The total heat supplied to the gas in the process is calculated by considering the two expansion processes separately.
Explanation:The total heat supplied to the gas in the process can be calculated by considering the two expansion processes separately.
In the first expansion, which is isobaric, the gas volume doubles. Since the pressure remains constant, the work done on the gas is given by W = PΔV, where P is the pressure and ΔV is the change in volume. The work done is equal to the heat supplied to the gas.
In the second expansion, which is adiabatic, the temperature returns to its initial value. In an adiabatic process, the work done on the gas is given by W = (γ - 1)ΔU, where γ is the ratio of specific heats and ΔU is the change in internal energy. Since the temperature returns to its initial value, the change in internal energy is zero, and therefore, no heat is supplied to the gas in this process.
Asteroid Ida was photographed by the Galileo spacecraft in 1993, and the photograph revealed that the asteroid has a small moon, which has been named Dactyl. Assume the mass of Ida is 4.4 x 1016 kg, the mass of Dactyl is 2.6 x 1012 kg, and the distance between the center of Dactyl and Ida is 95 km. G = 6.672x10-11 N-m2/kg2 Part Description Answer Save Status A. Assuming a circular orbit, what would be the orbital speed of Dactyl? (include units with
The orbital speed of Dactyl around Ida is approximately 161.6 m/s.
Explanation:The orbital speed of an object can be calculated using the formula v = √ (G * (M + m) / r), where v is the orbital speed, G is the gravitational constant (6.672x10-11 N-m2/kg2), M is the mass of the primary object (Ida), m is the mass of the secondary object (Dactyl), and r is the distance between the center of the two objects.
Plugging in the given values, we have:
v = √ ((6.672 x 10-11 N-m2/kg2) * ((4.4 x 1016 kg) + (2.6 x 1012 kg)) / (95,000 m))
Simplifying the equation, we get:
v = 161.6 m/s
Therefore, the orbital speed of Dactyl around Ida is approximately 161.6 m/s.
Final answer:
To calculate the orbital speed of Dactyl in a circular orbit around Ida, you can use the formula for orbital speed.
Explanation:
Orbital speed of Dactyl:
The orbital speed of a satellite in a circular orbit can be calculated using the formula: v = √(G * M / r), where v is the orbital speed, G is the gravitational constant, M is the mass of the central body, and r is the distance from the center of the central body.
Substitute the given values: G = 6.67²x10-11 N-m²/kg², M = 4.4 x 10¹⁶ kg, and r = 95 km (converted to meters).
A glass lens that has an index of refraction equal to 1.57 is coated with a thin layer of transparent material that has an index of refraction equal to 2.10. If white light strikes the lens at near-normal incidence, light of wavelengths 495 nm in air and 660 nm in air are absent from the reflected light. What is the thinnest possible layer of material for which this can be accomplished?
Answer:
Find the given attachment
Answer:
471.4 nm
Explanation:
Please kindly see attachment for the step by step and very detailed solution to the given problem.
The attached file gave an explicit solution
A machinist turns the power on to a grinding wheel, which is at rest at time t = 0.00 s. The wheel accelerates uniformly for 10 s and reaches the operating angular velocity of 59 rad/s. The wheel is run at that angular velocity for 26 s and then power is shut off. The wheel decelerates uniformly at 1.6 rad/s2 until the wheel stops. In this situation, the time interval of angular deceleration (slowing down) is closest to:
Answer:
Time interval;Δt ≈ 37 seconds
Explanation:
We are given;
Angular deceleration;α = -1.6 rad/s²
Initial angular velocity;ω_i = 59 rad/s
Final angular velocity;ω_f = 0 rad/s
Now, the formula to calculate the acceleration would be gotten from;
α = Change in angular velocity/time interval
Thus; α = Δω/Δt = (ω_f - ω_i)/Δt
So, α = (ω_f - ω_i)/Δt
Making Δt the subject, we have;
Δt = (ω_f - ω_i)/α
Plugging in the relevant values to obtain;
Δt = (0 - 59)/(-1.6)
Δt = -59/-1.6
Δt = 36.875 seconds ≈ 37 seconds
Which statements represent the rules of significant figures? Check all that apply.
The integers 1, 2, 3, 4, 5, 6, 7, 8, and 9 are always significant.
All numbers to the right of the decimal point are always significant
Zeros to the left of a decimal point and in a number greater than or equal to 10 are always significant.
All zeros are significant numbers.
Zeros in between numbers are always significant.
Answer:
its 1, 3, 5 :) Have a great day/night!!
Explanation:
Significant figures rules are: non-zero digits are always significant; zeros in between non-zero digits are significant; zeros to the right of a decimal point are significant if they follow non-zero digits; zeros to the left of a decimal point are significant only if between non-zero digits or at the end of decimals; trailing zeros without a decimal point are not significant.
Explanation:The following statements represent the rules of significant figures:
The integers 1, 2, 3, 4, 5, 6, 7, 8, and 9 are always significant. Zeros to the right of a decimal point in a number are considered significant only if they follow a non-zero number. Zeros to the left of a decimal point and in a number greater than or equal to 10 are not always significant. They are significant only if they are between non-zero digits or at the end of a decimal number. All zeros are not necessarily significant numbers. The zero(s) located at the end of a number without a decimal point are not significant. Zeros in between numbers (non-zero digits) are always significant.Learn more about Significant Figures here:https://brainly.com/question/33741100
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After a switch is thrown to replace the battery in a DC LR circuit with a conducting wire (so that the circuit is still left complete), the time constant represents:
a. the time rate of change of the current in the circuit.b. the time rate of change of the induced emf in the circuit.c. the magnitude of the ratio of the current to the time rate of change of the current.d. all of the above.e. only (a) and (b) above
Answer:
c. the magnitude of the ratio of the current to the time rate of change of the current
Explanation:
In a LR circuit where battery is connected , the expression for decay of current is given by
[tex]i = i_0e^{-\frac{t}{\lambda}[/tex] where i is instantaneous current at time t , i₀ is maximum current , λ is constant .
differentiating on both sides with respect to t
di / dt = [tex]- \frac{i_0}{\lambda} e^{-\frac{t}{\lambda}[/tex]
[tex]\lambda = - \frac{i}{\frac{di}{dt} }[/tex]
So time constant is equal to magnitude of ratio of current to time rate of change of current .
