To keep the seesaw balanced, the 3.7 kg cat must stand 2.06 meters to the left of the pivot.
Explanation:To keep the seesaw balanced, the torques (rotational forces) on both sides of the pivot must be equal. The torque is given by the product of the mass and the distance from the pivot. In this case, the 5.3 kg cat and the 2.5 kg bowl of tuna fish create a torque on one side, while the 3.7 kg cat creates a torque on the other side. Let's denote the distance from the pivot to the 5.3 kg cat as x:
Torque on one side = 5.3 kg * x
Torque on the other side = 2.5 kg * 1.5 m + 3.7 kg * (4 m - x)
Setting these two torques equal to each other and solving for x:
5.3 kg * x = 2.5 kg * 1.5 m + 3.7 kg * (4 m - x)
Simplifying the equation:
5.3x = 3.75 + 14.8 - 3.7x
9x = 18.55
x = 2.06 meters
Therefore, the 3.7 kg cat must stand 2.06 meters to the left of the pivot to keep the seesaw balanced.
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According to the Doppler affect what happens when a light surfaceMoves further away from an observer
Find the ratio of the new/old periods of a pendulum if the pendulum were transported from earth to the moon, where the acceleration due to gravity is 1.63 m/s2 .
The period of a pendulum on the Moon is roughly 2.45 times longer than the period of a similar pendulum on Earth. This is due to the difference in gravitational forces on the two celestial bodies.
Explanation:A pendulum's period is influenced by the acceleration due to gravity. On Earth, this is approximately 9.80 m/s². On the Moon, the gravitational pull is significantly weaker, approximating at about 1.63 m/s². Therefore, if a pendulum were taken from Earth to the Moon, its period would change due to the variance in gravitational forces.
The ratio of the new period (Tmoon) to the old period (Tearth) is determined by the square root of the ratio of the gravities: Tmoon/Tearth = sqrt(g earth / g moon)
Substituting values, we get Tmoon/Tearth = sqrt(9.80 / 1.63) = 2.45 (rounded).
So, the period of a pendulum on the Moon is about 2.45 times longer than that of the same pendulum on Earth.
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A magnetic field of 0.55 g is directed straight down, perpendicular to the plane of a circular coil of wire that is made up of 550 turns and has a radius of 20 cm. 1) if the coil is stretched, in a time of 35 ms, to a radius of 50 cm, calculate the emf induced in the coil during the process.
A certain part of the electromagnetic spectrum ranges from 200 nm to 400 nm. what is the lowest frequency associated with this portion of the spectrum? (c 3.00 108 m/s)
A calorie is the measurement of ____ required to raise a 1 gram of water 1 degree celsius.
Physics 1315, what is faster speed; red light or gamma rays?
What is the freezing point of radiator fluid that is 50% antifreeze by mass? k f for water is 1.86 ∘ c/m?
Answer:
-30.02 ºC
Explanation:
Assuming the antifreeze to be ethylene glycol (C₂H₆O₂) which is popular antifreeze.
Molar mass of ethylene glycol (C₂H₆O₂) = 62g/mol
Step 1: calculate the freezing point depression of the solution
ΔT = -Kf*M
where,
ΔT= depression in the freezing point.
M = the molarity of the solution (mol solute / Kg solvent)
Kf = molar freezing point constant of water = 1.86°C/m
To determine depression in the freezing point (ΔT), first we need to calculate;
molarity of solute (ethylene glycol) in mol mass of solvent (water) in kgmolarity of the solution (water +ethylene glycol)Step 2: calculate the molarity of the solute (ethylene glycol)
Molar mass ethylene glycol = 62 g/mol
molarity of ethylene glycol in mol = 50 g / 62g/mol = 0.807 mol
Step 3: calculate mass of solvent in kg
There is 1kg of ethylene glycol which is present in 1kg of water
mass of solvent (water) in kg= 50 g/ 1000 g/ Kg = 0.050 Kg
Step 4: calculate the molarity of the solution (M)
M = 0.807 mol / 0.050 Kg = 16.14 m
Step 5: calculate the freezing point depression of the solution (ΔT)
ΔT = - Kf*M = -1.86 ºC/m x 16.14 m
= -30.02 ºC
with a bar magnet where are the lines of force closest together
Answer: at the tips of the magnet
Explanation: charges are more at the tip of a substance so likewise for a magnet
Allison wants to calculate the speed of a sound wave. Which formula should she use?
distance traveled / amplitude
number of wavelengths / amplitude
number of wavelengths / time
distance traveled / time
Answer:
Option 4th is correct
distance traveled / time
Explanation:
Allison wants to calculate the speed of a sound wave.
