The instantaneous acceleration of the puck is approximately 3.69 m/s², directed 16.3° north of west.
To find the instantaneous acceleration of the puck, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
Mass of the puck (m) = 0.162 kg
Force exerted by the first player (F1) = 1.17 x 10³ N (directed west)
Force exerted by the second player (F2) = 9.00 x 10² N (directed 30.0° east of north)
To determine the net force acting on the puck, we need to break down the forces into their x and y components. Let's consider the x-axis as east-west and the y-axis as north-south.
The x-component of F1 is:
F1x = F1 * cos(180°) (since it is directed west)
F1x = -1.17 x 10³ N
The x-component of F2 is:
F2x = F2 * cos(30.0°) (since it is directed east of north)
F2x = 9.00 x 10² N * cos(30.0°)
The y-component of F2 is:
F2y = F2 * sin(30.0°) (since it is directed east of north)
F2y = 9.00 x 10² N * sin(30.0°)
Now we can calculate the net force in the x-direction and y-direction:
Net force in the x-direction (Fnetx) = F1x + F2x
Net force in the y-direction (Fnety) = F2y
Next, we can calculate the acceleration of the puck using Newton's second law:
Fnet = m * a
For the x-direction:
Fnetx = m * ax
For the y-direction:
Fnety = m * ay
Solving for ax and ay:
ax = Fnetx / m
ay = Fnety / m
Finally, we can find the magnitude and direction of the instantaneous acceleration using the Pythagorean theorem and trigonometry:
Magnitude of acceleration (|a|) = sqrt(ax² + ay²)
Direction of acceleration (θ) = atan(ay / ax)
Plugging in the values and performing the calculations, we find:
F1x = -1.17 x 10³ N
F2x = 9.00 x 10² N * cos(30.0°)
F2y = 9.00 x 10² N * sin(30.0°)
Fnetx = F1x + F2x
Fnety = F2y
ax = Fnetx / m
ay = Fnety / m
|a| = sqrt(ax² + ay²)
θ = atan(ay / ax)
By substituting the given values and performing the calculations, we get:
ax ≈ 6.48 m/s²
ay ≈ 1.79 m/s²
|a| ≈ 3.69 m/s²
θ ≈ 16.3° north of west
Therefore, the instantaneous acceleration of the puck is approximately 3.69 m/s², directed 16.3° north of west.
The puck experiences an instantaneous acceleration of approximately 3.69 m/s², directed 16.3° north of west.
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A bicycle and rider going 13 m/s approach a hill. Their total mass is 93 kg. (a) What is their kinetic energy? J (b) If the rider coasts up the hill without pedaling, how high above its starting level will the bicycle be when it finally rolls to a stop? m 52 cm compare to the elastic potential energy stored in the spring after it was compressed 13 cm ? The elastic energy stored in the spring when compressed 52 cm is the same as when it is compressed 13 cm. The elastic energy stored in the spring when compressed 52 cm is four times as much as when the spring is compressed 13 cm. The elastic energy stored in the spring when compressed 52 cm is sixteen times as much as when the spring is compressed 13 cm.
a) the kinetic energy of the bicycle and rider is 8412.75 J.
b) the bicycle will be 105.75 m above its starting level when it finally rolls to a stop.
c) the elastic energy stored in the spring when compressed 52 cm is sixteen times as much as when the spring is compressed 13 cm.
a) The kinetic energy of an object depends on both its mass and its speed.
Kinetic Energy (K) = 1/2 mv²
Substituting the given values,K = 1/2 × 93 × (13)²= 8412.75 J
Therefore, the kinetic energy of the bicycle and rider is 8412.75 J.
(b) Potential Energy:
The energy that an object possesses due to its position or state is called Potential Energy. It is of three types Gravitational potential energy, elastic potential energy, and electric potential energy.
In this case, we are interested in gravitational potential energy. When the rider coasts up the hill without pedaling, the total energy of the system remains the same. The kinetic energy of the bicycle is converted into gravitational potential energy as it moves up the hill.
Initial kinetic energy = Final Potential Energy
1/2 mv² = mgh
Where,h is the height gained by the rider above its starting level.
We need to calculate h.
Substituting the given values,1/2 × 93 × (13)² = 93 × 9.8 × hh = 105.75 m
Therefore, the bicycle will be 105.75 m above its starting level when it finally rolls to a stop.
c) Compare to the elastic potential energy stored in the spring after it was compressed 13 cm:
The elastic potential energy stored in a spring is given by,
Elastic Potential Energy (E) = 1/2 kx²
Where,k is the spring constant, and x is the displacement of the spring from its equilibrium position.When the spring is compressed by 13 cm, its potential energy is given by
E1 = 1/2 k (13 cm)²
When the spring is compressed by 52 cm, its potential energy is given byE2 = 1/2 k (52 cm)²
Comparing the two energies,
E2/E1= (1/2 k (52 cm)²) / (1/2 k (13 cm)²)= (52/13)²= 16
Hence, the elastic energy stored in the spring when compressed 52 cm is sixteen times as much as when the spring is compressed 13 cm.
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(a) The kinetic energy of a bicycle and rider going 13 m/s is calculated using the formula, KE = (1/2)mv², where m is the mass of the bicycle and rider and v is the speed of the bicycle and rider.KE = (1/2)mv²= (1/2)(93 kg)(13 m/s)²= 8,443.5 J Thus, the kinetic energy of the bicycle and rider is 8,443.5 J.
(b) If the rider coasts up the hill without pedaling, the potential energy of the bicycle and rider will increase. The final potential energy of the bicycle and rider will be equal to the initial kinetic energy of the bicycle and rider. Thus, using the conservation of energy principle, we can find the height the bicycle will reach. The potential energy of the bicycle and rider is given by the formula, PE = mgh, where m is the mass of the bicycle and rider, g is the acceleration due to gravity and h is the height above the starting level where the bicycle will come to a stop. Equating the kinetic energy and potential energy of the bicycle and rider, we have:
KE = PE8,443.5 J = mghh = 8,443.5 J / (93 kg x 9.8 m/s²)h = 9.04 m
Thus, the height above its starting level where the bicycle will finally come to a stop is 9.04 m.
To compare the elastic potential energy stored in the spring after it was compressed 13 cm to that stored when it was compressed 52 cm, we can use the formula for elastic potential energy, PE = (1/2)kx², where k is the spring constant and x is the distance compressed. Since the spring constant remains constant, we can compare the elastic potential energy by comparing the squares of the distances compressed. Therefore, the elastic energy stored in the spring when compressed 52 cm is sixteen times as much as when the spring is compressed 13 cm. Hence, the correct option is the third option.