A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative x direction and has a magnitude of 2.85 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 2840 m/s. At that instant, what is the magnitude of the net force acting on the proton if the electric field is (a) in the positive z direction and has a magnitude of 5.38 V/m, (b) in the negative z direction and has a magnitude of 5.38 V/m, and (c) in the positive x direction and has a magnitude of 5.38 V/m
Answer:
Explanation:
a ) Magnetic force on proton
= B q v , B is magnetic field , q is charge with velocity v
= 2.85 x 10⁻³ x 1.6 x 10⁻¹⁹ x 2840
= 12.95 x 10⁻¹⁹ N
Its direction will be towards positive z - direction according to Fleming's left hand rule.
force on proton due to electric field = charge x electric field.
= 1.6 x 10⁻¹⁹ x 5.38
= 8.608 x 10⁻¹⁹ N
this force will be along the field ie in positive z direction so both the forces are acting in the same direction so they will add up.
total force = (12.95 + 8.608)x 10⁻¹⁹
= 21.558 x 10⁻¹⁹ N .
b ) in this case , both the forces are acting in opposite direction . net force
= (12.95 - 8.608)x 10⁻¹⁹
= 4.342 x 10⁻¹⁹ N
c ) In this case both the forces are acting perpendicular to each other
resultant = √(12.95² + 8.608²) x 10⁻¹⁹ N
= 15.54 x 10⁻¹⁹ N .
A soccer ball of mass 0.4 kg is moving horizontally with a speed of 20 m/s when it is kicked by a player. The kicking force is so large that the ball flies up at an angle of 30 degrees above the ground. The player however claims (s)he aimed her/his foot at a 40 degree angle above the ground. First briefly but precisely explain how this is possible, based on what you heard in the lecture. Then calculate the average kicking force magnitude and the final speed of the ball, if you are given that the foot was in contact with the ball for one tenth of a second.
Answer:
Explanation:
The ball was moving with velocity of 20 m /s earlier in horizontal direction . Due to kicking, additional V velocity was added to it at 40° because he kicked it at this angle but the ball travelled in the direction of resultant which was making an angle of 30° with the horizontal .
From the relation of inclination of resultant
Tan θ = V sinα / (u + V cosα) where α is angle between u and V , θ is inclination of resultant
Tan30 = [tex]\frac{Vsin40}{20+ Vcos40}[/tex]
[tex]\frac{1}{\sqrt{3 } } =\frac{V\times .64}{20+ V\times.766}[/tex]
20 + .766 V = 1.11 V
20 = .344 V
V = 58 m /s
To know the force , we shall apply concept of impulse
F x t = mv , F is force for time t creating a change of momentum mv
F x .1 = .4 x 58
F = 232 N
What can help a scientist identify any object in a group of objects?
the object's size
the object's colors
the object's characteristics
the object's shape
Answer:The object characteristics
Explanation:the objects characteristics
Answer:
C- The objects characteristics
Explanation:
I just did it
A group of campers travels to a cabin which has no electrical power. In order to provide for a heater and lights, which device would be appropriate? a resistor, an insulator, a generator, or a voltmeter
Answer:A generator is the appropriate device to be used.
Explanation:
A generator is a device which transforms mechanical energy to electrical energy.
The different parts of a generator includes:
- Engine, alternator, fuel system, voltage regulator, lubricating system, cooling and exhaust system, Battery charger and control panel.
The generator works on the principle of electromagnetic induction which states that
that the above flow of electric charges could be induced by moving an electrical conductor, such as a wire that contains electric charges, in a magnetic field. This movement creates a voltage difference between the two ends of the wire or electrical conductor, which in turn causes the electric charges to flow, thus generating electric current.
Therefore, the use of a generator is the best device for the campers to provide electrical power source for their heaters and light in the cabin.
A generator would be the appropriate device to power a heater and lights in a cabin with no electrical power, as it can convert mechanical energy into electrical energy required to operate these devices.
To provide power for a heater and lights in a cabin with no electrical power, the appropriate device among the options provided is a generator. A generator converts mechanical energy into electrical energy, which can then be used to power electrical devices such as heaters and lights. It does not rely on external power sources, making it suitable for remote locations like a cabin.
Other devices mentioned, like a resistor, are components typically used within electronic circuits to control current flow or convert electrical energy into heat. An insulator is a material that prevents the transfer of electricity, and a voltmeter is a tool used to measure voltage across points in a circuit, rather than to provide power. None of these would be suitable to power a heater and lights in the absence of an existing power source.
Describe what ballistic stretching is and why it’s harmful. Then, provide at least two examples of how one should properly stretch? (Site 1)
Answer:
When doing ballistic stretching, it is using motion to bounce and stretch your body past its natural range of motion. In doing so you can harm yourself if you don't have a professional to help you as you might tear, damage, or pop your tendons, ligaments, or joints.
Explanation:
100% On Edge for 2021
The ballistic stretch is one of the dynamic stretching exercise, which can damage the tissues and ligaments if not performed properly and under the expert supervision.
The stretching activity that utilizes the momentum of body to achieve greater range of motion and flexibility, is known as Ballistic stretching. It is one of the intense stretching method that involves the bouncing movements to force the body beyond the normal range of motion.
This can be harmful if an athlete do not have a professional trainer to train for the cause. This may cause tear, damage of tendons, ligaments, or joints.
