by:
m = -di2/do2 = 0.279
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Explain the differences in the air seen at 30° N versus 60° N.
determine the total (equivalent) resistance of the circuit. b) what is the total current in the circuit? c) what is the voltage across r1
a)The total (equivalent) resistance of the circuit is 4.8 Ω
b) what is the total current in the circuit is 2.08 A
c) what is the voltage across r1 is 12.48 V
a) To determine the total (equivalent) resistance of the circuit, use the equation
R = 1/(1/R1 + 1/R2)
where R1 and R2 are the individual resistors.
The total resistance of the circuit is therefore R = 1/(1/R1 + 1/R2) = 1/(1/6 + 1/8) = 4.8 Ω
b) The total current in the circuit is determined by Ohm's Law:
I = V/R
where V is the voltage and R is the resistance.
In this circuit, V = 10 V and R = 4.8 Ω
so the total current is I = V/R = 10 V/4.8 Ω = 2.08 A.
c) The voltage across R1 is determined by Ohm's Law:
V = I * R
where I is the current and R is the resistance.
In this circuit, I = 2.08 A and R1 = 6 Ω
so the voltage across R1 is V = I * R = 2.08 A * 6 Ω = 12.48 V.
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2. a proton moves in a circular orbit of radius 65.00 cm perpendicular to a uniform magnetic field of magnitude 0.700 t. (a) what is the period for this motion? (b) find the speed of the proton. (c) find the kinetic energy of the proton.
(a) The period for this motion is 1.48 × 10^-7 s
(b) The speed of the proton is 2.79 × 10^6 m/s
(c) The kinetic energy of the proton is 6.86 × 10^-14 J.
a) To determine the period of motion, we can use the formula for the magnetic force on a charged particle in a circular motion F = qvB, where F is the magnetic field, q is the charge of the particle which is the proton in this case, v is the speed of the particle and B is the magnitude of the magnetic field.
The magnetic force is always perpendicular to both the direction of motion of the particle and the direction of the magnetic field, since the proton is moving in a circle, there must be a net force acting toward the center of the circle to keep it in its orbit.The force is the centripetal force, which is provided by the magnetic force in this case:
F = ma = mv^2/r, where is and m is the radius of the circle and mass of the proton.Comparing those equations we get qvB = mv^2/r.Now solving for the period T = 2πr/v, we get: T = 2πm/(qB), substituting the values we get:T = 2π(1.67 × 10^-27 kg)/(1.6 × 10^-19 C)(0.700 T) = 1.48 × 10^-7 s
b) we use the formula for the circumference of a circle :C = 2πr, as the proton completed one full revolution in the period T, its v is equal to the circumference divided by period v = C/T = 2πr/T, after substituting the values we get : v = (2π)(0.650 m)/(1.48 × 10^-7 s) = 2.79 × 10^6 m/s
(c) for calculating the kinetic energy, we use the formula:KE = 1/2 mv^2, plugging the know values we get:
KE = (1/2)(1.67 × 10^-27 kg)(2.79 × 10^6 m/s)^2 = 6.86 × 10^-14 J
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a 100 lb cabinet is placed on a smooth surface. if a force of a 100 lb is applied for 2 s, determine the net impulse on the cabinet during this time interval.
The net impulse on the cabinet during the 2-second time interval when a 100 lb force is applied is 200 lb·s.
The net impulse on an object is equal to the change in its momentum, which can be calculated using the equation,
Impulse = Force x Time
In this case, a force of 100 lb is applied to the cabinet for 2 s. Since the surface is smooth, there is no frictional force acting on the cabinet, and it will move with a constant velocity after the force is applied.
The initial momentum of the cabinet is zero, since it is initially at rest. The final momentum of the cabinet can be calculated using the equation:
Final Momentum = Mass x Velocity
Since the mass of the cabinet is 100 lb, and it moves with a constant velocity after the force is applied, its final momentum is:
Final Momentum = 100 lb x v
where v is the velocity of the cabinet.
Since the force is applied for 2 s, the impulse on the cabinet is:
Impulse = Force x Time = 100 lb x 2 s = 200 lb·s
Since there are no external forces acting on the cabinet, the net impulse on the cabinet during this time interval is equal to the impulse calculated above:
Net Impulse = Impulse = 200 lb·s
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how long does it take for a rotating object to speed up from 15.0 rad/s to 33.3 rad/s if it has a uniform angular acceleration of 3.45 rad/s2?
a. 4.35 s
b. 5.30 s
c. 10.6 s
d. 63.1 s
e. 9.57
The time taken by the rotating object to speed up from 15.0 rad/s to 33.3 rad/s if it has a uniform angular acceleration of 3.45 rad/s² is 5.30 s. Hence, option (b) is the correct answer.
We are given the initial angular velocity, ω1 = 15.0 rad/s, final angular velocity, ω2 = 33.3 rad/s, and angular acceleration, α = 3.45 rad/s². We are supposed to find the time, t, taken by the rotating object to speed up from ω1 to ω2.
We can use the following kinematic equation to solve this problem:
ω2 = ω1 + αt
Rearranging this equation, we get:
t = (ω2 - ω1) / α
Substituting the given values in the above equation, we get:
t = (33.3 - 15.0) / 3.45
t = 5.30 s
So, the correct option is B.
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A ball is thrown from the roof of a multi-story building, where is the acceleration of free fall greater, the first floor or the top floor?
Due to their free fall, the two balls are falling at the same acceleration. Gravity has an effect on the two balls that are thrown downward from the top of the structure. Because of gravity, both balls fall freely.
A falling item is subject to two forces: gravity and drag from air resistance. Everybody is subject to gravity, which exerts a force inversely proportionate to mass. With a force proportionate to the cross sectional area of the object, drag operates on objects of a comparable shape in the same way.This means that while the mass of the bowling ball will be more according to the cube of the size ratio, the drag force on the bowling ball will be greater according to the square of the size ratio.Although the mass of each is proportionate to the cube of the size, the density of the marble (glass) may be higher or lower than the density of the bowling ball. based on the disparity in density, off in either way.The relationship between drag force and mass will determine the actual acceleration brought on by drag. As a result, I believe the bowling ball will have less drag than the marble and will land first.To know more about acceleration
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Explain how the structure of DNA enables the molecule to be easily transcribed. Why is this important for genetic infomation
The nucleotides included in DNA allow for the joining of the two strands to form a double helix shape. The nucleotides are cytosine, adenine, guanine, and thiamine. The Strand of DNA unzips during transcription to enable the production of mRNA.
Why is knowledge of DNA's architecture as well as replication process crucial?
The plan for existence on this planet is included in it. enables humans and organisms to defend themselves against sickness and helps us grasp faults.
Biological physics makes use of physics' methods and instruments to comprehend the internal world of the mechanism present in living things on length different scales first from molecule to the macroscopic.
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How many seconds would a 300.0 W heater to consume 900,000.0 J?
Answer:
50 minutes / 3000 seconds
Explanation:
Energy = power x time
time = energy / power
time = 900,000 / 300
time = 3000s
This can also be written as 50 minutes