Answer:
Explanation:
velocity of proton v = 1.5 x 10³ i m /s
charge on proton e = 1.6 x 10⁻¹⁹ C
Let the magnetic field be B = Bx i + Bz k
force on charged particle ( proton )
F = e ( v x B )
2.06 x10⁻¹⁶ j = 1.6 x 10⁻¹⁹ [ 1.5 x 10³ i x ( Bx i + Bz k) ]
2.06 x10⁻¹⁶ j = - 1.6 x 10⁻¹⁹ x 1.5 x 10³ Bz j) ]
2.06 x10⁻¹⁶ = - 1.6 x 10⁻¹⁹ x 1.5 x 10³ Bz
Bz = - .8583
force on charged particle ( electron )
F = e ( v x B )
8.40 x10⁻¹⁶ j = -1.6 x 10⁻¹⁹ [ - 4.4 x 10³ k x ( Bx i + Bz k) ]
8.4 x10⁻¹⁶ j = 1.6 x 10⁻¹⁹ x 4.4 x 10³ Bx j ]
- 8.4 x10⁻¹⁶ = 1.6 x 10⁻¹⁹ x 4.4 x 10³ Bx
Bx = - 1.19
Magnetic field = - 1.19 i - .8583 k
magnitude = √ (1.19² + .8583²)
= 1.467 T
If it is making angle θ with x - axis in x -z plane
Tanθ = (.8583 / 1.19 )
36⁰ .
C )
v = - 3.7 x 10³j m /s
e = - 1.6 x 10⁻¹⁶ C
Force = F = e ( v x B )
= -1.6 x 10⁻¹⁹ [ -3.7 x 10³ j x ( Bx i + Bz k) ]
= - 1.6 x 10⁻¹⁹ x 3.7 x 10³ Bx k -1.6 x 10⁻¹⁹ x 3.7 x 10³Bzi ]
= 5.08 i - 7.04 k
Tanθ = 54 ° .
A physics student hurries through their lab, releasing the bob of a simple pendulum from a height, and allowing it to swing. He measures the period of the pendulum and calculates that g = 8 m/s2. What might the student have done wrong? Select all that apply. The length of the pendulum string was too long, so the equation for the period of a pendulum was no longer valid in this case. The string wasn't taut when he released the bob, causing the bob to move erratically. The student only timed one cycle, introducing a significant timing error. The angle of release was too large, so the equation for the period of a pendulum was no longer valid in this case. The mass of the bob was too large, so the equation for the period of a pendulum was no longer valid in this case. Instead of releasing the bob from rest, the student threw the bob downward.
Answer:
The string wasn't taut when he released the bob, causing the bob to move erratically.
The student only timed one cycle, introducing a significant timing error.
The angle of release was too large, so the equation for the period of a pendulum was no longer valid in this case.
Explanation:
The period of a simple pendulum is given by the formula
[tex]T=2\pi \sqrt{\frac{L}{g}}[/tex]
where
T is the period of oscillation
L is the length of the pendulum
g is the acceleration due to gravity
Therefore, it is possible to measure the value of [tex]g[/tex] in an experiment, by taking a pendulum, measuring its length L, and measuring its period of oscillation T. Re-arranging the equation above, we get the value of g as:
[tex]g=(\frac{2\pi}{T})^2 L[/tex]
Here the value of g measured in the experiment is [tex]8 m/s^2[/tex] instead of [tex]9.8 m/s^2[/tex]. Let's now analyze the different options:
The length of the pendulum string was too long, so the equation for the period of a pendulum was no longer valid in this case. --> FALSE. There is no constraint on the length of the pendulum.
The string wasn't taut when he released the bob, causing the bob to move erratically. --> TRUE. This is possible, as if the string is not taut, the pendulum would not start immediately its oscillation, so the period would be larger causing a smaller value measured for g.
The student only timed one cycle, introducing a significant timing error. --> TRUE. This is also impossible: in fact, we can get a more accurate measurement of the period if we measure several oscillations (let's say 10), and then we divide the total time by 10.
The angle of release was too large, so the equation for the period of a pendulum was no longer valid in this case. --> TRUE. The formula written above for the period of the pendulum is valid only for small angles.
The mass of the bob was too large, so the equation for the period of a pendulum was no longer valid in this case. --> FALSE. The equation that gives the period of the pendulum does not depend on the mass.
Instead of releasing the bob from rest, the student threw the bob downward. --> FALSE. In fact, this force would have been applied only at the very first moment, but then later the only force acting on the pendulum is the force of gravity, so the formula of the period would still be valid.
Answer:
The angle of release was too large, so the equation for the period of a pendulum was no longer valid in this case.
The string wasn't taut when he released the bob, causing the bob to move erratically.
The student only timed one cycle, introducing a significant timing error.
Instead of releasing the bob from rest, the student threw the bob downward.
Explanation:
The angle of release was too large, so the equation for the period of a pendulum was no longer valid in this case. your equation only works for angles > 20 degrees
The string wasn't taut when he released the bob, causing the bob to move erratically. this could lead to errors in either direction
The student only timed one cycle, introducing a significant timing error. unforseen events like wind could heavily influence one cycle.
Instead of releasing the bob from rest, the student threw the bob downward. this would artificially decrease your period (T) and therefor throw the other variables in your equation off. since L is a constant g would be forced to change.
State all four factors and explain how those factors affect the resistance of a wire.
Answer:
There are four factors affecting resistance which are Temperature,Length of wire,Area of the cross section of wire and nature of the material.When there is current in a conductive material,The free electrons move through the material and occasionally collide with atoms.
Explanation:
I HOPE ITS HELPFUL FOR YOUAnswer: its length, material, temperature, and cross section area which can also be considered as diameter.
Explanation:
1. A mass suspended from a spring oscillates vertically with amplitude of 15 cm. At what distance from the equilibrium position will the speed of the mass be 25% of its maximum speed?
Answer:
The value of the distance is [tex]\bf{14.52~cm}[/tex].
Explanation:
The velocity of a particle(v) executing SHM is
[tex]v = \omega \sqrt{A^{2} - x^{2}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`~(1)[/tex]
where, [tex]\omega[/tex] is the angular frequency, [tex]A[/tex] is the amplitude of the oscillation and [tex]x[/tex] is the displacement of the particle at any instant of time.
The velocity of the particle will be maximum when the particle will cross its equilibrium position, i.e., [tex]x = 0[/tex].
The maximum velocity([tex]\bf{v_{m}}[/tex]) is
[tex]v_{m} = \omega A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)[/tex]
Divide equation (1) by equation(2).
[tex]\dfrac{v}{v_{m}} = \dfrac{\sqrt{A^{2} - x^{2}}}{A}~~~~~~~~~~~~~~~~~~~~~~~~~~~(3)[/tex]
Given, [tex]v = 0.25 v_{m}[/tex] and [tex]A = 15~cm[/tex]. Substitute these values in equation (3).
[tex]&& \dfrac{1}{4} = \dfrac{\sqrt{15^{2} - x^{2}}}{15}\\&or,& A = 14.52~cm[/tex]
A research vessel is mapping the bottom of the ocean using sonar. It emits a short sound pulse called "ping" downward. The frequency of the sound is 5920 Hz. In water sound propagates at a speed of 1485 m/s. The sound pulse is then reflected back from the bottom of the ocean and it is detected by the vessel 5.63 s after it was emitted. How deep is the ocean just below the vessel? Submit Answer Tries 0/12 What is the wavelength of this sound wave?