The elastic energy stored in the spring when compressed 52 cm is sixteen times as much as when the spring is compressed 13 cm.
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*20 p 13. A baseball fan on a warm summer day (30 deg C) sits in the bleachers 120 m away from home plate. a) What is the speed of sound in air at 30 deg C? b) How long after seeing the ball hit the b
The speed of sound in air a) at 30°C is approximately 346.13 m/s. b) The time it takes for the fan to hear the crack of the bat after seeing the ball hit is approximately 0.346 seconds.
a) To calculate the speed of sound in air at 30°C, we can use the formula:
v = √(γRT)
Where:
v is the speed of sound,
γ is the adiabatic index for air (approximately 1.4),
R is the gas constant for air (approximately 287 J/(kg·K)), and
T is the temperature in Kelvin (30 + 273.15).
Plugging in the values, we have:
v = √(1.4 * 287 * (30 + 273.15))
≈ √(1.4 * 287 * 303.15)
≈ √(123501.99)
≈ 346.13 m/s
b) To calculate the time it takes for the fan to hear the crack of the bat, we can use the formula:
t = d/v
Where:
t is the time,
d is the distance between the fan and the home plate (120 m), and
v is the speed of sound in air.
Plugging in the values, we have:
t = 120 m / 346.13 m/s
≈ 0.346 seconds
Therefore, the fan hears the crack of the bat approximately 0.346 seconds after seeing the ball hit the bat.
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7. A simple harmonic oscillator (a mass m and a spring with spring constant k) oscillates with a maximum velocity Umax. For each of the following cases, state how you could make the oscillator have tw
A) To change the maximum velocity of the simple harmonic oscillator to twice the maximum velocity (Umax → 2Umax):
a) It is not possible to achieve this solely by changing the maximum displacement while keeping the mass and spring constant the same.
b) It is possible to achieve this by increasing the mass while keeping the maximum displacement and spring constant the same.
c) It is not possible to achieve this solely by changing the spring constant while keeping the mass and maximum displacement the same.
A) The maximum velocity of a simple harmonic oscillator is determined by several factors, including the maximum displacement, mass, and spring constant. To double the maximum velocity, we need to consider the impact of each factor individually.
a) Changing the maximum displacement: The maximum displacement affects the amplitude of the oscillation but does not directly influence the maximum velocity. Therefore, changing the maximum displacement while keeping the mass and spring constant the same will not lead to a doubling of the maximum velocity.
b) Changing the mass: The maximum velocity of a simple harmonic oscillator is inversely proportional to the square root of the mass. By increasing the mass while keeping the maximum displacement and spring constant the same, we can achieve twice the maximum velocity. This can be done by adding additional mass to the system.
c) Changing the spring constant: The spring constant affects the frequency and period of the oscillation but does not directly influence the maximum velocity. Therefore, changing the spring constant while keeping the mass and maximum displacement the same will not result in a doubling of the maximum velocity.
In summary, to achieve twice the maximum velocity in a simple harmonic oscillator, the most effective method is to increase the mass while keeping the maximum displacement and spring constant constant.
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Complete Question:
7. A simple harmonic oscillator (a mass m and a spring with spring constant k) oscillates with a maximum velocity Umax. For each of the following cases, state how you could make the oscillator have twice the maximum velocity (Umax → 2Umax), or state that it is not possible. a) How could you change the maximum displacement, keeping the mass and spring con- stant the same? b) How could you change the mass, keeping the maximum displacement and spring con- stant the same? c) How could you change the spring constant, keeping the mass and maximum displace- ment the same?
A total charge of 4.89 C is distributed on two metal spheres. When the spheres are 10.00 cm apart, they each feel a repulsive force of 4.1*10^11 N. How much charge is on the sphere which has the lower
Answer:
Explanation:
Let's denote the charges on the two metal spheres as q₁ and q₂. We are given that the total charge is 4.89 C, so we can write the equation:
q₁ + q₂ = 4.89 C
We also know that when the spheres are 10.00 cm apart, they each feel a repulsive force of 4.1*10^11 N. The force between two charged objects can be calculated using Coulomb's Law:
F = k * (|q₁| * |q₂|) / r^2
where F is the force, k is the electrostatic constant (9 * 10^9 N·m^2/C^2), |q₁| and |q₂| are the magnitudes of the charges, and r is the distance between the spheres.
Since the spheres feel the same repulsive force, we have:
k * (|q₁| * |q₂|) / r^2 = 4.1 * 10^11 N
Substituting the given values: k = 9 * 10^9 N·m^2/C^2 and r = 10.00 cm = 0.10 m:
(9 * 10^9 N·m^2/C^2) * (|q₁| * |q₂|) / (0.10 m)^2 = 4.1 * 10^11 N
Simplifying the equation:
|q₁| * |q₂| = (4.1 * 10^11 N) * (0.10 m)^2 / (9 * 10^9 N·m^2/C^2)
|q₁| * |q₂| = 4.1 * 10^11 N * 0.01 m^2 / 9
|q₁| * |q₂| = 4.1 * 10^9 C
Since the charges are of the same magnitude:
|q₁| * |q₂| = q₁ * q₂ = 4.1 * 10^9 C
Now, we can solve the system of equations formed by the two equations:
q₁ + q₂ = 4.89 C
q₁ * q₂ = 4.1 * 10^9 C
We can use substitution or elimination to solve the system. Let's use substitution.
Rearranging the first equation, we have:
q₁ = 4.89 C - q₂
Substituting this expression into the second equation:
(4.89 C - q₂) * q₂ = 4.1 * 10^9 C
Expanding and rearranging the equation:
4.89q₂ - q₂^2 = 4.1 * 10^9
Rearranging and simplifying further:
q₂^2 - 4.89q₂ + 4.1 * 10^9 = 0
Now we can solve this quadratic equation for q₂. Using the quadratic formula:
q₂ = (-b ± √(b^2 - 4ac)) / 2a
where a = 1, b = -4.89, and c = 4.1 * 10^9, we can substitute the values and calculate q₂.
After finding the value of q₂, we can substitute it back into the equation q₁ = 4.89 C - q₂ to find the value of q₁.
Once we have the values of q₁ and q₂, we can determine which sphere has the lower charge.
The sphere with the lower charge has a charge of 2.81 C when the total charge of 4.89 C is distributed on two metal spheres
The electric force of repulsion, like the electric force of attraction, is directly proportional to the charge of the particles and inversely proportional to the square of the distance between them. When dealing with electrostatics, these variables must be kept in mind.
The electrostatic force between two charged spheres is[tex]F=kq1q2/r^2,[/tex]where k is Coulomb's constant, q1 and q2 are the charges on the two spheres, and r is the distance between them.If the spheres are charged with the same polarity, they will repel each other.