Following are the ways to perform a perform a proper stretch:
One should balance its body weight by standing on its feet together.The bending of knees should be done in a steady manner, and before this proper warm ups are needed to be done.It is not required to start with higher intensity, one can go with 5-10 repetitions for initial days, after that the repetitions can be increased gradually.Thus, we can conclude that the ballistic stretch is one of the dynamic stretching exercise, which can damage the tissues and ligaments if not performed properly and under the expert supervision.
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The Smithsonain used to have a huge pendulum that was 21.0 m long. What was its period?
Answer:
T = 9.19 seconds
Explanation:
It is given that,
Length of the pendulum, l = 21 m
We need to find its time period. We know that the time period of the pendulum is given by :
[tex]T=2\pi \sqrt{\dfrac{L}{g}}[/tex]
g is acceleration due to gravity
So,
[tex]T=2\pi \sqrt{\dfrac{21}{9.8}}\\\\T=9.19\ s[/tex]
So, the period of the pendulum is 9.19 seconds.
Final answer:
The period of the Smithsonian's 21.0 m long pendulum is approximately 9.2 seconds, which is calculated using the formula for a simple pendulum's period T = 2π√(L/g).
Explanation:
The period of a pendulum primarily depends on its length and the acceleration due to gravity. Using the formula for the period of a simple pendulum T = 2π√(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity (approximately 9.81 m/s² on Earth), we can calculate the period of the Smithsonian's pendulum.
Given that the length L is 21.0 m and by taking the standard value for g, 9.81 m/s², we plug these values into the formula to find the period:
T = 2π√(21.0 m / 9.81 m/s²)
Calculating this we get:
T ≈ 2π√(2.14 s²)
T ≈ 2π√(2.14 s²)
T ≈ 2π(1.46 s)
T ≈ 9.2 s
This means the Smithsonian's pendulum would have taken approximately 9.2 seconds to complete one full swing, back and forth. This is the time it would take to go from one side to the other and back again, corresponding to one complete oscillation.
A bicyclist and a runner are waiting at a red light. When the light turns green they start to speed up and the bicyclist gets to a final speed of 20 mph in 5 seconds. The runner gets to a final speed of 11 mph in 3 seconds. Which one had the greater acceleration?
the bicyclist or the runner.
Answer:
The acceleration of bicyclist is greater than that of the runner.
Explanation:
It is given that,
Initial speed of both bicyclist and a runner is 0 as they both are waiting at a red light,
When the light turns green they start to speed up.
Final speed of the bicyclist is 20 mph in 5 seconds
The runner gets to a final speed of 11 mph in 3 seconds.
20 mph = 8.94 m/s
11 mph = 4.91 m/s
Acceleration of bicyclist is :
[tex]a_b=\dfrac{v}{t}\\\\a_b=\dfrac{8.94\ m/s}{5\ s}\\\\a_b=1.78\ m/s^2[/tex]
Acceleration of runner is :
[tex]a_r=\dfrac{v}{t}\\\\a_r=\dfrac{4.91\ m/s}{3\ s}\\\\a_r=1.63\ m/s^2[/tex]
It is clear that the acceleration of bicyclist is greater than that of the runner.
The acceleration of the bicyclist is 1.788m/s² and the acceleration of the runner is 1.639m/s².
Hence, the bicyclist has the greater acceleration.
Given the data in the question;
Since the runner and the bicyclist where initially at rest;
Initial velocity of bicyclist; [tex]u_b = 0[/tex]Final velocity of bicyclist; [tex]v_b = 20mph = 8.9408m/s[/tex]Time taken by the bicyclist; [tex]t_b = 5s[/tex]Acceleration is simply the rate at which velocity changes with respect to time. Formula for acceleration can be derived from the First Equation of Motion;
[tex]v = u + at\\\\at = v - u\\\\a = \frac{v - u}{t}[/tex]
Where a is acceleration, v is final velocity, u is initial velocity and t is time elapsed.
Now, to determine who has the greater acceleration, we substitute our given values into the expression above.
For the bicyclist;
[tex]a_b = \frac{v -u}{t} \\\\a_b = \frac{ 8.9408m/s - 0}{5s}\\ \\a_b = \frac{8.9408m/s}{5s}\\ \\a_b = 1.788m/s^2[/tex]
For the runner;
[tex]a_r = \frac{v -u}{t} \\\\a_r = \frac{ 4.91744m/s - 0}{3s}\\ \\a_r = \frac{4.91744m/s}{3s}\\ \\a_r = 1.639m/s^2[/tex]
The acceleration of the bicyclist is 1.788m/s² and the acceleration of the runner is 1.639m/s².
Therefore, the bicyclist has the greater acceleration.
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A wire loop with 60 turns is formed into a square with sides of length s . The loop is in the presence of a 1.20 T uniform magnetic field B⃗ that points in the negative y direction. The plane of the loop is tilted off the x-axis by θ=15∘ . If i=2.50 A of current flows through the loop and the loop experiences a torque of magnitude 0.0186 N⋅m , what are the lengths of the sides s of the square loop, in centimeters?