Answer:
d = 4180.3m
wavelengt of sound is 0.251m
Explanation:
Given that
frequency of the sound is 5920 Hz
v=1485m/s
t=5.63s
let d represent distance from the vessel to the ocean bottom.
an echo travels a distance equivalent to 2d, that is to and fro after it reflects from the obstacle.
[tex]velocity=\frac{distance}{time}\\\\ v=\frac{2d}{t} \\\\vt=2d\\\\d=\frac{vt}{2}[/tex]
[tex]d=\frac{1485*5.63}{2}\\d= 4180.3m[/tex]
wavelengt of sound is [tex]\lambda[/tex] = v/f
= (1485)/(5920)
= 0.251 m
To what temperature (in °C) must a cylindrical rod of one metal 10.083 mm in diameter and a plate of second metal having a circular hole 9.987 mm in diameter have to be heated for the rod to just fit into the hole? Assume that the initial temperature is 27°C and that the linear expansion coefficient values for metals one and two are 4.0 x 10-6 (°C)-1 and 16 x 10-6 (°C)-1, respectively.
Answer:
834°C
Explanation:
By setting the equation of final diameter of one metal and another metal equal to one another in order to determine final temperature, the equation for final diameter is given by,
[tex]d_{f}[/tex] =[tex]d_{o[/tex](1 + α([tex]t_{f}[/tex]-[tex]t_{o[/tex]))
[tex]d_{f1[/tex]=[tex]d_{f2[/tex]
9.987(1 + 16 x [tex]10^{-6}[/tex]([tex]t_{f}[/tex]-27)) = 10.083( 1 + (4 x [tex]10^{-6}[/tex])([tex]t_{f}[/tex]-27))
9.987(1 + 16 x [tex]10^{-6}[/tex][tex]t_{f}[/tex] - 432 x [tex]10^{-6}[/tex]) = 10.083( 1 + 4 x [tex]10^{-6}[/tex][tex]t_{f}[/tex] - 108 x [tex]10^{-6}[/tex])
9.987 + 1.59 x [tex]10^{-4[/tex][tex]t_{f}[/tex] - 4.31 x [tex]10^{-3[/tex] = 10.083 + 4.033 x [tex]10^{-5[/tex] [tex]t_{f}[/tex] - 1.088 x [tex]10^{-3[/tex]
1.59 x [tex]10^{-4[/tex][tex]t_{f}[/tex] - 4.033 x [tex]10^{-5[/tex] [tex]t_{f}[/tex] = 10.083-1.088 x [tex]10^{-3[/tex]- 9.987+4.31 x [tex]10^{-3[/tex]
1.187 x [tex]10^{-4[/tex][tex]t_{f}[/tex]= 0.099
[tex]t_{f}[/tex] = 0.099/ 1.187 x [tex]10^{-4[/tex]
[tex]t_{f}[/tex] = 834°C
A hawk flew 600 meters in 60 seconds. A sparrow flew 400 meters in 30 seconds. Which bird flew faster? How fast did each bird fly?
Answer:
A sparrow flew faster. the sparrow flew 10 meters per second. The sparow flew 13.(3) meters per second
Explanation:
A sparrow flew faster than the hawk as it completes more distance in 60 seconds than that of hawk which is about 1200 meters. Speed is the distance travelled per unit time.
What is Speed?Speed is the measure of the distance travelled by an object per unit time taken. Speed is a vector quantity. It has both magnitude and direction.
Speed of an object can be calculated as: Distance travelled divided by time taken.
Speed of hawk is 600 meters/ 60 seconds
Speed of hawk = 10 m/s
Speed of sparrow is 400 meters/ 30 seconds
Speed of Sparrow = 13.33 m/s
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Assuming that the tungsten filament of a lightbulb is a blackbody, determine its peak wavelength if its temperature is 3 200 K.
Answer:
the peak wavelength when the temperature is 3200 K = [tex]9.05625*10^{-7} \ m[/tex]
Explanation:
Given that:
the temperature = 3200 K
By applying Wien's displacement law ,we have
[tex]\lambda _m[/tex]T = 0.2898×10⁻² m.K
The peak wavelength of the emitted radiation at this temperature is given by
[tex]\lambda _m[/tex] = [tex]\frac{0.2898*10^{-2} m.K}{3200 K}[/tex]
[tex]\lambda _m[/tex]= [tex]9.05625*10^{-7} \ m[/tex]
Hence, the peak wavelength when the temperature is 3200 K = [tex]9.05625*10^{-7} \ m[/tex]
A straight fin fabricated from 2024 Aluminum alloy (k=185 W/mK) has a base thickness of t=3 mm and a length of L=15 mm. Its base temperature is Tb=100oC, and it is exposed to a fluid for which T[infinity] =20oC and h=50 W/m2K. For the foregoing conditions and a fin of unit width, compare the fin heat rate, efficiency, and volume for rectangular, triangular, and parabolic profiles.
The heat rate, efficiency, and volume of a straight fin made from 2024 Aluminum alloy can be calculated using relevant formulas considering its physical dimensions, the transferred heat, and the profile of the fin. Comparisons across different profiles (rectangular, triangular, parabolic) are commonly done using numerical or graphical solution methods.
Explanation:To compare the heat rate, efficiency, and volume for a straight fin made of 2024 Aluminum alloy with a rectangular, triangular, and parabolic profile, we first need to convert all known variables into SI units. The fin has a base thickness (t) of 3 mm or 0.003 m, a length (L) of 15 mm or 0.015 m, and it's exposed to a fluid at a temperature (T infinity) of 20C. The base temperature (Tb) of the fin is 100oC, and the heat transfer coefficient (h) is 50 W/m2K.
Heat Transfer CalculationWe can estimate the heat transfer rate (Q) by applying the formula Q = hA(Tb - T infinity), where A represents the surface area of the fin which would depend on the fin's profile. For a unit width, the area of a rectangular fin is A = wt = unit width*L, for a triangular profile A = 0.5*wt, and for a parabolic profile A = (2/3)*wt.
Fin EfficiencyFin efficiency can be calculated by dividing the actual heat transferred by the fin by the maximum possible heat transfer. Since these are dependent on the fin's profile (shape), numerical or graphical solution methods are commonly used for calculations.
Fin VolumeVolume can be calculated as the product of the surface area and thickness, which again would depend on the fin's profile.
Understanding these differences among fin profiles is important in heat transfer management and finding ways to increase fin efficiency.
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Compare the fin heat rate using equation: Qfin = √(hPkA) (Tb - T∞), efficiency using equation: η = Qfin / (hAt (Tb - T∞)) , and volumes are found to be bLt, (1/2)bLt and (2/3)bL for rectangular, triangular, and parabolic profiles, respectively.