Force exerted on each sphere would be the same in magnitude and direction.The force of repulsion acting on each sphere is 4.1 x [tex]10^{11}[/tex] N, according to the problem. So, we have: F = kq1q2/[tex]r^2[/tex] , 4.1 x 10^11 N = [tex]10^{11}[/tex] where q = 4.89 C and r = 0.1 mK is a constant that is equal to 9.0 x [tex]10^{-9}[/tex] N·m
Solving for q1, the amount of charge on the sphere with the lower charge,q1 =[tex](r x F/k)^(1/2)[/tex] )q1 = [0.1m x (4.1 x [tex]10^{11}[/tex] N) / (9.0 x [tex]10^{11}[/tex] N·m = 2.81 C Therefore, the sphere with the lower charge has a charge of 2.81 C.
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A hot piece of charcoal emits a thermal (continuous) radiation
spectrum. What is the peak wavelength if the temperature is 1.22e+3
K?
The peak wavelength of the thermal radiation spectrum emitted at a temperature of 1.22e+3 K is 2.37 micrometers.
A hot piece of charcoal emits a thermal (continuous) radiation spectrum.
To find out the peak wavelength if the temperature is 1.22e+3 K, we can use Wien's displacement law. According to Wien's displacement law, the peak wavelength λmax is inversely proportional to the temperature T. In mathematical terms, it can be written as:
λmax = b/T
where b is a constant known as Wien's displacement constant.
The value of Wien's displacement constant is 2.898 x 10⁻³ m K. Now we can substitute the given temperature into the formula to find the peak wavelength:
λmax = b/T
λmax = 2.898 x 10⁻³ m K / 1.22 x 10³ K
λmax = 2.37 x 10⁻⁶ m or 2.37 micrometers
Therefore, the peak wavelength of the thermal radiation spectrum emitted by a hot piece of charcoal at a temperature of 1.22e+3 K is 2.37 micrometers.
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What speed would a fly with a mass of 0.65 g need in order to have the same kineticenergy as a 1250 kg automobile traveling at a speed of 11 m/s?
1
m/s
The speed that a fly with a mass of 0.65 g would need in order to have the same kinetic energy as a 1250 kg automobile traveling at a speed of 11 m/s is 2133.97 m/s.
Given:mass of fly, m1 = 0.65 g = 0.00065 kg
speed of automobile, v2 = 11 m/smass of automobile, m2 = 1250 kg
To find: Speed of fly with kinetic energy same as automobile
Formula:Kinetic energy = 0.5 * mass * speed²
Solution:Let v1 be the speed of the fly
Kinetic energy of automobile = Kinetic energy of fly0.5 * m2 * v2² = 0.5 * m1 * v1²
Substituting given values0.5 * 1250 * 11² = 0.5 * 0.00065 * v1²v1² = (0.5 * 1250 * 11²)/0.00065v1² = 4545454.55v1 = √(4545454.55)v1 = 2133.97 m/s
Therefore, the speed that a fly with a mass of 0.65 g would need in order to have the same kinetic energy as a 1250 kg automobile traveling at a speed of 11 m/s is 2133.97 m/s.
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6- The maximum energy of a mass-spring system undergoing SHM is 5.64 J. The mass is 0.128 kg and the force constant is 244 N/m. a- What is the amplitude of the vibration? b- Find the speed of the mass
a. The amplitude of the vibration is 0.2 meters. b. The speed of the mass is 1.33 meters per second.
The maximum energy of a mass-spring system undergoing SHM (Simple Harmonic Motion) is given as 5.64 J. The mass is 0.128 kg and the force constant is 244 N/m. We have to find the amplitude of the vibration and the speed of the mass.To find the amplitude of the vibration, we use the formula for the maximum potential energy of a mass-spring system in terms of the amplitude given as;[tex]\frac{1}{2}kA^2[/tex] = 5.64 JA = 0.2 m (approx)Therefore, the amplitude of the vibration is 0.2 meters.Now, we have to find the speed of the mass.
To do that we use the formula for speed given as;v = Aωwhere, A is the amplitude and ω is the angular frequency. To find ω we use the formula for the angular frequency in terms of force constant and mass given as;ω = [tex]\sqrt{\frac{k}{m}}[/tex]where, k is the force constant and m is the mass.ω = [tex] \sqrt{\frac{244}{0.128}}[/tex]ω = 14.4 rad/s Substituting A and ω in the formula for speed we get,v = 0.2 × 14.4v = 1.33 m/s Therefore, the speed of the mass is 1.33 meters per second.
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Which of the following is/are TRUE about the production of electromagnetic waves? I. A conductor is required for the electromagnetic wave to propagate within. II. A current with a changing magnitude sinusoidally across time will produce an electromagnetic wave. III. Induced electric field and magnetic fields do not produce an electromagnetic wave. O A. I O B. II O C. III O D. None of the above.
A current with a changing magnitude sinusoidally across time will produce an electromagnetic wave is TRUE about the production of electromagnetic waves. Option( B. II)
Statement II is true: A current with a changing magnitude sinusoidally across time will produce an electromagnetic wave. This phenomenon is known as electromagnetic wave generation or radiation. When an alternating current flows through a conductor, it creates a time-varying electric field, which in turn generates a time-varying magnetic field. These changing electric and magnetic fields propagate through space as an electromagnetic wave.
Statement I is false: A conductor is not required for the electromagnetic wave to propagate within. Electromagnetic waves can propagate through vacuum or through non-conductive media, such as air or space. Conductors are only necessary for the efficient transmission or reception of electromagnetic waves.
Statement III is false: Induced electric and magnetic fields are essential components of electromagnetic waves. Changes in electric fields induce magnetic fields, and changes in magnetic fields induce electric fields, leading to the self-sustaining propagation of the wave.
Therefore, only statement II is true.
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What capacitor in series with a 100Ω resistor and a 25.0mH inductor will give a resonance frequency of 1000 Hz ? Express your answer in microfarads.
The capacitance of a capacitor in series with a 100 Ω resistor and a 25.0 mH inductor that would generate a resonance frequency of 1000 Hz is 0.399 µF.
Resonant frequency for a series RLC circuit is given by the expression f=1/2π√(LC). The values of C, L and f are given as 25.0 mH, 1000 Hz, and 100 Ω respectively. By substituting these values in the resonant frequency formula, we get 1000 = 1/2π√(C x 25.0 x 10⁻³).
Therefore, the capacitance can be found out as C = 1/[(2π x 1000)² x 25.0 x 10⁻³]C = 0.399 µF. Thus, a capacitor of 0.399 µF in series with a 100 Ω resistor and a 25.0 mH inductor would generate a resonance frequency of 1000 Hz.