Answer:
the length of the sides s is [tex]s = 1.998 \ cm[/tex]
Explanation:
From the question we are told that
The number of turns is [tex]N = 60 \ turn[/tex]
The magnetic field is [tex]B = 1.20 \ T[/tex]
The angle the loop makes with the x-axis [tex]\theta = 15 ^o[/tex]
The current flowing through the loop is [tex]I = 2.50 A[/tex]
The magnitude of the torque is [tex]\tau = 0.0186 \ N[/tex]
the length of the sides of the square is [tex]s[/tex]
Generally, we can represent the torque magnitude as
[tex]\tau = N I A B sin \theta[/tex]
Where A is the area of the square which is mathematically represented as
[tex]A = s^2[/tex]
Substituting this into the formula for torque
[tex]\tau = N I s^2 B sin \theta[/tex]
making s the subject
[tex]s = \sqrt{\frac{\tau }{NIB sin \theta } }[/tex]
Substituting values
[tex]s = \sqrt{\frac{0.0186}{(60) * (2.50) * (1.20) * (sin (15))} }[/tex]
[tex]s = 0.01998 m[/tex]
Converting to centimeters
[tex]s = 0.01998 * 100[/tex]
[tex]s = 1.998 \ cm[/tex]
Answer:
2 cm
Explanation:
To fins the lengths of the sides of the loop you use the following formula for the calculation of the torque experienced by the loop in a magnetic field:
[tex]\tau=NiABsin\theta[/tex]
N: turns of the loop = 60
i: current in the loop = 2.50A
A: area of the loop = s*s
B: magnitude of the magnetic field = 1.20T
θ: angle between the plane of the lop and the direction of B = 15°
BY replacing the values of the torque and the other parameters in (1) you can obtain the area of the loop:
[tex]A=\frac{\tau}{NiBsin\tetha}=\frac{0.0186Nm}{(60)(2.5A)(1.2T)sin15\°}=3.99*10^{-4}m^2[/tex]
but the area is s*s:
[tex]A=\sqrt{s}=\sqrt{3.99*10^{-4}m^2}=0.019m\approx2cm[/tex]
hence, the sides of the square loop have a length of 2cm
A small airplane is sitting at rest on the ground. Its center of gravity is 2.58 m behind the nose of the airplane, the front wheel (nose wheel) is 0.800 m behind the nose, and the main wheels are 3.02 m behind the nose. What percentage of the airplane's weight is supported by the nose wheel?
Answer:
The percentage of the weight supported by the front wheel is A= 19.82 %
Explanation:
From the question we are told that
The center of gravity of the plane to its nose is [tex]z = 2.58 m[/tex]
The distance of the front wheel of the plane to its nose is [tex]l = 0.800\ m[/tex]
The distance of the main wheel of the plane to its nose is [tex]e = 3.02 \ m[/tex]
At equilibrium the Torque about the nose of the airplane is mathematically represented as
[tex]mg (z- l) - G_B *(e - l) = 0[/tex]
Where m is the mass of the airplane
[tex]G_B[/tex] is the weight of the airplane supported by the main wheel
So
[tex]G_B =\frac{mg (z-l)}{(e - l)}[/tex]
Substituting values
[tex]G_B =\frac{mg (2.58 -0.8 )}{(3.02 - 0.80)}[/tex]
[tex]G_B = 0.8018 mg[/tex]
Now the weight supported at the frontal wheel is mathematically evaluated as
[tex]G_F = mg - G_B[/tex]
Substituting values
[tex]G_F = mg - 0.8018mg[/tex]
[tex]G_F = (1 - 0.8018) mg[/tex]
[tex]G_F = 0.1982 mg[/tex]
Now the weight of the airplane is = mg
Thus percentage of this weight supported by the front wheel is [tex]A = 0. 1982 *100 =[/tex] 19.82 %
You attach a 1.90 kg mass to a horizontal spring that is fixed at one end. You pull the mass until the spring is stretched by 0.400 m and release it from rest. Assume the mass slides on a horizontal surface with negligible friction. The mass reaches a speed of zero again 0.600 s after release (for the first time after release). What is the maximum speed of the mass (in m/s)
Answer:
2.09 m/s
Explanation:
As the spring is stretched initially , and the mass released from rest i.e v=0. Also, The next time the speed becomes zero again is when the spring is fully compressed, and the mass is on the opposite side of the spring with respect to its equilibrium position, after a time t=0.600 s. This illustrates half oscillation of the system.
Therefore, for the period of a full oscillation of the system
T= 2t => 2(0.6)=> 1.2 s
As the frequency is the reciprocal of the period, we have
f= 1/T => 1/1.2
f= 0.833 Hz
The angular frequency'ω' is given by,
ω= 2πf => 2π x 0.833=>5.23 rad/s
For the maximum velocity of the object in a spring-mass system:
V[tex](_{max} )[/tex]= Aω
where A is the amplitude of the oscillation. As here, the amplitude of the motion corresponds to the initial displacement of the object (A=0.400 m)
V[tex](_{max} )[/tex]= 0.4 x 5.23 =>2.09 m/s
A spring gun with a 75 N/m spring constant is loaded with a 5 g foam dart and isaimed vertically. When the spring is compressed by 10 cm and then released, the fireddart rises to a max height of 5 m above the end of the spring gun. Assuming the dartexperiences a constant friction force due to the air, how fast is it traveling when ithas fallen 2 m from its maximum height