1.) Heat Transfer Analysis of a Straight Fin
To compare the fin heat rate, efficiency, and volume for different fin profiles (rectangular, triangular, and parabolic), we need to perform the following calculations:
Step 1: Identify Knowns and Convert to SI Units
Given:
Thermal conductivity, k = 185 W/mKBase thickness, t = 3 mm = 0.003 mLength, L = 15 mm = 0.015 mBase temperature, Tb = 100°CAmbient temperature, T∞ = 20°CHeat transfer coefficient, h = 50 W/m²KStep 2: Determine the Fin Parameter, m
The fin parameter, m, is calculated using the formula:
m = sqrt(hP / kA)For a rectangular fin:
P = 2(bt + L) and A = btWe adjust the perimeter and area for parabolic and triangular profiles likewise.
Step 3: Heat Transfer Rate
For each fin profile, the heat transfer rate, Qfin, is given by:
Qfin = √(hPkA) (Tb - T∞)where A is the cross-sectional area, and P is the perimeter of the respective fin.
Step 4: Fin Efficiency
Fin efficiency, η, is determined as:
η = Qfin / (hAt (Tb - T∞))where At is the total surface area of the fin.
Step 5: Compare Fin Volume
The volume, V, can be found by:
Rectangular: V = bLtTriangular: V = (1/2)bLtParabolic: V = (2/3)bLAfter calculating these values, you can compare the heat rate, efficiency, and volume between the fin profiles to determine the most efficient design.
A specimen of aluminum having a rectangular cross section 9.5 mm × 12.9 mm (0.3740 in. × 0.5079 in.) is pulled in tension with 35000 N (7868 lbf) force, producing only elastic deformation. The elastic modulus for aluminum is 69 GPa (or 10 × 106 psi). Calculate the resulting strain.
Answer:
The resultant strain in the aluminum specimen is [tex]4.14 \times {10^{ - 3}}[/tex]
Explanation:
Given that,
Dimension of specimen of aluminium, 9.5 mm × 12.9 mm
Area of cross section of aluminium specimen,
[tex]A=9.5\times 12.9=123.84\times 10^{-6}\ mm^2A=122.55\times 10^{-6}\ m^2[/tex]
Tension acting on object, T = 35000 N
The elastic modulus for aluminum is,[tex]E=69\ GPa=69\times 10^9\ Pa[/tex]
The stress acting on material is proportional to the strain. Its formula is given by :
[tex]\epsilon=\dfrac{\sigma}{E}[/tex]
[tex]\sigma[/tex] is the stress
[tex]\epsilon=\dfrac{F}{EA}\\\epsilon=\dfrac{35000}{69\times 10^9\times 122.5\times 10^{-6}}\\\epsilon=4.14\times 10^{-3}[/tex]
Thus, The resultant strain in the aluminum specimen is [tex]4.14 \times {10^{ - 3}}[/tex]
Answer:
Strain = 4.139 x 10^(-3)
Explanation:
We are given;
Dimension; 9.5 mm by 12.9 mm = 0.0095m by 0.0129m
Elastic Modulus; E = 69GPa = 69 x 10^(9)N/m²
Force = 35,000N
Now, Elastic modulus is given by;
E = σ/ε
Where
E is elastic modulus
σ is stress
ε is strain
Now, stress is given by the formula;
σ = F/A
Area = 0.0095m x 0.0129m = 0.00012255 m²
Thus, σ = 35000/0.00012255 = 285597715.218 N/m²
Now, we are looking for strain.
Let's make ε the subject;
E = σ/ε, Thus, ε = σ/E = 285597715.218/69 x 10^(9) = 0.00413909732 = 4.139 x 10^(-3)
A student has 67-cm-long arms. What is the minimum angular velocity (in rpm) for swinging a bucket of water in a vertical circle without spilling any? The distance from the handle to the bottom of the bucket is 35 cm .
Answer:
29.61 rpm.
Explanation:
Given,
student arm length, l = 67 cm
distance of the bucket, r = 35 m
Minimum angular speed of the bucket so, the water not fall can be calculated by equating centrifugal force with weight.
Now,
[tex]mg = m r \omega^2[/tex]
[tex]\omega = \sqrt{\dfrac{g}{R}}[/tex]
R = 67 + 35 = 102 cm = 1.02 m
[tex]\omega = \sqrt{\dfrac{9.81}{1.02}}[/tex]
[tex]\omega = 3.101\ rad/s[/tex]
[tex]\omega = \dfrac{3.101}{2\pi} = 0.494\ rev/s[/tex]
[tex]\omega = 0.494 \times 60 = 29.61\ rpm[/tex]
minimum angular velocity is equal to 29.61 rpm.
Answer:
29.6 rpm
Explanation:
length of arm = 67 cm
distance of handle to the bottom = 35 cm
radius of rotation, R = 67 + 35 = 102 cm = 1.02 m
The centripetal force acting on the bucket is balanced by the weight of the bucket.
mRω² = mg
R x ω² = g
[tex]\omega = \sqrt\frac{g}{R}[/tex]
[tex]\omega = \sqrt\frac{9.8}{1.02}[/tex]
ω = 3.1 rad/s
Let f is the frequency in rps
ω = 2 x 3.14 x f
3.1 = 2 x 3.14 xf
f = 0.495 rps
f = 29.6 rpm
1.- An elevator is being lowered at a constant speed by a steel cable attached to an electric motor. Which statement is correct? A. The cable does positive work on the elevator, and the elevator does positive work on the cable. B. The cable does positive work on the elevator, and the elevator does negative work on the cable. C. The cable does negative work on the elevator, and the elevator does positive work on the cable. D. The cable does negative work on the elevator, and the elevator does negative work on the cable
Answer:
the correct one is C
Explanation:
For this exercise we must use the work definition
W = F. s
Where the bold characters indicate vectors and the point is the scalar producer
W = F s cos θ
Where θ is the angles between force and displacement.
Let us support this in our case. The cable creates an upward tension and with the elevator going down the angle between them is 180º so the work of the cable on the elevator is negative.
The evade has a downward force, its weight so the force goes down and the displacement goes down, as both are in the same direction the work is positive
When examining the statements the correct one is C
Final answer:
The correct statement for an elevator being lowered at constant speed is that the steel cable does negative work on the elevator, and the elevator does negative work on the cable, illustrating the principle that work can be negative if force and displacement are in opposite directions.
Explanation:
The question pertains to the work done by an elevator cable while lowering an elevator at constant speed. According to the principles of work and energy in physics, work done is defined as the force applied in the direction of motion times the distance moved. If an elevator is being lowered at a constant speed, the steel cable exerts an upward force to counteract gravity but the elevator moves downward. Therefore, the displacement of the elevator is in the opposite direction to the force exerted by the cable, resulting in the cable doing negative work on the elevator. Conversely, because the elevator is moving downwards (in the direction opposite to the force exerted by the cable), we can interpret this as the elevator doing negative work on the cable as well, due to the concept that positive work adds energy to a system while negative work removes it.
Thus, the correct statement is: D. The cable does negative work on the elevator, and the elevator does negative work on the cable. This illustrates the application of the definition of work in physics, particularly in scenarios involving opposite directions of force and motion.