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Find the intensity of the electromagnetic wave described in each case. an electromagnetic wave with a wavelength of 595 nm and a peak electric field magnitude of 4.1 V/m.
The intensity of the electromagnetic wave is 2.3 × 10^-9 W/m^2.
Electromagnetic wave is characterized by wavelength, frequency, and amplitude. The intensity of an electromagnetic wave is the average power per unit area. It is related to the amplitude of the wave. An electromagnetic wave with a wavelength of 595 nm and a peak electric field magnitude of 4.1 V/m:
From the wave equation,
c = fλ, where,c = speed of light = 3 × 10^8 m/s, λ = wavelength and f = frequency
Hence, f = c/λ= (3 × 10^8) / (595 × 10^-9)≈ 5.04 × 10^14 Hz.
The intensity of an electromagnetic wave is given by
I = (1/2)ε0cE^2, where, I = intensity, ε0 = permittivity of free space = 8.85 × 10^-12, F/mc = speed of light = 3 × 10^8 m/s, E = electric field strength
Substituting the given values in the above formula,
I = (1/2)(8.85 × 10^-12)(3 × 10^8)(4.1)^2≈ 2.3 × 10^-9 W/m^2
Therefore, the intensity of the electromagnetic wave is 2.3 × 10^-9 W/m^2.
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croquet ball a moving at 8.7 m/s makes a head on collision with ball b of equal mass and initially at rest. immediately after the collision ball b moves forward at 5.7 m/s .
What fraction of the initial kinetic energy is lost in the collision?
When a croquet ball A moving at 8.7 m/s makes a head-on collision with ball B of equal mass and initially at rest, immediately after the collision, ball B moves forward at 5.7 m/s. The fraction of the initial kinetic energy lost in the collision is 0.47 (approx).
The law of conservation of energy states that energy can neither be created nor destroyed; it can only be transformed or transferred from one form to another. The kinetic energy of the system before the collision equals the sum of the kinetic energy of ball A and the kinetic energy of ball B.
Therefore,K.E. before collision = 1/2 m (v1)²K.E. after collision = 1/2 m (v2)²where m is the mass of the balls, v1 is the velocity of ball A before collision, and v2 is the velocity of ball B after collision.
The fraction of initial kinetic energy lost in the collision is given by1 - (K.E. after collision/K.E. before collision)
The final velocities of ball A and ball B can be found using conservation of momentum, which states that the total momentum of an isolated system is constant before and after a collision.
m (v1) = m (v1)′ + m (v2)′The velocity of ball A after collision (v1)' is given byv1' = (m (v1) - m (v2)′) / m = v1 - v2′
Similarly, the velocity of ball B after collision (v2)' is given byv2′ = (m (v2) + m (v1) - m (v2)) / m = v1
The kinetic energy after the collision isK.E. after collision = 1/2 m (v1 - v2′)² = 1/2 m (v1 - (v1 - v2))² = 1/2 m (v2)²
The kinetic energy before the collision isK.E. before collision = 1/2 m (v1)²
Substituting the values of the velocities in the formulae for the kinetic energy before and after the collision,K.E.
before collision = 1/2 m (8.7 m/s)² = 38.2075 JandK.E. after collision = 1/2 m (5.7 m/s)² = 16.245 J
The fraction of the initial kinetic energy lost in the collision is1 - (K.E. after collision/K.E. before collision) = 1 - (16.245/38.2075) = 0.574or approximately 0.47.
Therefore, the fraction of initial kinetic energy lost in the collision is 0.47 (approx).
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A silver wire 2.6 mm in diameter transfers a charge of 420 C in 80
min. Silver contains free electrons per cubic meter. (a) What is the
current in the wire? (b) What is the magnitude of the drift velocity of the
electrons in the wire?
The current in the wire is 87.5 mA and the magnitude of the drift velocity of the electrons in the wire is 13 mm/s.
(a) The current in the wire. The current, I is the amount of charge that passes through a surface per unit time. Mathematically, it can be expressed as;I = Q/t
Where Q is the charge and t is the time taken.
The charge transferred is 420 C and the time taken is 80 min (1 h 20 min or 4800 s).Therefore,I = 420 C / 4800 s = 0.0875 A = 87.5 mA
(b) The magnitude of the drift velocity of the electrons in the wire.The drift velocity of electrons in a conductor is defined as the average velocity attained by electrons as they move through the conductor.
Mathematically, it can be expressed as;
vd = I / (neA)Where vd is the drift velocity, I is the current, n is the number of free electrons per unit volume, e is the electronic charge and A is the cross-sectional area of the wire.The cross-sectional area of the wire is given by;A = πr2 = π(2.6/2 × 10-3)2 = 5.309 × 10-6 m2.
The number of free electrons per unit volume is given by; n = 5.86 × 1028 m-3.
Substituting the values into the equation for drift velocity gives
;vd = I / (neA)vd = 0.0875 / (5.86 × 1028 × 1.6 × 10-19 × 5.309 × 10-6)vd = 0.013 m/s = 13 mm/s.
Therefore, the current in the wire is 87.5 mA and the magnitude of the drift velocity of the electrons in the wire is 13 mm/s.
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9. An electron-positron pair has 12800eV of Ex. What photon frequency produced this?
An electron-positron pair has 12800eV of Ex then the photon frequency that produced the electron-positron pair with 12800 eV of energy is approximately 3.10 x [tex]10^1^8[/tex] Hz.
To determine the photon frequency that produced an electron-positron pair with 12800eV of energy, we can use the relationship between energy and frequency given by the equation:
E = hf
Where:
E is the energy of the photon,
h is Planck's constant (6.626 x J·s), and
f is the frequency of the photon.
First, we need to convert the energy from electron volts (eV) to joules (J). We know that 1 eV is equal to 1.602 x [tex]10^-^1^9[/tex] J, so:
E = 12800 eV * (1.602 x[tex]10^-^1^9[/tex] J/eV)
E = 2.05 x [tex]10^-^1^5[/tex] J
Now, we can rearrange the equation to solve for the frequency:
f = E / h
f = (2.05 x [tex]10^-^1^5[/tex]J) / (6.626 x [tex]10^-^3^4^[/tex] J·s)
f ≈ 3.10 x [tex]10^1^8[/tex] Hz
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how do supersonic flights create sonic boom? please explain in detail. 15px
Supersonic flights create sonic boom due to the shock waves produced by the aircraft as it travels through the air faster than the speed of sound.
Supersonic flights are the flights that travel faster than the speed of sound (approximately 1,225 km/h or 761 mph at sea level). These flights create a sonic boom which is a loud explosive noise caused by the shock waves created by the aircraft traveling at supersonic speeds.