The speed of the sound wave is the distance traveled divided by the time.
[tex]v = \dfrac{d}{t}[/tex]
Where, v = speed of sound
d = distance
t= time
Therefore, The formula should she use is, distance traveled / time
The potential energy due to the strong interaction, between two quarks or a quarkantiquark pair, goes something like v (r) = −α/r + βr, where α and β are positive constants. is there any point rv where the potential energy goes to zero? is there any point
An electron in the n = 5 level of an h atom emits a photon of wavelength 1282.17 nm. to what energy level does the electron move?
The electron moves to energy level n = 3
[tex]\texttt{ }[/tex]
Further explanationThe term of package of electromagnetic wave radiation energy was first introduced by Max Planck. He termed it with photons with the magnitude is :
[tex]\large {\boxed {E = h \times f}}[/tex]
E = Energi of A Photon ( Joule )
h = Planck's Constant ( 6.63 × 10⁻³⁴ Js )
f = Frequency of Eletromagnetic Wave ( Hz )
[tex]\texttt{ }[/tex]
The photoelectric effect is an effect in which electrons are released from the metal surface when illuminated by electromagnetic waves with large enough of radiation energy.
[tex]\large {\boxed {E = \frac{1}{2}mv^2 + \Phi}}[/tex]
[tex]\large {\boxed {E = qV + \Phi}}[/tex]
E = Energi of A Photon ( Joule )
m = Mass of an Electron ( kg )
v = Electron Release Speed ( m/s )
Ф = Work Function of Metal ( Joule )
q = Charge of an Electron ( Coulomb )
V = Stopping Potential ( Volt )
Let us now tackle the problem !
[tex]\texttt{ }[/tex]
Given:
initial shell = n₁ = 5
wavelength = λ = 1282.17 nm = 1.28217 × 10⁻⁶ m
Unknown:
final shell = n₂ = ?
Solution:
We will use this following formula to solve this problem:
[tex]\Delta E = R (\frac{1}{(n_2)^2} - \frac{1}{(n_1)^2})[/tex]
[tex]h \frac{c}{\lambda} = R (\frac{1}{(n_2)^2} - \frac{1}{(n_1)^2})[/tex]
[tex]6.63 \times 10^{-34} \times \frac{3 \times 10^8}{1.28217 \times 10^{-6}} = 2.18 \times 10^{-18} \times ( \frac{1}{(n_2)^2} - \frac{1}{5^2})[/tex]
[tex]1.55128 \times 10^{-19} = 2.18 \times 10^{-18} \times ( \frac{1}{(n_2)^2} - \frac{1}{5^2})[/tex]
[tex]( \frac{1}{(n_2)^2} - \frac{1}{5^2}) = \frac{16}{225}[/tex]
[tex]\frac{1}{(n_2)^2} = \frac{1}{25} + \frac{16}{225}[/tex]
[tex]\frac{1}{(n_2)^2} = \frac{1}{9}[/tex]
[tex](n_2)^2 = 9[/tex]
[tex]n_2 = \sqrt{9}[/tex]
[tex]\boxed{n_2 = 3}[/tex]
[tex]\texttt{ }[/tex]
Learn morePhotoelectric Effect : https://brainly.com/question/1408276Statements about the Photoelectric Effect : https://brainly.com/question/9260704Rutherford model and Photoelecric Effect : https://brainly.com/question/1458544[tex]\texttt{ }[/tex]
Answer detailsGrade: College
Subject: Physics
Chapter: Quantum Physics
We can use the Rydberg formula to find out to which energy level an electron will move after emitting a certain wavelength of photon. Inserting the given values of photon wavelength and initial energy level will provide us the final energy level to which the electron moves. In this case, it moves to energy level n = 2.
Explanation:The subject of this question involves examining the transition of an electron in a hydrogen atom after emitting a photon. Given we know the wavelength of the emitted photon, we can determine the initial and final energy levels, then calculate the transition the electron undergoes. We can use the Rydberg formula, R = 1.097373 x 107 m-1, which relates the wavelength of light emitted to the energy levels in an atom.
We find:
1/λ = R x (1/nf2 - 1/ni2)
Where:
λ = wavelength of lightR = Rydberg constantnf = final energy levelni = initial energy level
By inserting the given data, we see:
1/1282.17 x 10-9 m = R x (1/nf2 - 1/52)
Solving this equation gives nf = 2. Therefore, the electron moves to the energy level n = 2 after emitting the photon.