Astronaut Jennifer's lifeline to her spaceship comes loose and she finds herself stranded, "floating" 100 m from the mothership. She suddenly throws her 2.00-kg wrench at 20 m/s in a direction away from the ship. If she and her spacesuit have a combined mass of 200 kg, how long does it take her to coast back to her spaceship
Answer:
The time it will take the astronaut to coast back to her ship is 500 seconds
Explanation:
Mass of wrench, m₁ = 2 kg
Mass of astronaut with spacesuit, m₂ = 200 kg
Velocity of wrench, v₁ = 20 m/s
Velocity of astronaut with spacesuit v₂
Distance between the astronaut and the ship is 100 m
As linear momentum is conserved
[tex]m_1v_1+m_2v_2=0\\\\\Rightarrow v_2=\frac{m_1v_1}{m_2}\\\\\Rightarrow v_2=\frac{2\times 20}{200}\\\\\Rightarrow v_2=0.2\ m/s[/tex]
[tex]Time = \frac{ Distance}{Speed}[/tex]
[tex]Time=\frac{100}{0.2}=500\ s[/tex]
Hence, The time it will take the astronaut to coast back to her ship is 500 seconds
By conserving momentum, we find that Astronaut Jennifer will move at a velocity of 0.2 m/s after throwing the wrench. It will take her 500 seconds to travel the 100 meters back to her spaceship.
Explanation:To determine the time it takes for Astronaut Jennifer to return to her spaceship, we will use the conservation of momentum principle. Since there are no external forces acting on the system in space, the momentum before and after Jennifer throws the wrench is conserved.
Let's calculate the velocity of Jennifer after she throws the wrench using the following equation:
Momentum before = Momentum after
(mass of Jennifer) × (initial velocity of Jennifer) + (mass of wrench) × (initial velocity of wrench) = (mass of Jennifer) × (final velocity of Jennifer) + (mass of wrench) × (final velocity of wrench)
Since Jennifer and the wrench start at rest, their initial velocities are 0, so:
0 + 0 = 200 kg × (final velocity of Jennifer) + 2.00 kg × (-20 m/s)
Therefore, the final velocity of Jennifer v_j is:
200 kg × v_j = -2.00 kg × (-20 m/s)
v_j = (2.00 kg × 20 m/s) / 200 kg
v_j = 0.2 m/s
Now, we need to find out how long it takes her to cover 100 meters at this velocity:
Time = Distance / Velocity
Time = 100 m / 0.2 m/s
Time = 500 s
It would take Jennifer 500 seconds to coast back to her spaceship.
The roads are icy, and you observe a head-on collision on Summit, at the corner with Rhodes: a 1ton car swerves out of his lane and slides through a stop sign at 34mph straight into a 3ton SUV traveling at 13mph in the other direction. The car and the SUV crumple from the collision, and stick together. What is the final velocity, in MPH (you don't need to enter MPH in your answer) , of the SUV/car entanglement (the positive direction is the direction the car was initially going)?
Answer:
v = -1.3 mph
Explanation:
Assuming no external forces acting during the collision, total momentum must be conserved, so the following condition must be met:[tex]p_{init} = p_{final} (1)[/tex]
The initial momentum is the sum of the momenta of both vehicles, taking into account their relative velocities as they are going in opposite directions. If we take as positive the direction the car was initially going, we can write the following expression:[tex]p_{init} = m_{car} * v_{car} - m_{SUV} * v_{SUV}[/tex]
Replacing by the values of the masses of both vehicles and their speeds, we have:[tex]p_{init} = 1t*34 mph - 3t*13 mph = - 5t*mph (2)[/tex]
This must be equal to the final momentum of the car/SUV entanglement, as follows:[tex]p_{final} =( m_{car} + m_{SUV} )* v_{final} (3)[/tex]
Replacing in (3) for the masses, and equating (1) and (3), we can solve for vfinal, as follows:[tex]v_{final} = \frac{p_{init}}{(m_{car} + m_{SUV} } = \frac{-5t*mph}{4t} = -1.3 mph[/tex]
This means that the car/SUV entanglement will move in the opposite direction that the car was initially going.A 63.0 \mu F capacitor is connected to a generator operating at a low frequency. The rms voltage of the generator is 4.00 V and is constant. A fuse in series with the capacitor has negligible resistance and will burn out when the rms current reaches 15.0 A. As the generator frequency is increased, at what frequency will the fuse burn out
Answer:
[tex]f=9.5\ KHz[/tex]
Explanation:
AC Circuit
When connected to an AC circuit, the capacitor acts as an impedance of module
[tex]\displaystyle Z=\frac{1}{wC}[/tex]
Where w is the angular frequency of the power source and C is the capacitance.
If the capacitor is the only element connected to a circuit, then the Ohm's law establishes that
[tex]V=Z.I[/tex]
Where V and I are the rms voltage and current respectively. Replacing the value of Z, we have
[tex]\displaystyle V=\frac{I}{wC}[/tex]
Solving for w
[tex]\displaystyle w=\frac{I}{VC}[/tex]
The question provides us the following values
[tex]C=63\ \mu F=63\cdot 10^{-6}\ F[/tex]
[tex]V=4\ Volt[/tex]
[tex]I=15\ A[/tex]
Plugging in the values
[tex]\displaystyle w=\frac{15}{4\cdot 63\cdot 10^{-6}}[/tex]
[tex]w=59523.81\ rad/s[/tex]
Since
[tex]w=2\pi f[/tex]
Then
[tex]\displaystyle f=\frac{59523.81}{2\pi}=9473.51\ Hz[/tex]
[tex]f=9.5\ KHz[/tex]
Final answer:
The fuse in the AC circuit with a 63.0 µF capacitor will burn out when the generator frequency is increased to approximately 1,001 Hz.
Explanation:
To find at what frequency the fuse will burn out in the circuit with a 63.0 µF capacitor and a 4.00 V rms voltage generator, we will need to calculate the capacitive reactance (XC) and then use the relationship between current, voltage, and reactance for an AC circuit.
The capacitive reactance is given by the formula XC = 1 / (2πfC), where f is the frequency in hertz (Hz), and C is the capacitance in farads. The rms current I in the circuit is given by the rms voltage V divided by XC:
I = V / XC
Setting I to the maximum allowable current of 15.0 A, we have:
15.0 A = 4.00 V / (1 / (2πf × 63.0 × 10⁻⁶ F))
Solving for f, we get:
f = 1 / (2π × 63.0 × 10⁻⁶ F × (4.00 V / 15.0 A))
f ≈ 1,001 Hz
Therefore, the fuse will burn out when the generator frequency is increased to approximately 1,001 Hz.
A 11.0 kg satellite has a circular orbit with a period of 1.80 h and a radius of 7.50 × 106 m around a planet of unknown mass. If the magnitude of the gravitational acceleration on the surface of the planet is 20.0 m/s2, what is the radius of the planet?
Answer:
Explanation:
Expression for times period of a satellite can be given as follows
Time period T = 1.8 x 60 x 60
= 6480
T² = [tex]\frac{4\times \pi^2\times r^3}{GM}[/tex] where T is time period , r is radius of orbit , G is gravitational constant and M is mass of the satellite.