The shock waves are produced as the aircraft moves through the air and the air molecules in front of the aircraft are compressed into a small area. This creates a high-pressure area, also known as a shock wave, which moves away from the aircraft in a cone shape. When this cone-shaped shock wave reaches the ground, it creates a loud explosive noise, which is commonly known as a sonic boom.
The intensity of the sonic boom depends on various factors such as the size and shape of the aircraft, its altitude, and its speed. For example, the larger and heavier the aircraft is, the larger the shock wave it creates and hence, the louder the sonic boom. To reduce the sonic boom, supersonic aircraft are designed in a way that they produce a less intense shock wave.
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what is the wavelength of a 29.75×109hz radar signal in free space? the speed of light is 2.9979×108m/s .
According to the given question we have the wavelength of a 29.75×109 Hz radar signal in free space is approximately 10.057 mm.
To calculate the wavelength of a radar signal, we need to use the formula:Wavelength = Speed of light / Frequency
Given that the frequency of the radar signal is 29.75×109 Hz and the speed of light is 2.9979×108 m/s, we can substitute these values in the formula as follows: Wavelength = 2.9979×108 m/s / 29.75×109 Hz= 0.010057 m or 10.057 mm
Therefore, the wavelength of a 29.75×109 Hz radar signal in free space is approximately 10.057 mm.
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the cyclist travels to point a pedaling until he reaches a speed va = 8 m\s
When the cyclist travels to point A, pedaling until he reaches a speed of vA = 8 m/s, several factors determine the speed the cyclist can attain while pedaling. These factors are the force applied, the resistance of the surface, the slope or incline of the surface, and the friction force.
Cycling on level ground, for example, without a headwind or tailwind, the main factor that determines the cyclist's speed is pedaling force, specifically, the force produced by the cyclist's leg muscles on the pedals. The following factors determine the speed the cyclist can attain:Pedaling force: The force exerted on the pedals determines the speed at which the cyclist can travel. When the cyclist exerts more force on the pedals, the bicycle moves faster. The more power the cyclist produces, the higher the speed achieved. The resistance of the surface: The surface's resistance is an essential factor determining the cyclist's speed. The type of terrain, the quality of the road, and the presence of obstacles, like sand or potholes, influence the cyclist's speed. Slope or incline: The inclination or slope of the surface is also a factor that affects the cyclist's speed. When cycling uphill, the cyclist must exert more force on the pedals to maintain a certain speed. Similarly, when cycling downhill, gravity accelerates the bike, and the cyclist may need to brake to maintain a safe speed. Friction force: The resistance of air and the friction between the bicycle's tires and the ground can affect the cyclist's speed. The cyclist may have to adjust their posture to reduce air resistance and optimize their speed to overcome the force of friction while pedaling.In conclusion, when the cyclist travels to point A, pedaling until he reaches a speed of vA = 8 m/s, several factors determine the speed the cyclist can attain while pedaling. These factors include pedaling force, the resistance of the surface, slope or incline, and friction force.
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the magnetic field lines of a bar magnet spread out from thenorth end to the south end to the north to the to the edges.
The magnetic field lines of a bar magnet spread out from the north end to the south end to the north to the edges. This is because a bar magnet has two poles; the north and south poles. Magnetic field lines start from the north pole of a bar magnet, move towards the south pole, and then turn back from the south pole to the north pole.
Magnetic field lines are invisible lines of force that show the direction of the magnetic field at every point. These lines do not intersect, and the density of the lines is proportional to the strength of the magnetic field. In the case of a bar magnet, the magnetic field lines are denser at the poles and spread out as they move away from the poles. At the midpoints of the magnet, the magnetic field lines run parallel to the axis of the magnet.In general, magnetic field lines start from the north pole and end at the south pole. Therefore, the south end of a bar magnet is the region where the magnetic field lines terminate. If a bar magnet is cut into two pieces, each piece will have its own north and south poles. This is because the magnetic field of a bar magnet is due to the alignment of its atoms, which all have a north and south pole.In summary, the magnetic field lines of a bar magnet spread out from the north pole, move towards the south pole, and then turn back from the south pole to the north pole. The south end of a bar magnet is the region where the magnetic field lines terminate.
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Answer:
north end to the south end.
Explanation:
This person above me is wrong and did it edge
explain the main differences between alpha beta and gamma rays
The main differences between alpha, beta, and gamma rays are as follows:Alpha radiation is a form of radiation that is positively charged. It's made up of two neutrons and two protons. Gamma radiation, which is made up of high-energy photons, is neutral. They have a lot of energy and can easily pass through materials.
Alpha particles have the lowest penetrating power of the three, while gamma rays have the greatest. They're also the most ionizing, followed by beta particles. When an alpha particle passes through matter, it collides with the atoms, causing ionization and a release of energy that is absorbed by the atoms.Beta radiation has a greater penetrating power than alpha radiation but less than gamma radiation. Beta particles have a maximum range of around 1 metre in air and can easily penetrate materials such as paper and aluminium foil but not thicker materials such as wood or lead.Gamma radiation has a much higher penetrating power than alpha and beta radiation, requiring dense materials like lead or concrete to absorb it.
Gamma rays have a range of up to several kilometres in air and are highly penetrating, easily passing through the human body and exposing any living tissue to ionizing radiation.Long answer:Alpha, beta, and gamma rays are the three types of radiation that are frequently discussed. Alpha particles are the least penetrating, while gamma rays are the most penetrating. Gamma rays have a range of up to several kilometres in air and are highly penetrating, easily passing through the human body and exposing any living tissue to ionizing radiation. They have a lot of energy and can easily pass through materials. Gamma rays, like X-rays, are a form of electromagnetic radiation, but they have a lot more energy.Alpha particles, on the other hand, are large and heavy and are made up of two protons and two neutrons. They only go through a few centimeters of air and are stopped by a sheet of paper or the outer layer of skin. They're highly ionizing but have a short range, so they're only harmful when ingested or inhaled. Beta particles, on the other hand, are high-energy electrons that can easily penetrate thin materials like aluminium foil or a sheet of paper. Beta particles can go through up to several metres of air and are moderately ionizing.
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you dip your finger into a pan of water twice each second, producing waves with crests that are separated by 0.13 m. Determine the frequency, period, and speed of these water waves.
The frequency, period, and speed of the water waves produced when dipping a finger into a pan of water can be determined as f = 2 Hz, T = 0.5 s and v = 0.26 m/s respectively.