Answer:
The speed is [tex]v = 4.425 m/s[/tex]
Explanation:
From the question we are told that
The spring constant is [tex]k = 75 \ N /m[/tex]
The mass of the foam dart is [tex]m = 5 g = \frac{5}{100} = 0.05 \ kg[/tex]
The compression distance is [tex]d = 10 cm = 0.1 m[/tex]
The height which the gun raised the dart is [tex]h = 5 m[/tex]
The change in height is [tex]\Delta h = 2 m[/tex]
The new height is [tex]h_2 = 5 -2 = 3 m[/tex]
Generally from the law of conservation of energy
[tex]E_s = KE[/tex]
Where [tex]E_s[/tex] is the energy stored in spring and it is mathematically represented as
[tex]E_s = \frac{1}{2} k d^2[/tex]
KE is the kinetic energy possessed by the dart when it is being shut and this is mathematically represented as
[tex]KE = \frac{1}{2} mv^2_r[/tex]
So
[tex]\frac{1}{2} k d^2 = \frac{1}{2} mv^2_r[/tex]
Substituting values
[tex]0.5 * 75 * 0.1 = 0.5 * 0.0005 * v^2_r[/tex]
=> [tex]v_r = \sqrt{\frac{0.5 * 75 * 0.1}{0.5 * 0.0005 } }[/tex]
[tex]v_r = 12.25 m/s[/tex]
When the dart is at the maximum height the
let it acceleration due air resistance be z
So by equation of motion
[tex]v^2 = u^2 - 2ah[/tex]
Where v is the velocity at maximum height which is equal to zero
and u is it initial velocity before reaching maximum height which we calculated as [tex]v_r = 12.25 m/s[/tex]
and a is the acceleration due to gravity + the acceleration due to air resistance
So
a = z+g
= 9.8 + z
=> [tex]v^2 = u^2 - 2(9.8 +z)h[/tex]
Substituting values
[tex]0 = 12.25^2 - 2(9.8 +z)h[/tex]
Making z the subject
[tex]z = \frac{ 12.25}{2 * 5} - 9.8[/tex]
[tex]z = \frac{ 12.25}{2 * 5} - 9.8[/tex]
[tex]z = 5 m/s[/tex]
When the dart is moving downward we can mathematically represent the motion as
[tex]v^2 = u^2 + 2ah[/tex]
Since the motion is downward and air resistance is upward we have that
a = g - z
and the the initial velocity u becomes the velocity at maximum height
i.e u = 0
And v is the velocity the dart has when it is moving downward
So
[tex]v^2 = 0 + 2 * (g -z )h[/tex]
Substituting values
[tex]v = \sqrt{0+ 2 (10 - 5) * 2}[/tex]
[tex]v = 4.425 m/s[/tex]
An electron is confined in a harmonic oscillator potential well. A photon is emitted when the electron undergoes a 3→1 quantum jump. What is the wavelength of the emission if the net force on the electron behaves as though it has a spring constant of 3.6 N/m? (m el = 9.11 × 10-31 kg, c = 3.00 × 108 m/s, 1 eV = 1.60 × 10-19 J, ħ = 1.055 × 10-34 J · s, h = 6.626 × 10-34 J · s)
Answer:
4.74*10^-7 m
Explanation:
TO find the wavelength of the photon you calculate the energy of the photon emitted in a harmonic oscillator:
[tex]E_{m-n}=\hbar \omega(m+\frac{1}{2})-\hbar \omega(n+\frac{1}{2})=\hbar \omega (m-n)\\\\\omega=\sqrt{\frac{k}{m}}\\\\E_{m-n}=\hbar \sqrt{\frac{k}{m}}(m-n)[/tex]
m: initial state = 3
n: final state = 1
k = 3.6N/m
By replacing the values of m for the electron, m,n and ħ you obtain:
[tex]E_{m-n}=(1.055*10^{-34}Js)\sqrt\frac{3.6N/m}{9.11*10^{-31}kg}}(3-1)=4.19*10^{-19}J[/tex]
Furthermore, this energy is equivalent to the expression:
[tex]E_{3-1}=\hbar \omega=\hbar 2\pi \nu=2\pi \hbar\frac{c}{\lambda}\\\\\lambda=\frac{2\pi \hbar c}{E_{3-1}}[/tex]
By replacing you obtain:
[tex]\lambda=\frac{2\pi (1.055*10^{-34}Js)(3*10^8m/s)}{4.19*10^{-19}J}=4.74*10^{-7}m[/tex]
hence, the wavelength of the photon is 4.74*10^-7 m
The wavelength of the emission if the net force on the electron behaves as though it has a spring constant of 3.6 N/m - [tex]4.74\times10^-7 m[/tex]
Formula:The energy of the photon emitted in a harmonic oscillator:
[tex]E_{m-n}=\hbar \omega(m+\frac{1}{2})-\hbar \omega(n+\frac{1}{2})=\hbar \omega (m-n)\\\\\omega=\sqrt{\frac{k}{m}}\\\\E_{m-n}=\hbar \sqrt{\frac{k}{m}}(m-n)[/tex]
Given:m: initial state = 3
n: final state = 1
k = 3.6N/m
Solution:By replacing the values of m for the electron, m, n and ħ you obtain:
this energy is equivalent to the expression:[tex]E_{3-1}=\hbar \omega=\hbar 2\pi \nu=2\pi \hbar\frac{c}{\lambda}\\\\\lambda=\frac{2\pi \hbar c}{E_{3-1}}[/tex]
By replacing you obtain:Thus, the wavelength of the photon is [tex]4.74\times10^-7 m[/tex]
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A record turntable is rotating at 33 rev/min. A watermelon seed is on the turntable 2.3 cm from the axis of rotation. (a) Calculate the acceleration of the seed, assuming that it does not slip. (b) What is the minimum value of the coefficient of static friction between the seed and the turntable if the seed is not to slip? (c) Suppose that the turntable achieves its angular speed by starting from rest and undergoing a constant angular acceleration for 0.36 s. Calculate the minimum coefficient of static friction required for the seed not to slip during the acceleration period.