6480² = 4 x 3.14² x 7.5³ x 10¹⁸ / GM
GM = 4 x 3.14² x 7.5³ x 10¹⁸ / 6480²
= 3.96 X 10¹⁴
Expression for acceleration due to gravity
g = GM / R² where R is radius of satellite
20 = 3.96 X 10¹⁴ / R²
R² = 3.96 X 10¹⁴ / 20
= 1.98 x 10¹³ m
R= 4.45 x 10⁶ m
A thin uniform film of refractive index 1.750 is placed on a sheet of glass with a refractive index 1.50. At room temperature ( 18.8 ∘C), this film is just thick enough for light with a wavelength 580.9 nm reflected off the top of the film to be canceled by light reflected from the top of the glass. After the glass is placed in an oven and slowly heated to 170 ∘C, you find that the film cancels reflected light with a wavelength 588.2 nm .
Complete Question
A thin uniform film of refractive index 1.750 is placed on a sheet of glass with a refractive index 1.50. At room temperature ( 18.8 ∘C), this film is just thick enough for light with a wavelength 580.9 nm reflected off the top of the film to be canceled by light reflected from the top of the glass. After the glass is placed in an oven and slowly heated to 170 ∘C, you find that the film cancels reflected light with a wavelength 588.2 nm .
What is the coefficient of linear expansion of the film? (Ignore any changes in the refractive index of the film due to the temperature change.) Express your answer using two significant figures.
Answer:
the coefficient of linear expansion of the film is [tex]\alpha = 7.93 *10^{-5} / ^oC[/tex]
Explanation:
From the question we are told that
The refractive index of the film is [tex]n_f = 1.750[/tex]
The refractive index of the glass is [tex]n_g = 1. 50[/tex]
The wavelength of light reflected at 18°C is [tex]\lambda _r = 580.9nm = 580.9*10^{-9}m[/tex]
The wavelength of light reflected at 170°C is [tex]\lambda_h = 588.2 nm = 588.2 * 10^{-9}m[/tex]
For destructive interference the condition is
[tex]2t = \frac{m \lambda }{n_f}[/tex]
Where m is the order of interference
t is the thickness
For the smallest thickness is when m= 1 and this is represented as
[tex]t = \frac{\lambda }{2n_f }[/tex]
At 18°C the thickness would be
[tex]t_{r} = \frac{580.9 *10^{-9}}{2 * 1.750}[/tex]
[tex]t_{r} = 166nm[/tex]\
At 170° the thickness is
[tex]t_h = \frac{588.2 *10^{-9}}{2 * 1.750}[/tex]
[tex]t_h = 168 nm[/tex]
The coefficient of linear expansion f the film is mathematically represented as
[tex]\alpha = \frac{t_h - t_r}{t_r \Delta T}[/tex]
Substituting value
[tex]\alpha = \frac{168 *10^{-9} - 166 *10^{-9} }{166*10^{-9} * (170 -18)}[/tex]
[tex]\alpha = 7.93 *10^{-5} / ^oC[/tex]
Answer:
α=8.37 x [tex]10^{-5}[/tex] °[tex]C^{-1}[/tex]
Explanation:
Complete question : What is the coefficient of linear expansion of the film?
SOLUTION:
There is a net ( λ/2 ) phase change due to reflection for this film, therefore, destructive interference is given by
2t = m( λ/n) where n=1.750
for smallest non-zero thickness
t= λ/2n
At 18.8°C, [tex]t_{o[/tex]=580.9 x [tex]10^{-9}[/tex]/(2 x 1.750)
[tex]t_{o[/tex]= 165.9nm
At 170°C, t= 588.2x [tex]10^{-9}[/tex]/(2x1.750)
t=168nm
t=[tex]t_{o[/tex](1 + αΔT)
=>α= (t-[tex]t_{o[/tex])/ ([tex]t_{o[/tex]ΔT) [ΔT= 170-18.8 =151.2°C]
α= (168 x [tex]10^{-9}[/tex] - 165.9 x [tex]10^{-9}[/tex])/ (165.9 x [tex]10^{-9}[/tex] x 151.2)
α= 2.1 x [tex]10^{-9}[/tex]/ 2.508 x [tex]10^{-5}[/tex]
α=8.37 x [tex]10^{-5}[/tex] °[tex]C^{-1}[/tex]
Therefore, the coefficient of linear expansion of the film is 8.37 x [tex]10^{-5}[/tex] °[tex]C^{-1}[/tex]
Evaluate the solution:
A friend proposes a solution for the folowing problem.
A violin A string is 0.33 m long and has mass 0.30 x 10^-3 kg. It vibrates a fundamental frequency of 440 Hz (concert A). What is the tension in the string?
Proposed solution: Speed depends on the tension and string mass (v = [T/m]^1/2). Thus:
T = v^2 m = (340 m/s)^2 (0.30g) = 34.680 N.
a. Evaluate the solution and identify any errors.
b. Provide a corrected solution if you find errors.
Answer:
Tension, T = 105.09 N
Explanation:
Given that,
Length of the string, l = 0.33 m
Mass of the string, [tex]m=0.3\times 10^{-3}\ kg[/tex]
Fundamental frequency, f = 440 Hz
The expression for the speed in terms of tension is given by :
[tex]v=\sqrt{\dfrac{T}{(m/l)}}[/tex]
[tex]v^2=\dfrac{Tl}{m}\\\\T=\dfrac{v^2m}{l}\\\\T=\dfrac{(340)^2\times 0.3\times 10^{-3}}{0.33}\\\\T=105.09\ N[/tex]
So, the tension in the string is 105.09 N.
If the opening to the harbor acts just like a single-slit which diffracts the ocean waves entering it, what is the largest angle, in degrees relative to the incident direction, that a boat in the harbor would be protected from the wave action
Answer:
The angle that the wave would be [tex]\theta = sin ^{-1}\frac{2 \lambda}{D}[/tex]
Explanation:
From the question we are told that the opening to the harbor acts just like a single-slit so a boat in the harbor that at angle equal to the second diffraction minimum would be safe and the on at angle greater than the diffraction first minimum would be slightly affected
The minimum is as a result of destructive interference
And for single-slit this is mathematically represented as
[tex]D sin \ \theta =m \lambda[/tex]
where D is the slit with
[tex]\theta[/tex] is the angle relative to the original direction of the wave
m is the order of the minimum j
[tex]\lambda[/tex] is the wavelength
Now since in the question we are told to obtain the largest angle at which the boat would be safe
And the both is safe at the angle equal to the second minimum then
The the angle is evaluated as
[tex]\theta = sin ^{-1}[\frac{m\lambda}{D} ][/tex]
Since for second minimum m= 2
The equation becomes
[tex]\theta = \frac{2 \lambda}{D}[/tex]
A parallel-plate capacitor is charged until it carries charge + q +q on one plate and charge − q −q on the other plate. The capacitor is then disconnected from the power supply and isolated. What is the direction of the magnetic field that surrounds the charged capacitor? The magnetic field is directed counterclockwise as viewed from the plate with positive charge + q +q toward the plate with negative charge − q −q. There is no magnetic field. The magnetic field is directed counterclockwise as viewed from the plate with negative charge − q −q toward the plate with positive charge + q +q. The magnetic field is directed toward the plate with positive charge + q +q inside the capacitor and toward the plate with negative charge − q −q outside of the capacitor. The magnetic field is directed toward the plate with negative charge − q −q inside the capacitor and toward the plate with positive charge + q +q outside of the capacitor.
Answer:
There is no magnetic field.
Explanation:
Since th capacitor is charged and isolated, magnetic field doesn't exist.