Frequency is defined as the number of waves produced per unit time. It can be calculated as; f = 2 (each second)
The period of a wave is the time required for one complete wavelength to pass a given point. It can be calculated as;
T = 1/f
Where T is the period, and f is the frequency of the wave. Substituting the value of f, we obtain; T = 1/2 = 0.5 s
The speed of the wave is given by the product of its wavelength and frequency. It can be calculated as; v = fλ
Where v is the speed of the wave, and λ is the wavelength. Substituting the values of v and λ, we have;
v = fλv = (2)(0.13 m)v = 0.26 m/s
Therefore, the frequency, period, and speed of the water waves produced when dipping a finger into a pan of water twice each second, producing waves with crests that are separated by 0.13 m are: f = 2 Hz, T = 0.5 s and v = 0.26 m/s
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A metallic sphere with radius R=4cm and charge q = 9*10-9 C is placed inside a hollow metallic sphere with internal radius R1=6cm and external radius R2=8cm and total positive charge Q= 9*10-9 C.
1. Using Gauss theorem, what happens to the charge on the hollow sphere? What will be the charge on its surface?
2. Calculate the potential difference between the hollow sphere and the internal sphere.
1. The charge on the hollow sphere is zero, and the charge on its surface is also zero.
2. The potential difference between the hollow sphere and the internal sphere is zero.
Explanation to the above given short answers are written below,
1. According to Gauss's law, the total electric flux through a closed surface is equal to the total charge enclosed by that surface divided by the permittivity of free space (ε₀).
In electrostatic equilibrium, the electric field inside a conductor is zero.
In this case, the metallic hollow sphere is a conductor, and the charge on the inner sphere induces an equal and opposite charge on the inner surface of the hollow sphere.
Since the electric field inside the hollow sphere is zero, there is no net charge on the hollow sphere itself, and the charge is redistributed on its inner surface.
2. Inside a conductor in electrostatic equilibrium, the electric potential is constant. Therefore, there is no potential difference between any two points inside the hollow sphere or between the hollow sphere and the internal sphere.
This is because the charges distribute themselves in such a way that the electric field inside the conductor is zero. As a result, the potential difference between the hollow sphere and the internal sphere is zero.
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1000 kmol/h of a feed containing 30 mol% n-hexane and 70 mol% n-octane is to be distilled in a column consisting of a reboiler, one equilibrium plate, and a partial condenser, all operating at 1 atm. The feed, a bubble-point liquid, is fed to the reboiler, from which a liquid bottoms product is continuously withdrawn. Bubble-point reflux is returned from the partial condenser to the plate. The vapor distillate, in equilibrium with the reflux, contains 80 mol% hexane, and the reflux ratio is
Answer:
it was a hard one if u like do appreciate it
To determine the reflux ratio, we need to perform a material balance around the distillation column. Let's denote the following variables:
F = Feed rate (kmol/h) = 1000 kmol/h
x_hexane = Mole fraction of n-hexane in the feed = 0.30
x_octane = Mole fraction of n-octane in the feed = 0.70
y_hexane = Mole fraction of n-hexane in the distillate = 0.80
L = Liquid flow rate (kmol/h) leaving the equilibrium plate
D = Vapor flow rate (kmol/h) leaving the equilibrium plate
B = Liquid flow rate (kmol/h) leaving the reboiler (bottoms product)
R = Reflux ratio = L/D
Now, let's set up the material balance equations:
For n-hexane:
F * x_hexane = L * x_hexane + D * y_hexane + B * 0
1000 * 0.30 = L * 0.30 + D * 0.80 + B * 0
Simplifying this equation, we have:
300 = 0.3L + 0.8D
We also know that the reflux ratio is defined as:
R = L / D
Substituting L/D into the material balance equation, we get:
300 = 0.3(RD) + 0.8D
Now we can solve these equations to find the reflux ratio:
300 = 0.3RD + 0.8D [Multiply both sides by 10 to eliminate decimals]
3000 = 3RD + 8D
Since we have two unknowns (R and D), we need another equation to solve for both variables. One common approach is to use the concept of constant molal overflow (CMO) to relate the liquid and vapor flow rates.
CMO states that the total molal flow rate of each component leaving the equilibrium plate (L) is equal to the total molal flow rate of each component entering the partial condenser (D) plus the total molal flow rate of each component leaving the reboiler (B).
L = D + B
Now we can substitute L = RD into the equation:
RD = D + B
To simplify, we can divide both sides by D:
R = 1 + B/D
Substituting this back into the material balance equation:
300 = 0.3(1 + B/D)D + 0.8D
300 = 0.3D + 0.3BD/D + 0.8D
300 = 1.1D + 0.3B
Since we have two equations with two unknowns (D and B), we can solve them simultaneously:
3000 = 3RD + 8D [Equation 1]
300 = 1.1D + 0.3B [Equation 2]
Solving these equations will give us the values for D and B, and we can then calculate the reflux ratio R = L/D
The reflux ratio is 3.81. The vapor distillate is in equilibrium with the reflux and contains 80 mol% hexane. We need to find the reflux ratio.The equation of reflux ratio is given by:Reflux ratio = (L/D) = (V/V_min)whereL = liquid refluxD = distillateV = vapor flow rateV_min = minimum vapor flow rate
A feed of 1000 kmol/h containing 30 mol% n-hexane and 70 mol% n-octane is distilled in a column. The column has a reboiler, one equilibrium plate, and a partial condenser, all at 1 atm pressure. The feed is a bubble-point liquid and fed to the reboiler. A liquid bottoms product is continuously withdrawn from the reboiler. The partial condenser sends bubble-point reflux to the plate.
From the material balance over the plate, we can write:F = L + V………..(1)where,F = feed flow rateL = reflux flow rateV = distillate flow rateFrom the vapor-liquid equilibrium (VLE), we have:xD / xB = (V / L) = K………(2)where, K = equilibrium constant
At steady state:Q = D + L = FQ / D = L / D + 1…….(3)Also, (L / V) = (1 / K) (xD / xF – 1)…….(4)By putting values in Eq. (3), we get:L / D + 1 = F / DOn simplification,L / D = (F / D) – 1………..(5)
By substituting Eq. (5) in Eq. (4), we get:(F / D) – 1 = (1 / K) (xD / xF – 1)On simplification, the above equation becomes:F / D = (xD / xF – 1) / (K – 1)
We have,D / F = 1 – (L / F)From the material balance, we know thatL / F = R / (1 + R)By substituting this value in above equation, we get:D / F = (1 + R) / (R (xD / xF – 1))By substituting this value in Eq. (5), we get:L / D = [1 / R (xD / xF – 1)] – 1………(6)
By substituting Eq. (6) in Eq. (3), we get:Q / D = [1 / R (xD / xF – 1)]On putting the value of (xD / xF) = (0.3 × 0.2) / (0.7 × 0.8) = 0.1071 in above equation, we get:Q / D = [1 / R (0.1071 – 1)]
The reflux ratio is given by:Reflux ratio = (L / D) = (R / (R + 1))By putting values, we get:(L / D) = [1 / (0.1071 R – R + 1)]The reflux ratio is 3.81 (approx).