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What keeps an inflated balloon from falling down if you rub it against your hair and place it against a wall? 1. rubbing distorts the atoms inside the ballon and polarizes it. 2. rubbing leaves a balloon electrically charged; the charged balloon polarizes the wall. 3. when you rub the balloon against your hair, it will remove some m?
The answer is: rubbing leaves a balloon electrically charged; the charge polarizes the wall
there is no friction here
A balloon sticks to a wall after being rubbed on your hair because the friction causes electrons to move from your hair to the balloon, giving it a negative charge that attracts to the slightly positively charged surface.
When you rub a balloon against your hair, the balloon becomes electrically charged due to the transfer of electrons. This process is known as the triboelectric effect, where friction causes electrons to be transferred from your hair to the balloon. As a result, the balloon holds a negative charge while your hair holds a positive charge.
Due to this, the charged balloon can polarize the wall's molecules, causing the opposite charges to attract, which keeps the balloon stuck to the wall temporarily.
Acceleration is best defined as the rate of change of ________ of an object.
Answer:
Velocity
Explanation:
Remember that acceleration is given by the next formula:
Acceleration: Vf-Vo/Time
Where Vf final velocity and Vo is initial velocity, so acceleration of an object is best defined as the change in velocity the an object experiences in a certain period of time, it is often described in M/s^2 and those are the units in the international system.
What is the wavelength λ of light in glass, if its wavelength in air is λ0, its speed in air is c, and its speed in the glass is v? express your answer in terms of λ0, c, and v?
The wavelength λ of light in glass, if its wavelength in air is λ0, its speed in air is c, and its speed in the glass is v will be
[tex]\lambda=\dfrac{v}{c}\lambda_o[/tex]
What is wavelength?
The wavelength of any wave is defined as the distance between two max adjacent amplitudes, or the distance between two successive troughs or crest.
When an electromagnetic wave moves from a medium to another medium, its frequency remains constant. In this problem, we have light (which is an electromagnetic wave) moving from air to glass (or vice-versa), so we can write
[tex]f_o=f(1)[/tex] (1)
where is the frequency of the light in air, while f is the frequency in glass.
Using the relationship between frequency, wavelenght and speed of a wave, we have
[tex]f=\dfrac{v}{\lambda}[/tex]
where v is the speed of light in glass and the wavelength in glass, and
[tex]f_o=\dfrac{c}{\lambda_o}[/tex]
where c is the speed of light in air and the wavelength in air. Using these two relationships, we can rewrite eq.(1) as
[tex]\dfrac{c}{\lambda_o}=\dfrac{v}{\lambda}[/tex]
and by re-arranging it, we find
[tex]\lambda=\dfrac{v}{c} \lambda_o[/tex]
Hence the wavelength λ of light in glass, if its wavelength in air is λ0, its speed in air is c, and its speed in the glass is v will be
[tex]\lambda=\dfrac{v}{c}\lambda_o[/tex]
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What method of forecasting uses the effects of past weather conditions as part of its forecasting method
the method is analog forecasting method
analog forecasting method uses the past weather data and compares it with the existing one to forecast. it is a complex method since it involves remembering previous data for whether that can be triggered by a specific event.
A 0.50-kg block attached to an ideal spring with a spring constant of 80 n/m oscillates on a horizontal frictionless surface. the total mechanical energy is 0.12 j. the greatest extension of the spring from its equilibrium length is:
The greatest extension of the spring from its equilibrium length is calculated through the formula for total energy in harmonic motion, given as 0.5 * k * x^2. When rearranged for displacement(x), and values substituted, the extension is found to be 0.155 meters or 15.5 cm.
Explanation:In this problem, we are dealing with a block attached to an ideal spring and on a frictionless surface. The total mechanical energy, E, is given as 0.12 joules. We can use this information to find the peak displacement or the greatest extension, x, of the spring (amplitude) from its equilibrium length, using the formula E = 0.5*k*x^2.
The total energy of the system is the potential energy at the position of the greatest extension or amplitude, which means the block is momentarily at rest and has zero kinetic energy. Therefore, E = 0.5 * k * x^2, where E is the total energy, k is the spring constant, and x is the displacement.
Here, rearranging the formula to solve for the extension (x) we get x = sqrt((2*E)/k).