For magnetic field to exist, there must be flow of charge.
A salad spinner is used to dry lettuce that has been recently washed. For the salad spinners to operate properly, the lettuce should be evenly distributed in the basket of the salad spinner. Otherwise, after you spin the basket and remove your hand, the salad spinner will wobble and slip on the kitchen counter.
You're curious how unbalanced your salad spinner can be before it starts to slip on the counter.
-You measure the dimensions of your salad spinner and determine that the basket has a diameter of 32.2 cm and a height of 21.0 cm.
-You place the salad spinner on a kitchen scale and determine that the mass of the salad spinner is 449 g.
-The bottom of the salad spinner is rubberized. Consulting friction coefficient tables, you estimate that the coefficient of friction between the bottom of the salad spinner and your kitchen counter is 0.83.
-You search online for information about the rotational speed of your spinner. You come across a journal article about students at Rice University who modified a salad spinner to create an inexpensive medical centrifuge to diagnose anemia in remote or low-resource areas. The article states that the maximum rotational speed of their salad spinner is 600 rpm. Your salad spinner is similar to the one featured in the article.
Suppose you place some lettuce into one side of the basket, pump the salad spinner until the basket is spinning at top speed and then remove your hand.
What is the maximum amount of lettuce that can be in the basket such that the salad spinner does not slip when you remove your hand?
The maximum amount of lettuce that dimensions can be in the basket such that the salad spinner does not slip when you remove your hand is 62.58 kg.
What is dimensions?Dimensions are the physical measurement of an object in terms of length, width, and height. They are used in various fields such as mathematics, engineering, and architecture. Dimensions are also used to describe the shape of an object, like a cube or a sphere.
The maximum amount of lettuce that can be in the basket is determined, Where Ff is the maximum static friction force (3470.54 N) and r is the radius of the salad spinner's basket (16.1 cm).
τ = 3470.54 x 0.161 = 559.21 Nm
To calculate the maximum amount of lettuce that can be in the basket, we need to calculate the rotational inertia (I) of the salad spinner
I = mr2
Where m is the mass of the salad spinner (449 g) and r is the radius of the salad spinner's basket (16.1 cm).
I = 449 x 0.1612 = 7.51 kg m2
The rotational inertia is the measure of how difficult it is to change the angular velocity of a spinning object.
α = τ / I
Where τ is the maximum rotational torque (559.21 Nm) and I is the rotational inertia (7.51 kg m2).
α = 559.21 / 7.51 = 74.37 rad/s2
The maximum amount of lettuce that can be in the basket without causing the spinner to slip is determined by the equation m = I/t, where m is the maximum mass of lettuce and I is the rotational inertia of the spinner (7.51 kg m2).
m = I/t
Where I is the rotational inertia of the spinner (7.51 kg m2) and t is the time it takes for the spinner to reach its final angular velocity (0.12 s).
m = 7.51 / 0.12 = 62.58 kg
Therefore, the maximum amount of lettuce that can be in the basket such that the salad spinner does not slip when you remove your hand is 62.58 kg.
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You leave the doctor's office after your annual checkup and recall that you weighed 688 N in her office. You then get into an elevator that, conveniently, has a scale. Find the magnitude and direction of the elevator's acceleration if the scale readsQ: Find the magnitude of the elevator's acceleration if the scale reads 726 NQ: Find the direction of the elevator's acceleration if the scale reads 726 NQ: Find the magnitude of the elevator's acceleration if the scale reads 598 NQ: Find the direction of the elevator's acceleration if the scale reads 598 N
Answer:
Explanation:
weight of the person = 688 N
a) reading while in the elevator using the scale was 726 N
since when the elevator is going upward the floor of the elevator and the the scale pushes against the person leading to the person experiences a normal force greater than the weight as a result of the acceleration of the elevator and also
F = ma and W, weight = mg
mass of the body = weight / g = 688 / 9.8 = 70.20 kg
net force on the body = force of normal - weight of the body = 726 - 688 = 38 N
ma = 38 N
70.20 kg × a ( acceleration) = 38 N
a = 38 / 70.20 = 0.54 m/s² and the elevator is moving upward
b) net force, ma = force of normal - weight of the body = 598 - 688 = -90 N
a = -90 / 70.20 = -1.282 m/s² and the elevator is coming downward
In order to determine the mass moment of inertia of a flywheel of radius 600 mm, a 12-kg block is attached to a wire that is wrapped around the flywheel. The block is released and is observed to fall 3 m in 4.6 s. To eliminate bearing friction from the computation, a second block of mass 24 kg is used and is observed to fall 3 m in 3.1 s. Assuming that the moment of the couple due to friction remains constant, determine the mass moment of inertia of the flywheel.
Answer:
Explanation:
Given that,
When Mass of block is 12kg
M = 12kg
Block falls 3m in 4.6 seconds
When the mass of block is 24kg
M = 24kg
Block falls 3m in 3.1 seconds
The radius of the wheel is 600mm
R = 600mm = 0.6m
We want to find the moment of inertia of the flywheel
Taking moment about point G.
Then,
Clockwise moment = Anticlockwise moment
ΣM_G = Σ(M_G)_eff
M•g•R - Mf = I•α + M•a•R
Relationship between angular acceleration and linear acceleration
a = αR
α = a / R
M•g•R - Mf = I•a / R + M•a•R
Case 1, when y = 3 t = 4.6s
M = 12kg
Using equation of motion
y = ut + ½at², where u = 0m/s
3 = ½a × 4.6²
3 × 2 = 4.6²a
a = 6 / 4.6²
a = 0.284 m/s²
M•g•R - Mf = I•a / R + M•a•R
12 × 9.81 × 0.6 - Mf = I × 0.284/0.6 + 12 × 0.284 × 0.6
70.632 - Mf = 0.4726•I + 2.0448
Re arrange
0.4726•I + Mf = 70.632-2.0448
0.4726•I + Mf = 68.5832 equation 1
Second case
Case 2, when y = 3 t = 3.1s
M= 24kg
Using equation of motion
y = ut + ½at², where u = 0m/s
3 = ½a × 3.1²
3 × 2 = 3.1²a
a = 6 / 3.1²
a = 0.6243 m/s²
M•g•R - Mf = I•a / R + M•a•R
24 × 9.81 × 0.6 - Mf = I × 0.6243/0.6 + 24 × 0.6243 × 0.6
141.264 - Mf = 1.0406•I + 8.99
Re arrange
1.0406•I + Mf = 141.264 - 8.99
1.0406•I + Mf = 132.274 equation 2
Solving equation 1 and 2 simultaneously
Subtract equation 1 from 2,
Then, we have
1.0406•I - 0.4726•I = 132.274 - 68.5832
0.568•I = 63.6908
I = 63.6908 / 0.568
I = 112.13 kgm²
maria was riding her bike at a velocity of 3 m/s to the north. Her velocity changed to 11 m/s to the north. What was her change in velocity
Maria's change in velocity while riding her bike is 8 meters per second to the north.
Since the motion is along the same direction (to the north), we do not need to consider the direction as negative or positive. Here's the calculation:
Change in velocity = Final velocity - Initial velocity
= 11 m/s - 3 m/s
= 8 m/s.
So, Maria's change in velocity is 8 meters per second to the north.