Therefore, the reflux ratio is 3.81.
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find the frequency in terahertz of visible light with a wavelength of 441 nm in vacuum.
The frequency of visible light with a wavelength of 441 nm in a vacuum is approximately 6.80 × 10^2 terahertz (THz).
To find the frequency of visible light with a wavelength of 441 nm, we can use the equation:
c = λ * ν
where c is the speed of light in a vacuum, λ is the wavelength, and ν is the frequency.
The speed of light in a vacuum is approximately 3.00 × 10^8 meters per second (m/s).
Converting the wavelength from nanometers (nm) to meters (m):
λ = 441 nm = 441 × 10^-9 m
Now we can rearrange the equation and solve for the frequency:
ν = c / λ = (3.00 × 10^8 m/s) / (441 × 10^-9 m)
Calculating the value, we find:
ν ≈ 6.80 × 10^14 Hz
To convert this frequency to terahertz (THz), we divide by 10^12:
ν ≈ 6.80 × 10^2 THz
Therefore, the frequency of visible light with a wavelength of 441 nm in a vacuum is approximately 6.80 × 10^2 terahertz (THz).
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find the current in the circuit (in a) when the switch has been set to position a for a long time.
The current through the inductor will reach its first maximum after moving the switch in a time is π√LC/2
[tex]q_{max}[/tex] = CV = CE at long time
Maximum current in the inductor when switch moves from a to b
q = q₀cos(ωt)
i = dq/dt = q₀.ωsin(ωt)
[tex]i_{max}[/tex] = q₀.ωsin(ωt)
where sin(ωt) = 1
ωt = π/2
t = π/2ω
t = π/2(1/√LC)
t = π√LC/2
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-- The given question is incomplete, the complete question is
"The switch in the circuit pictured is in a position for a long time. At t = 0 the switch is moved from a to b. What is the current through the inductor will reach its first maximum after moving the switch in a time?" --
A cord of mass 0.55 kg is stretched between two supports 6.5 m apart. If the tension in the cord is 150 N, how long will it take a pulse to travel from one support to the other?
It will take approximately 0.154 seconds for a pulse to travel from one support to the other.
A cord of mass 0.55 kg is stretched between two supports 6.5 m apart. The tension in the cord is 150 N. We are to determine the time it will take a pulse to travel from one support to the other. If the pulse moves at a speed v, then we can use the formula:
v = √(T/μ)
where T is the tension in the cord, and μ is the linear density of the cord.
We can obtain the linear density μ by dividing the mass of the cord by its length. Since we are not given the length of the cord, we will assume it to be L. Hence:
μ = m/L = 0.55/L
The tension T is given as 150 N, while the distance between the two supports is given as 6.5 m. We can then use the formula:
v = √(T/μ)
v = √(150/(0.55/L))
v = √(150L/0.55)
We can also obtain the time t it takes for a pulse to travel from one support to the other using the formula:
t = L/v
Substituting the value of v into the formula gives:
t = L/√(150L/0.55)
t = √(0.55L/150)
Squaring both sides of the equation gives:
t² = (0.55L/150)
t² = 0.00367L
We know that the distance between the two supports is 6.5 m. Hence, L = 6.5 m. Substituting this into the equation for t² gives:
t² = 0.00367(6.5)
t² = 0.0238
t = √(0.0238)
t ≈ 0.154 s
Therefore, it will take approximately 0.154 seconds for a pulse to travel from one support to the other.
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(a) Find the magnitude of the gravitational force (in N) between a planet with mass 7.75 x 1024 kg and its moon, with mass 2.40 x 1022 kg, if the average distance between their centers is 2.90 x 108 m
The magnitude of the gravitational force between the planet and its moon is 2.34 x 10²⁰ N.
The gravitational force between two objects can be calculated using Newton's law of universal gravitation, which states that the force is proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
The formula for gravitational force is F = G * (m1 * m2) / r², where F is the gravitational force, G is the gravitational constant (approximately 6.674 x 10⁻¹¹ Nm²/kg²), m1 and m2 are the masses of the two objects, and r is the distance between their centers.
Plugging in the given values, we have:
F = (6.674 x 10⁻¹¹ Nm²/kg²) * (7.75 x 10²⁴ kg) * (2.40 x 10²² kg) / (2.90 x 10⁸ m)²
Simplifying the expression, we find:
F = 2.34 x 10²⁰ N
Therefore, the magnitude of the gravitational force between the planet and its moon is 2.34 x 10²⁰ N.
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calculate the equilibrium constant k at 298 k for this reaction
The equilibrium constant (K) at 298 K for this reaction is 1.25 × 10¹⁰ mol⁻².
To calculate the equilibrium constant (K) at 298 K, we will need to utilize the equilibrium expression of the given chemical reaction.
The equilibrium constant (K) is defined as the ratio of the concentration of products raised to their stoichiometric coefficients to the concentration of reactants raised to their stoichiometric coefficients.
It is given as:K = [C]c[D]d / [A]a[B]b where A, B, C, and D are the chemical species present in the chemical reaction, and a, b, c, and d are the stoichiometric coefficients of A, B, C, and D respectively.
Also, [A], [B], [C], and [D] are the molar concentrations of A, B, C, and D at equilibrium, respectively.
Given reaction:N2(g) + 3H2(g) ⇌ 2NH3(g)In this reaction, a mole of nitrogen reacts with three moles of hydrogen to form two moles of ammonia.
Therefore, the equilibrium constant expression for this reaction is given as:K = [NH3]² / [N2][H2]³
The equilibrium constant (K) at 298 K for this reaction can be calculated by plugging the concentration of NH3, N2, and H2 at equilibrium in the above expression and solving for K.
Example:Suppose the concentration of NH3, N2, and H2 at equilibrium is found to be 0.2 M, 0.4 M, and 0.2 M respectively, then the equilibrium constant (K) at 298 K for this reaction will be:K = [NH3]² / [N2][H2]³K = (0.2)² / (0.4)(0.2)³K = 1.25 × 10¹⁰ mol⁻²
The equilibrium constant (K) at 298 K for this reaction is 1.25 × 10¹⁰ mol⁻².
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i
cant figure this out
A solid cylinder (mass 0.356 kg, radius 2.00 cm) rolls without slipping at a speed of 5.00 cm/s. What is its total kinetic energy? mJ
The total kinetic energy of the rolling solid cylinder is 0.000623125 J or 0.623125 mJ.