Answer:
a) [tex]a_{r} = 0.275\,\frac{m}{s^{2}}[/tex], b) [tex]\mu_{s} = 0.028[/tex], c) [tex]\mu_{s} = 0.036[/tex]
Explanation:
a) The linear acceleration of the watermelon seed is:
[tex]a_{r} = \omega^{2}\cdot r[/tex]
[tex]a_{r} = \left[\left(33\,\frac{rev}{min} \right)\cdot \left(2\pi\,\frac{rad}{rev} \right)\cdot \left(\frac{1}{60}\,\frac{min}{s} \right)\right]^{2}\cdot (0.023\,m)[/tex]
[tex]a_{r} = 0.275\,\frac{m}{s^{2}}[/tex]
b) The watermelon seed is experimenting a centrifugal acceleration. The coefficient of static friction between the seed and the turntable is calculated by the Newton's Laws:
[tex]\Sigma F = \mu_{s}\cdot m\cdot g = m\cdot a[/tex]
[tex]a = \mu_{s}\cdot g[/tex]
[tex]\mu_{s} = \frac{a}{g}[/tex]
[tex]\mu_{s} = \frac{0.275\,\frac{m}{s^{2}} }{9.807\,\frac{m}{s^{2}} }[/tex]
[tex]\mu_{s} = 0.028[/tex]
c) Angular acceleration experimented by the turntable is:
[tex]\alpha = \frac{\omega-\omega_{o}}{\Delta t}[/tex]
[tex]\alpha = \frac{3.456\,\frac{rad}{s}-0\,\frac{rad}{s} }{0.36\,s}[/tex]
[tex]\alpha = 9.6\,\frac{rad}{s^{2}}[/tex]
The tangential acceleration experimented by the watermelon seed is:
[tex]a_{t} = \left(9.6\,\frac{rad}{s^{2}} \right)\cdot (0.023\,m)[/tex]
[tex]a_{t} = 0.221\,\frac{m}{s^{2}}[/tex]
The linear acceleration experimented by the watermelon seed is:
[tex]a = \sqrt{a_{t}^{2}+a_{r}^{2}}[/tex]
[tex]a = \sqrt{\left(0.221\,\frac{m}{s^{2}} \right)^{2}+\left(0.275\,\frac{m}{s^{2}} \right)^{2}}[/tex]
[tex]a = 0.353\,\frac{m}{s^{2}}[/tex]
The minimum coefficient of static friction is:
[tex]\mu_{s} = \frac{0.353\,\frac{m}{s^{2}} }{9.807\,\frac{m}{s^{2}} }[/tex]
[tex]\mu_{s} = 0.036[/tex]
A point source of light is submerged 3.3 m below the surface of a lake and emits rays in all directions. On the surface of the lake, directly above the source, the area illuminated is a circle. What is the maximum radius that this circle could have? Take the refraction index of water to be 1.333.
Answer:
Maximum Radius = 2.89m
Explanation:
The maximum radius will be determined by the angle of incidence which is equal to the critical angle. Now, any angle larger than that will make the light to be totally internally reflected. Hence, we can figure out that angle from Snell’s law where the refracted angle is 90°, and then use the tangent function.
From Snell's law;
n_air*sin90° = n_water*sin(θ _c)
Where;
θ_c is the critical angle
Refractive index of water; n_water = 1.333
Refractive index of air;n_air = 1
Thus;
1*1 = 1.33sinθ_c
sinθ_c = 1/1.33
θ_c = sin^(-1)0.7519
θ_c = 48.76°
Like I said earlier, we'll use tangent to find the radius.
Thus;
tanθ_c = d/R
From the question, d = 3.3m
Thus;
3.3/tan48.76 = R
So, R = 2.89m
Kim holds a pinwheel in the air and says it can be used to model a source of energy. Identify the type of energy Kim's pinwheel models. Explain the type of energy source used.
Answer:
it is kenetic
Explanation: its in motion:D
A 225-kg object and a 525-kg object are separated by 3.80 m. (a) Find the magnitude of the net gravitational force exerted by these objects on a 61.0-kg object placed midway between them. 1 .0000003383 Correct: Your answer is correct. N (b) At what position (other than an infinitely remote one) can the 61.0-kg object be placed so as to experience a net force of zero from the other two objects? 2 1.824 Incorrect: Your answer is incorrect.
Answer:[tex]F_{net}=3.383\times 10^{-7}\ N[/tex]
Explanation:
Given
Mass of first object [tex]m_1=225\ kg[/tex]
Mass of second object [tex]m_2=525\ kg[/tex]
Distance between them [tex]d=3.8\ m[/tex]
[tex]m_3=61\ kg[/tex] object is placed between them
So force exerted by [tex]m_1[/tex] on [tex]m_3[/tex]
[tex]F_{13}=\frac{Gm_1m_3}{1.9^2}[/tex]
[tex]F_{13}=\frac{6.674\times 10^{-11}(225\times 61)}{1.9^2}[/tex]
[tex]F_{13}=2.5374141274×10^{−7}\ N[/tex]
Force exerted by [tex]m_2\ on\ m_3[/tex]
[tex]F_{23}=\frac{Gm_2m_3}{1.9^2}[/tex]
[tex]F_{23}=\frac{6.674\times 10^{-11}(525\times 61)}{1.9^2}[/tex]
[tex]F_{23}=5.920632964\times 10^{-7}\ N[/tex]
So net force on [tex]m_3[/tex] is
[tex]F_{net}=F_{23}-F_{13}[/tex]
[tex]F_{net}=5.920632964\times 10^{-7}-2.5374141274\times 10^{-7}[/tex]
[tex]F_{net}=3.383\times 10^{-7}\ N[/tex]
i.e. net force is towards [tex]m_2[/tex]
(b)For net force to be zero on [tex]m_3[/tex], suppose
So force exerted by [tex]m_1[/tex] and [tex]m_2[/tex] must be equal
[tex]F_{13}=F_{23}[/tex]
[tex]\Rightarrow \frac{Gm_1m_3}{x^2}=\frac{Gm_2m_3}{(3.8-x)^2}[/tex]
[tex]\Rightarrow \frac{m_1}{x^2}=\frac{m_2}{(3.8-x)^2}[/tex]
[tex]\Rightarrow (\frac{3.8-x}{x})^2=\frac{m_2}{m_1}[/tex]
[tex]\Rightarrow \frac{3.8-x}{x}=\sqrt{\frac{525}{225}}[/tex]
[tex]\Rightarrow 3.8-x=1.52752x[/tex]
[tex]\Rightarrow 3.8=2.52x[/tex]
[tex]\Rightarrow x=1.507\ m[/tex]
A block with mass M attached to a horizontal spring with force constant k is moving with simple harmonic motion having amplitude A1. At the instant when the block passes through its equilibrium position, a lump of putty with mass m is dropped vertically onto the block from a very small height and sticks to it. Part APart complete What should be the value of the putty mass m so that the amplitude after the collision is one-half the original amplitude? Express your answer in terms of the variables M, A1, and k. m = 3M Previous Answers Correct Part B For this value of m, what fraction of the original mechanical energy is converted into heat? Express your answer in terms of the variables M, A1, and k.