Moving from boron to carbon, the intensity of the bulb Moving from boron to carbon, the intensity of the bulb blank because Z increases from blank to blank. The thickness of the frosting blank because the core electron configuration is the same for both atoms. because Z increases from Moving from boron to carbon, the intensity of the bulb blank because Z increases from blank to blank. The thickness of the frosting blank because the core electron configuration is the same for both atoms. to Moving from boron to carbon, the intensity of the bulb blank because Z increases from blank to blank. The thickness of the frosting blank because the core electron configuration is the same for both atoms.. The thickness of the frosting Moving from boron to carbon, the intensity of the bulb blank because Z increases from blank to blank. The thickness of the frosting blank because the core electron configuration is the same for both atoms. because the core electron configuration is the same for both atoms.
Complete Question
The complete Question is shown on the first and second uploaded image
Answer:
The underlined words are the answers
Part A
Moving from boron to carbon, the intensity of the bulb Increases because Z increases from 5 to 6 , The thickness of the frosting stays the same because the core electron configuration is the same for both atoms
Part B
Moving from boron to aluminum the intensity of the bulb Increases because Z increases from 5 to 13 . The thickness of the frosting also increases because Al has the core configuration of Ne, while B has the core configuration of He
Explanation:
Here Z denotes the atomic number
Ne denoted the element called Neon and its electronic configuration is
[tex]1s^2 \ 2s^2 \ 2p^6[/tex]
He denoted the element called Helium and its electronic configuration is
[tex]1s^2[/tex]
B denoted the element called Boron and its electronic configuration is
[tex]1s^2 \ 2s^2\ 2p^1[/tex]
Looking at its electronic configuration we can see that the core is He
I,e [tex][He]\ 2s^2 2p^1[/tex]
Al denoted the element called Aluminium and its electronic configuration is
[tex]1s^2 \ 2s^2 \ 2p^2 \ 3s^2 \ 3p^1[/tex]
Looking at its electronic configuration we can see that the core is Ne
I,e [tex][Ne]\ 3s^2 \ 3p^1[/tex]
An AC generator has an output rms voltage of 100.0 V at a frequency of 42.0 Hz. If the generator is connected across a 45.0-mH inductor, find the following. (a) inductive reactance Ω (b) rms current A (c) maximum current in the circuit
Answer:
(a) 11.8692 ohm
(b) 12.447 A
(c) 17.6 A
Explanation:
a) inductive reactance Z = L Ω
= L x 2π x F
= 45.0 x 10⁻³ x 2(3.14) x 42
= 11.8692 ohm
b) rms current
= 100 / 8.034
= 12.447 A
c) maximum current in the circuit
= I eff x rac2
= 12.447 x 1.414
= 17.6 A
Suppose that the coefficient of kinetic friction between Zak's feet and the floor, while wearing socks, is 0.250. Knowing this, Zak decides to get a running start and then slide across the floor.
a) If Zak's speed is 3.00 when he starts to slide, what distance will he slide before stopping? d=1.84
b) Now, suppose that Zak's younger cousin, Greta, sees him sliding and takes off her shoes so that she can slide as well (assume her socks have the same coefficient of kinetic friction as Zak's). Instead of getting a running start, she asks Zak to give her a push. So, Zak pushes her with a force of 125 over a distance of 1.00 . If her mass is 20.0 , what distance does she slide after Zak's push ends? Remember that the frictional force acts on Greta during Zak's push and while she is sliding after the push.
a) 1.84 m
b) 1.55 m
Explanation:
a)
In this problem, the only force acting on Zak along the direction of motion (horizontal direction) is the force of friction, which is
[tex]F_f=-\mu mg[/tex]
where
[tex]\mu=0.250[/tex] is the coefficient of friction
m is Zak's mass
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
According to Newton's second law of motion, the net force acting on Zak is equal to the product between its mass (m) and its acceleration (a), so we have
[tex]F=ma[/tex]
Here the only force acting is the force of friction, so this is also the net force:
[tex]-\mu mg = ma[/tex]
Therefore we can find Zak's acceleration:
[tex]a=-\mu g=-(0.250)(9.8)=-2.45 m/s^2[/tex]
Since Zak's motion is a uniformly accelerated motion, we can now use the following suvat equation:
[tex]v^2-u^2=2as[/tex]
where
v = 0 is the final velocity (he comes to a stop)
u = 3.00 m/s is the initial velocity
[tex]a=-2.45 m/s^2[/tex] is the acceleration
s is the distance covered before stopping
Solving for s,
[tex]s=\frac{v^2-u^2}{2a}=\frac{0^2-3.0^2}{2(-2.45)}=1.84 m[/tex]
b)
In this second part, Zak gives a push to Greta.
We can find Greta's velocity after the push by using the work-energy theorem, which states that the work done on her is equal to her change in kinetic energy:
[tex](F-F_f)d =\frac{1}{2}mv^2-\frac{1}{2}mu^2[/tex]
where
F = 125 N is the force applied by Zak
d = 1.00 m is the distance
[tex]F_f=\mu mg[/tex] is the force of friction, where
[tex]\mu=0.250[/tex]
m = 20.0 kg is Greta's mass
[tex]g=9.8 m/s^2[/tex]
v is Greta's velocity after the push
u = 0 is Greta's initial velocity
Solving for v, we find:
[tex]v=\sqrt{\frac{2(F-\mu mg)d}{m}}=\sqrt{\frac{2(125-(0.250)(20.0)(9.8))(1.00)}{20.0}}=2.76 m/s[/tex]
After that, Zak stops pushing, so Greta will slide and the only force acting on her will be the force of friction; so the acceleration will be:
[tex]a=-\mu g = -(0.250)(9.8)=-2.45 m/s^2[/tex]
And so using again the suvat equation, we can find the distance she slides after Zak's push ends:
[tex]s=\frac{v'^2-v^2}{2a}[/tex]
where
v = 2.76 m/s is her initial velocity
v' = 0 when she stops
Solving for s,
[tex]s=\frac{0-(2.76)^2}{2(-2.45)}=1.55 m[/tex]
(a) The distance traveled by Zack before stopping is 1.84 m.
(b) The distance traveled by Greta after Zack's push ends is 1.56 m.
The given parameters;
coefficient of kinetic friction, [tex]\mu_k[/tex] = 0.25initial speed of Zack, u = 3 m/sThe distance traveled by Zack before stopping is calculated as follows;
The acceleration of Zack;
[tex]-F_k = ma\\\\-\mu_k mg = ma\\\\-\mu_k g = a\\\\-(0.25 \times 9.8) = a\\\\- 2.45 \ m/s^2 = a[/tex]
The distance traveled by Zack;
[tex]v^2 = u^2 + 2as\\\\when \ Zack \ stops \ v = 0\\\\0 = u^2 + 2as\\\\0 = (3)^2 +2(-2.45)s\\\\0 = 9 - 4.9s\\\\4.9s = 9\\\\\s = \frac{9}{4.9} \\\\s = 1.84 \ m[/tex]
The distance traveled by Greta is calculated as follows;
Apply law of conservation of energy to determine the velocity of Greta after the push.