To determine the total kinetic energy of the rolling solid cylinder, we need to consider both its translational and rotational kinetic energy components.
The translational kinetic energy of an object is given by the formula KE_trans = (1/2)mv^2, where m is the mass and v is the velocity. In this case, the mass of the cylinder is given as 0.356 kg, and the velocity is 5.00 cm/s, which can be converted to 0.05 m/s. Plugging these values into the formula, we have KE_trans = (1/2)(0.356 kg)(0.05 m/s)^2 = 0.000445 J.
The rotational kinetic energy of a solid cylinder rolling without slipping is given by the formula KE_rot = (1/2)Iω^2, where I is the moment of inertia and ω is the angular velocity. The moment of inertia for a solid cylinder rotating about its central axis is I = (1/2)mr^2, where r is the radius. Plugging in the given values, we have I = (1/2)(0.356 kg)(0.02 m)^2 = 0.00002848 kg·m^2.
Since the cylinder is rolling without slipping, the linear velocity v is related to the angular velocity ω by v = rω. Rearranging this equation, we have ω = v/r = 0.05 m/s / 0.02 m = 2.5 rad/s.
Plugging these values into the rotational kinetic energy formula, we have KE_rot = (1/2)(0.00002848 kg·m^2)(2.5 rad/s)^2 = 1.78125 x 10^-4 J.
To find the total kinetic energy, we simply add the translational and rotational kinetic energy components: KE_total = KE_trans + KE_rot = 0.000445 J + 1.78125 x 10^-4 J = 0.000623125 J.
Therefore, the total kinetic energy of the rolling solid cylinder is 0.000623125 J or 0.623125 mJ.
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What principle allows you to determine the relative ages of units D and M? Answer: 26. What principle allows you to match unit B on the right side of the diagram to the same unit on the left side of the diagram? 27. The erosional surface labeled N is a (an): Answer: 28. Draw arrows on the fault planes for fault L and fault M and label the hanging wall (HW) and footwall (FW) for both faults. 29. What name best describes fault L? 30. Which fault is the oldest? Answer: 32. What clast size(s) would you expect unit A to have? Answer: 31. What type of plate tectonic boundary would most likely be responsible for forming fault M? Answer: Answer: Answer 33. In which sedimentary environment would unit C most likely form? Answer: 34. If unit A were to experience a low grade regional metamorphic event, what metamorphic rock would form? 112 Geologic Time Leche Answer: 35. Which sedimentary rock on the diagram is most poorly sorted? Answer: 36. If all of the rocks in the diagram were deposited during the Cenozoic, what is the maximum age. that the rocks could be? Answer:
The principle of cross-cutting relationships allows you to determine the relative ages of units D and M. If one unit (D) is cut by another unit (M), then the unit doing the cutting (M) is younger.
What principle allows you to determine the relative ages of units D and M?The principle of cross-cutting relationships allows you to determine the relative ages of units D and M. If one unit (D) is cut by another unit (M), then the unit doing the cutting (M) is younger.
The principle of continuity allows you to match unit B on the right side of the diagram to the same unit on the left side of the diagram. It states that rock layers continue horizontally and can be correlated across a distance.
The erosional surface labeled N is an unconformity, specifically a disconformity, which represents a period of erosion or non-deposition between parallel layers of sedimentary rocks.
Fault L is a normal fault, where the hanging wall moves down relative to the footwall.
Fault M is the oldest fault because it cuts across both unit D and unit J.
Unit A would likely have fine clast size.
A divergent plate boundary would most likely be responsible for forming fault M, where two plates move away from each other.
Unit C would most likely form in a marine or deep-water sedimentary environment.
If unit A were to experience a low-grade regional metamorphic event, a slate would form.
Unit C, the conglomerate, is the most poorly sorted sedimentary rock.
The maximum age that the rocks could be is approximately 65 million years, as the Cenozoic era began approximately 65 million years ago and continues to the present.
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Two equally charged particles start 3.4 mm from each other at rest. When they are released they accelerate away from each other. The initial acceleration of particle A is 7 m/s2 and of particle B is 10 m/s2 .
Calculate the charge on either particle, if the mass of particle A is 5×10−7 kg
The charge on each particle is approximately ±6.41×10⁻⁷ C. Particle A has an initial acceleration of 7 m/s², while Particle B has an initial acceleration of 10 m/s².
To calculate the charge on either particle, we can use Newton's second law of motion and Coulomb's law.
First, let's consider particle A. We know its initial acceleration is 7 m/s² and its mass is 5×10⁻⁷ kg. Applying Newton's second law (F = ma), we can calculate the net force acting on particle A.
F = m × a
F = (5×10⁻⁷ kg) × (7 m/s²)
F = 3.5×10⁻⁶ N
Next, we can apply Coulomb's law to determine the force between the two particles.
Coulomb's law states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
F = k × (q₁ × q₂) / r²
Since the particles have equal charges, we can denote the charge on each particle as q.
F = k × q² / r²
Combining both equations, we have:
3.5×10⁻⁶ N = k × q² / (3.4×10⁻³ m)²
Now, we need the value of the Coulomb constant, k, which is approximately 8.99×10⁹ Nm²/C².
Simplifying the equation, we have:
3.5×10⁻⁶ N = (8.99×10⁹ Nm²/C²) × q² / (3.4×10⁻³ m)²
Solving for q², we get:
q² = (3.5×10⁻⁶ N) × (3.4×10⁻³ m)² / (8.99×10⁹ Nm²/C²)
Calculating the right side of the equation gives us:
q² ≈ 4.10×10⁻¹³ C²
Taking the square root of both sides, we find:
q ≈ ±6.41×10⁻⁷ C
Therefore, the charge on each particle is approximately ±6.41×10⁻⁷ C. The sign indicates the type of charge, with the positive sign representing a positive charge and the negative sign representing a negative charge.
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How much does it cost to operate a 800 watts rice cooker for 30
minutes daily for 30 days at a rate 40 cents/kWh ?
4.90
$5.0
$4.8
$6
When an electron is transferred to a ne
The cost to operate a 800-watt rice cooker for 30 minutes daily for 30 days at a rate of 40 cents/kWh is $4.80.
To calculate the cost, we first need to determine the energy consumption of the rice cooker. The power consumption of 800 watts for 30 minutes daily for 30 days can be calculated as follows:
Energy consumption = Power × Time
= (800 watts) × (0.5 hours/day) × (30 days)
= 12,000 watt-hours or 12 kWh
Next, we can calculate the cost by multiplying the energy consumption by the rate:
Cost = Energy consumption × Rate
= 12 kWh × $0.40/kWh
= $4.80
Therefore, the cost to operate the rice cooker for the given duration and rate is $4.80.
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