Substitute E = 0.12 joules and k = 80 N/m, into the equation. Therefore the greatest extension of the spring from its equilibrium length is x = sqrt((2*0.12 J)/80 N/m) = 0.155 meters or 15.5 cm.
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The greatest extension of the spring from its equilibrium length, which correlates to the amplitude A, is calculated from the given mechanical energy and spring constant. This can be determined by solving the equation for total mechanical energy, ETotal = (1/2)kA² for A. The calculation gives a maximum extension of approximately 15.5 cm.
Explanation:The total mechanical energy of a block and spring system is equal to the potential energy stored in the spring at the maximum extension/compression because at these points all of the energy is potential and none is kinetic. The total mechanical energy is given by ETotal = (1/2)kA², where k is the spring constant and A is the amplitude, which is the maximum extension or compression of the spring from its equilibrium length.
In this circumstance, the total mechanical energy is 0.12 Joules, and the spring constant 'k' is 80 N/m. So we can solve the above equation for A, the maximum extension:
A = sqrt( (2 * ETotal) / k ) = sqrt( (2 * 0.12 J) / 80 N/m ) = 0.15494 meters,
which is about 15.5 cm (since 1 m = 100 cm).
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A silver wire has a cross sectional area a = 2.0 mm2. a total of 9.4 × 1018 electrons pass through the wire in 3.0 s. the conduction electron density in silver is 5.8 × 1028 electrons/m3 and e = 1.60 × 10-19
c. what is the drift velocity of these electrons?
The drift velocity of electrons in a silver wire given an electron density of 5.8 x 10^28 electrons/m3, a cross-sectional area of 2.0 mm2, and a total of 9.4 x 10^18 electrons passing through in 3.0 s, is approximately 1.08 x 10^-4 m/s.
Explanation:The drift velocity of electrons is a term used in physics to describe the average velocity that free charges such as electrons traverse in a material due to an externally applied electric field. Given the cross-sectional area (A), the number of electrons (Q), the time (t), the charge of an electron (e), and the electron density (n), you can determine the drift velocity (v) using the formula:
v = Q / (n * A * t)
Firstly, convert the cross-sectional area to m² since the electron density is given in 1/m³. So, the area would be 2.0 mm² = 2.0 x 10^-6 m².
To determine the number of electrons passing through during the given time period (3.0 s), we substitute what we know into the formula giving:
v = (9.4 x 10^18) / (5.8 x 10^28 x 2.0 x 10^-6 x 3.0)
This yields a drift velocity of 1.08 x 10^-4 m/s.
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Which of these is a cultural service—a non material benefit people get from the environment
Answer: providing national parks
Explanation:
Why are conductors and insulators both required to construct the electrical wiring in our home
When light propagates from a material with a given index of refraction into a material with a smaller index of refraction, the speed of the light vie?
When light moves from a material with a higher index of refraction to one with a lower value, the speed of light increases, potentially causing the light to refract. The index of refraction ratio determines the extent of this change in speed.
When light travels from a material with a higher index of refraction to one with a lower index of refraction, the speed of light in the material increases. The index of refraction, denoted as n, is defined by the ratio n = c/v, where c is the speed of light in a vacuum (approximately 3.0 * 10⁻⁸ m/s), and v is the observed speed of light in the material.
Since the index of refraction for a vacuum is exactly 1 (because the speed of light in a vacuum is c), and since n is always greater than or equal to 1, when light enters a material with a smaller n, it means that v must increase."
For example, when light moves from water (where it travels slower due to a higher n) into air (with a smaller n), the light speeds up. This change in speed can cause the light to bend or refract. The extent to which this bending occurs depends on the difference in the indices of refraction between the two materials.
Recall that the differential equation for the instantaneous charge q(t) on the capacitor in an lrc-series circuit is l d 2q dt 2 + r dq dt + 1 c q = e(t). see this excerpt about lrc-series circuits. use the laplace transform to find q(t) when l = 1 h, r = 20 ω, c = 0.005 f, e(t) = 160 v, t > 0, q(0) = 0, and i(0) = 0. q(t) =
The Laplace transform can be used to solve the differential equation for the charge on the capacitor in an LRC-series circuit.
Explanation:The Laplace transform can be used to solve the differential equation for the charge on the capacitor in an LRC-series circuit.
The Laplace transform transforms the differential equation into an algebraic equation, which can be solved to find q(t). First, we substitute the given values of L, R, C, and e(t) into the differential equation.