[tex]Fd - F_kd = \frac{1}{2} mv^2\\\\125\times 1 - (0.25 \times 20 \times 9.8 \times 1) = (0.5 \times 20)v^2\\\\76 = 10v^2\\\\v^2 = \frac{76}{10} \\\\v ^2 = 7.6\\\\v = \sqrt{7.6} \\\\v = 2.76 \ m/s[/tex]
The acceleration of Greta;
[tex]a = -\mu_k g\\\\a = 0.25 \times 9.8\\\\a = -2.45 \ m/s^2[/tex]
The distance traveled by Greta;
[tex]v_f^2 = v^2 + 2as\\\\when \ Greta \ stops \ v_f = 0\\\\0 = v^2 + 2as\\\\-2as = v^2\\\\-2(-2.45)s = (2.76)^2\\\\4.9s = 7.62\\\\s = \frac{7.62}{4.9} \\\\s = 1.56 \ m[/tex]
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A 60.0kg skier starts from rest at the top of a ski slope of height 62.0 . Part A If frictional forces do −1.06×10 4 of work on her as she descends, how fast is she going at the bottom of the slope?
Answer:
v = 29.35 m /s
Explanation:
potential energy at the height of 62m
= m g h , m is mass , g is acceleration due to gravity and h is height
= 60 x 9.8 x 62
= 36456 J
negative work done by friction = -10600 J
energy at the bottom = 36456 - 10600 = 25856 J
This energy will be in the form of kinetic energy . If v be velocity at the bottom
1/2 m v² = 25856
1/2 x 60 x v² = 25856
v = 29.35 m /s
At a given instant, a 3.9-A current flows in the wires connected to a parallel-plate capacitor. Part APart complete What is the rate at which the electric field is changing between the plates if the square plates are 1.60 cm on a side? Express your answer using two significant figures.
Answer:
[tex]1.7\cdot 10^{15} V m^{-1} s^{-1}[/tex]
Explanation:
The electric field between the plates of a parallel-plate capacitor is given by
[tex]E=\frac{V}{d}[/tex] (1)
where
V is the potential difference across the capacitor
d is the separation between the plates
The potential difference can be written as
[tex]V=\frac{Q}{C}[/tex]
where
Q is the charge stored on the plates of the capacitor
C is the capacitance
So eq(1) becomes
[tex]E=\frac{Q}{Cd}[/tex] (2)
Also, the capacitance of a parallel-plate capacitor is
[tex]C=\frac{\epsilon_0 A}{d}[/tex]
where
[tex]\epsilon_0[/tex] is the vacuum permittivity
A is the area of the plates
Substituting into (2) we get
[tex]E=\frac{Q}{\epsilon_0 A}[/tex] (3)
Here we want to find the rate of change of the electric field inside the capacitor, so
[tex]\frac{dE}{dt}[/tex]
If we calculate the derivative of expression (3), we get
[tex]\frac{dE}{dt}=\frac{1}{\epsilon_0 A}\frac{dQ}{dt}[/tex]
However, [tex]\frac{dQ}{dt}[/tex] corresponds to the definition of current,
[tex]I=\frac{dQ}{dt}[/tex]
So we have
[tex]\frac{dE}{dt}=\frac{I}{\epsilon_0 A}[/tex]
In this problem we have
I = 3.9 A is the current
[tex]A=(0.0160 m)\cdot (0.0160 m)=2.56\cdot 10^{-4} m^2[/tex] is the area of the plates
Substituting,
[tex]\frac{dE}{dt}=\frac{3.9}{(8.85\cdot 10^{-12})(2.56\cdot 10^{-4})}=1.7\cdot 10^{15} V m^{-1} s^{-1}[/tex]
The rate at which the electric field is changing in the parallel-plate capacitor is approximately 1.7 x 10¹⁴ V/m·s.
Current (I) = 3.9 ASide length of square plates (L) = 1.60 cm = 0.016 m1. The area (A) of the plates:
A = L² = (0.016 m)² = 0.000256 m²2. The displacement current (I_D) is given by:
I_D = ε₀ dΦ/dtwhere
ε₀ (the permittivity of free space) ≈ 8.85 x 10⁻¹² F/mdΦ/dt = rate of change of the electric flux.3. The electric flux (Φ) is related to the electric field (E) by:
Φ = E ASince:
dΦ/dt = A dE/dt4. Substitute this into the displacement current equation:
I_D = ε₀ A dE/dtRearranging for dE/dt:
dE/dt = I / (ε₀ A)5. Plugging in the values:
dE/dt = 3.9 A / (8.85 x 10⁻¹² F/m * 0.000256 m²) dE/dt ≈ 1.73 x 10¹⁴ V/m·sThus, the rate at which the electric field is changing is approximately 1.7 x 10¹⁴ V/m·s.
A horse shoe magnet is placed on a mass balance such that a uniform magnetic field of magnitude B runs between it from North to South. A coil of resistance R is connected to a battery which supplies a potential difference of V across the coil and is suspended such that a section of the coil of length L meters lies between it with current running from East to West. The mass balance measures a mass of M. • What is the measured change in mass due to the effect of Fmag? • What is the total measured mass of the magnets? Keep in mind the effect of Newton's third law.
Answer:
(a) Measured change in mass (Δm) = BVL/Rg
(b) Total measured mass M' = M - BVL/Rg
Explanation:
Current (I) across is coil is given by the formula;
I = V/R ------------------------1
The magnetic force is given by the formula;
Fb = B*I*L -------------------2
Putting equation 1 into equation 2, we have;
Fb = B*V*L/R -------------------3
Change in mass (Δm) is given as:
Δm = Fb/g -----------------------4
Putting equation 3 into equation 4, we have;
Δm = BVL/Rg
Therefore, change in mass (Δm) = BVL/Rg
2. Since B runs from North to South and current running from East to West, then the magnetic force is directed upward.
Therefore,
Total measure mass M' = M - BVL/Rg
The simplest form of a traveling electromagnetic wave is a plane wave. For a wave traveling in the x direction whose electric field is in the y direction, the electric and magnetic fields are given by the picture shown above. This wave is linearly polarized in the y direction.
1. In these formulas, it is useful to understand which variables are parameters that specify the nature of the wave. The variables E_0and B_0 are the __________ of the electric and magnetic fields.
a. maximas
b. wavelenghts
c. amplitudes.
d. velocities
2. The variable omega is called the __________ of the wave.
a. wavenumber
b. wavelength
c. velocity
d. frequency
3. What is the mathematical expression for the electric field at the point x=0, y=0, z at time t?
4. For a given wave, what are the physical variables to which the wave responds?
Answer:
please the answer below
Explanation:
A general electromagnetic plane wave, traveling in the x direction, can be expressed in the form:
[tex]\vec{E}=E_0e^{-i(k\cdot x-\omega t)}\hat{j}\\\\\vec{B}=B_0e^{-i(k\cdot x-\omega t)}\hat{k}\\\\[/tex] (1)
1.
a. amplitudes
from (1) we can observe that E_0 and B_0 are the amplitudes.
2. frequency
3.
By replacing (1) we obtain:
[tex]\vec{E}=E_0e^{-i(k(0)-\omega t)}\hat{j}=E_0[cos\omega t+sin\omega t]\hat{j}[/tex]
4.
the wave respond to the followinf physical variables: amplitude, frequency, time and position, as we can see in (1).
hope this helps!!