Answer:
Explanation:
Given
Mass of block is [tex]M[/tex]
spring constant [tex]=k[/tex]
Amplitude is [tex]A_1[/tex]
when putty is placed then amplitude decreases to [tex]\frac{A_1}{2}[/tex]
Initially [tex]\frac{1}{2}kA^2=\frac{1}{2}Mv^2\quad \ldots(i)[/tex]
Conserving momentum
[tex]Mv_o=(m+M)v[/tex]
where [tex]v_o[/tex]=initial velocity
[tex]v=\frac{M}{M+m}v_o[/tex]
Now
[tex]\frac{1}{2}k(\frac{A_1}{2})^2=\frac{1}{2}(M+m)v^2[/tex]
[tex]\frac{1}{2}k(\frac{A_1}{2})^2=\frac{1}{2}(M+m)(\frac{M}{M+m}v_o)^2\quad \ldots(ii)[/tex]
divide (i) and (ii) we get
[tex]\frac{4}{1}=\frac{M}{M+m}\times (\frac{m+M}{m})^2[/tex]
[tex]4=\frac{m+M}{M}[/tex]
[tex]m=3M[/tex]
Fraction of energy converted into heat[tex]=\frac{1}{2}kA_1^2-\frac{1}{2}k(\frac{A_1}{2})^2[/tex]
[tex]=\frac{1}{2}kA_1^2[1-\frac{1}{4}][/tex]
[tex]=\frac{1}{2}kA_1^2[0.75][/tex]
So, [tex]\frac{3}{4}[/tex] fraction is converted into heat energy
Unpolarized light of intensity I0 is incident on a series of three polarizing filters. The axis of the second filter is oriented at 45o to that of the first filter, while the axis of the third filter is oriented at 90o to that of the first filter. What is the intensity of the light transmitted through the third filter
Answer:
The intensity of the light transmitted through the third filter is [tex]I_3 = \frac{I_o}{8}[/tex]
Explanation:
From the question we are told
The intensity of the unpolarised light [tex]I_o[/tex]
The angle between the first and second polarizer is [tex]\theta _1 = 45^o[/tex]
The angle between the first and third polarizer is [tex]\theta _2 = 90^o[/tex]
Generally the intensity of light emerging from the first polarizer is mathematically represented as
[tex]I_1 = \frac{I_o}{2}[/tex]
According to Malus law the intensity of light emerging from the second polarizer is mathematically represented as
[tex]I_2 = I_1 cos^2 (\theta_1)[/tex]
Substituting for [tex]I_1[/tex] and [tex]\theta _1[/tex]
[tex]I_2 = \frac{I_o}{2} cos^2 (45)[/tex]
[tex]I_2 = \frac{I_o}{4 }[/tex]
According to Malus law the intensity of light emerging from the third polarizer is mathematically represented as
[tex]I_3 = I_2 cos ^2 (\theta_2 - \theta_1)[/tex]
Substituting for [tex]I_2[/tex] and [tex]\theta _1 \ and \ \theta _2[/tex]
[tex]I_3 = \frac{I_o}{4} cos ^2 (90 - 45)[/tex]
[tex]I_3 = \frac{I_o}{8}[/tex]
Suppose a 2.0×10−62.0×10−6-kgkg dust particle with charge −1.0×10−9C−1.0×10−9C is moving vertically up a chimney at speed 6.0 m/sm/s when it enters the +2000-N/CN/C E⃗ E→ field pointing away from a metal collection plate of an electrostatic precipitator. The particle is 4.0 cmcm from the plate at that instant. Find the time needed for the particle to hit the plate. Express your answer with the
Answer:
Explanation:
mass of particle m = 2 x 10⁻⁶ kg
charge q = 1 x 10⁻⁹ C
electric field E = 2000 N/C
force on charge = E q
= 2000 x 1 x 10⁻⁹
acceleration = force / mass
= 2000x10⁻⁹ / 2 x 10⁻⁶
= 1 m /s²
initil velocity u = 6 m /s
distance s = 4 x 10⁻²
time = t
s = ut + .5 t²
4 x 10⁻² = 6t + .5 x 1 x t²
t² + 12t - .08 = 0
= .0067 s .
= 6.7 ms .
(a) The roof of a large arena, with a weight of 410 kN, is lifted by 34 cm so that it can be centered. How much work is done on the roof by the forces making the lift? (b) In 1960 a Tampa, Florida, mother reportedly raised one end of a car that had fallen onto her son when a jack failed. If her panic lift effectively raised 4000 N (about 1/4 of the car's weight) by 5 cm, how much work did her force do on the car?
Answer:
a) W = 139.4 kJ
b) W = 200J
Explanation:
a) given;
Force F = 410kN
distance d = 34cm = 0.34m
Workdone = force × distance = Fd
Substituting the values;
Workdone = 410kN × 0.34m = 139.4 kJ
b) given;
Force F = 4000N
distance d = 5cm = 0.05m
Workdone = force × distance = Fd
Substituting the values;
Workdone = 4000N × 0.05m = 200J
A helicopter, which starts directly above you, lands at a point that is 4.50 km from your present location and in a direction that is 25° north of east. You want to meet the helicopter at it's landing site, however, you must travel along streets that are oriented either east-west or north-south. What is the minimum distance you must travel to reach the helicopter?
Answer:
5.98 km
Explanation:
This question can be easily solved by using the trigonometric properties of a right angled triangle.