Then, we take the Laplace transform of both sides of the equation. Finally, we solve for Q(s) and take the inverse Laplace transform to find q(t).
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Calculate the absolute pressure at the bottom of a freshwater lake at a point whose depth is 27.8 m. assume the density of the water is 1.00 103 kg/m3 and the air above is at a pressure of 101.3 kpa.
Two astronauts are playing catch with a ball in space. The first astronaut throws the ball; and A) the ball moves, but the astronaut doesn’t. B) the ball moves, and so does the astronaut. C) the ball doesn’t move, but the astronaut does. D) the ball doesn’t move, and neither does the astronaut.
B) the ball moves, and so does the astronaut.
Two astronauts are playing catch with a ball in space. The first astronaut throws the ball, and the ball moves, and so does the astronaut because of the action-reaction law. When the ball goes forward, the astronaut goes backward.
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Two astronauts are playing catch with a ball in the space. One astronaut throws the ball, then the ball moves, and so does the astronaut. Hence, option B is correct.
What is Newton's third law?Newton's third law states that if two bodies meet, they exert forces on each other that are comparable in size and directed in the opposite direction.
Action and reaction law is another name for the third law. This law is critical in understanding issues involving static equilibrium, in which all the forces are in balance, but it also holds true for entities moving uniformly or quickly.
A game of catch is being played in space by two astronauts. A ball is thrown by the first astronaut, and thanks to the action-reaction rule, both the ball and the astronaut move. The astronaut is forced to move backwards as the ball advances.
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What form does the signal take while it is broadcast from a transmitting station to your radio or television, regardless of whether it is a digital or analog wave?
The answer is transverse waves.
I need help finding the answer:
When a guitar string plays the note "A", the string vibrates at 440 Hz. What is the period of the vibration?
The period of the guitar string's vibration when it plays the note 'A' is approximately 0.00227 seconds.
Explanation:In physics, the period of vibration refers to the time it takes for a single complete cycle of vibration to occur. It is usually represented by the symbol T and is measured in seconds. To find the period of the guitar string's vibration, you can use the formula T = 1/f, where f is the frequency of the vibration.
In this case, the frequency of the guitar string playing the note 'A' is given as 440 Hz. So, the period can be calculated as T = 1/440 = 0.00227 seconds.
Therefore, the period of the guitar string's vibration when it plays the note 'A' is approximately 0.00227 seconds.
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a 10.0 kg ball falling at 10.0 m/s hits a mattress and comes to a complete stop in 1 s. what force does the mattress exert to stop the ball
Which of these statements would best explain the problem encountered with nuclear waste disposal?
Answer: A
Explanation:
A parallel circuit has two 8.0-ohm resistors and a power source of 9.0 volts. If a 12.5-ohm resistor is added to the circuit in parallel, how will the current be affected and what value will it have?
Answer:
1/R1 + 1/R2 + ... = 1/Re
So...
1/17.2 + 1/22.4 = 1/Re
0.1021 = 1/Re
Re = 9.792 Ohms
Now use the Voltage equation V = IR
6 = I * 9.792
I = 6/9.792 = 0.613 Amps.Pato 0.61
Explanation:
A particle is attached to a spring and is pushed so that the spring is compressed more and more. As a result, the spring exerts a greater and greater force on the particle. Similarly, a charged particle experiences a greater and greater force when pushed closer and closer to another particle that is fixed in position and has a charge of the same polarity. In spite of the similarity, the charged particle will not exhibit simple harmonic motion on being released, as will the particle on the spring. Explain why not.
To what potential should you charge a 3.0 μf capacitor to store 1.0 j of energy?
To store 1.0 J of energy in a 3.0 μF capacitor, you should charge it to approximately 816.5 V.
To determine the potential (V) needed to charge a 3.0 μF capacitor to store 1.0 J of energy, we can use the formula for the energy stored in a capacitor:
Energy (E) = 0.5 × C × V²
Where:
E is the energy stored (1.0 J in this case)C is the capacitance (3.0 μF or 3.0 × 10⁻⁶F)V is the potential difference we need to findRearranging the formula to solve for V:
V² = (2E)/C
Substituting the given values:
V² = (2 × 1.0 J) / (3.0 × 10⁻⁶ F)
V² = 2.0 J / 3.0 × 10⁻⁶ F
V² = 666666.67 V²
Taking the square root of both sides:
V ≈ 816.5 V
Therefore, you should charge the 3.0 μF capacitor to approximately 816.5 V to store 1.0 J of energy.