See attachment for pictorial explanation
To get x we have
Sinθ = opp / hyp
Sin25 = x / 4.5
x = 4.5 sin 25
x = 4.5 * 0.423
x = 1.9 km
To get y, we have
Cosθ = adj / hyp
Cos25 = y / 4.5
y = 4.5 cos 25
y = 4.5 * 0.906
y = 4.08 km
x + y = 1.9 + 4.09 = 5.98 km
Thus, the minimum distance required is 5.98 km
Final answer:
To find the minimum distance to the helicopter, calculate the eastward and northward components using trigonometry and sum them.
Explanation:
The minimum distance you must travel to reach the helicopter is determined by decomposing the direct diagonal path into two perpendicular paths that correspond to the grid of streets running east-west and north-south. Since the direction to the helicopter is 25° north of east, you can use trigonometry to find the lengths of the east and north legs of your journey. Using the cosine function for the eastward distance (cos(25°) × 4.50 km) and the sine function for the northward distance (sin(25°) × 4.50 km), you can calculate the exact distances you need to travel east and north:
Eastward distance = cos(25°) × 4.50 km
Northward distance = sin(25°) × 4.50 km
Sum these two distances to get the total minimum distance you need to travel.
calcular la longitud de un péndulo que oscila a 10 Hz en santa fe de bogota, sabiendo que en esta ciudad la aceleracion de la gravedad es de 978 cm/s2.
Answer:
[tex]L=2.48*10^{-3} m[/tex]
Explanation:
The period equation for a pendulum is given by:
[tex]T=2\pi \sqrt{\frac{L}{g}}[/tex]
and we know that T = 1/f, where f is the frequency, so we will have:
[tex]\frac{1}{f}=2\pi \sqrt{\frac{L}{g}}[/tex]
Now, we just need to solve this equation for L.
[tex]\frac{1}{2\pi f}=\sqrt{\frac{L}{g}}[/tex]
[tex]L=\frac{g}{(2\pi f)^{2}}[/tex]
g is the gravity in Bogota (g=9.78 m/s^{2})f is 10 HzL is the lenght of the pendulum[tex]L=\frac{9.78}{(2\pi*10)^{2}}[/tex]
[tex]L=2.48*10^{-3} m[/tex]
I hope it helps you!
a star is observed to have a temperature of 3000 K and then luminosity of 105.
What is the color of the star?
Answer: the first box is 'red' and the second box is 'supergiants'
Explanation: just did it edg. And it was correct
A star is observed to have a temperature of 3000 K and then luminosity of 105. The color of the star is red.
What is spectra of star?A spectrum is a collection of all visible light. The region of the electromagnetic spectrum that is visible to the human eye is the light that we see, which includes the rainbow's hues.
All electromagnetic energy emits some kind of radiation, whether it is in the form of visible light or another type, and it also radiates heat. Other stars emit heat and light, just like our sun does. Numerous star measurements have revealed a strong correlation between star temperature and star hue.
Given parameters:
Temperature of the star: T = 3000 K.
luminosity of the star = 105.
For this temperature of the star, the color of the star is red.
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A child on a 2.4 kg scooter at rest throws a 2.2 kg ball. The ball is given a speed of 3.1 m/s and the child and scooter move in the opposite directions at 0.45 m/s. Find the child's mass.
Answer:
The child's mass is 14.133 kg
Explanation:
From the principle of conservation of linear momentum, we have;
(m₁ + m₂) × v₁ + m₃ × v₂ = (m₁ + m₂) × v₃ - m₃ × v₄
We include the negative sign as the velocities were given as moving in the opposite directions
Since the child and the ball are at rest, we have;
v₁ = 0 m/s and v₂= 0 m/s
Hence;
0 = m₁ × v₃ - m₂ × v₄
(m₁ + m₂)× v₃ = m₃ × v₄
Where:
m₁ = Mass of the child
m₂ = Mass of the scooter = 2.4 kg
v₃ = Final velocity of the child and scooter = 0.45 m/s
m₃ = Mass of the ball = 2.4 kg
v₄ = Final velocity of the ball = 3.1 m/s
Plugging the values gives;
(m₁ + 2.4)× 0.45 = 2.4 × 3.1
(m₁ + 2.4) = 16.533
∴ m₁ + 2.4 = 16.533
m₁ = 16.533 - 2.4 = 14.133 kg
The child's mass = 14.133 kg.
A small spotlight mounted in the bottom of a swimming pool that is 4.5 m deep emits light in all directions. On the surface of the pool, directly above the light, the area illuminated is a circle. Determine the maximum radius of this circle. The index of refraction of water is 1.33.
Answer:
Maximum radius = 5.1m
Explanation:
For us to get the radius of the circle of light, we have to first calculate the critical angle which is the angle of incidence above which total internal reflection occurs, i.e. the angle of incidence when θ2 = 90° .
At the point where the total internal reflection occurs, the light ray doesn't pass through the interface, thus, this point is on the edge of the circle of light.
From Snell's law, we have:
n_water * sin (θ1) = n_air * sin(θ2)
Thus;
sin(θ1) = n_air/(n_water * sin(θ2))
Since the critical angle is the value of θ1 when θ2 = 90° and that sin(90°) =1, thus;
sin(θ1) = (n_air/n_water) * sin(90°) = n_air/n_water
We are told the Refractive index of water is 1.33. meanwhile the Refractive index of air is not given but it has a constant value of 1.
Thus, we can determine θ1:
sin(θ1) = (nair/nwater) = 1/1.33
sin(θ1) = 1/1.33
(θ1) = sin^(-1)(1/1.33)
(θ1) = 48.75°
The question when looked at critically, depicts a right triangle with vertices including the light and the extremity of the circle, and we know one of its angles(θ1 = 48.75°) and one of its sides(4.5 m).
Thus, from trigonometric ratio, we can determine the radius as;
r/4.5 = tan(θ1)
r = 4.5tan(48.75°) = 5.1 m