Explanation:
We have,
Initial speed of an automobile, u = 71 km/h = 19.72 m/s
Diameter of the tie, d = 60 cm
Radius, r = 30 cm
(a) The angular speed of the tires about their axles is given by :
[tex]\omega=\dfrac{v}{r}\\\\\omega=\dfrac{19.72}{0.3}\\\\\omega=65.73\ rad/s[/tex]
(b) Final angular velocity of the wheel is equal to 0 as its stops. The angular acceleration of the wheel is given by :
[tex]\omega_f^2-\omega_i^2=2\alpha \theta[/tex]
[tex]\theta=40\ rev\\\\\theta=251.32\ rad[/tex]
[tex]0-\omega_i^2=2\alpha \theta\\\\\alpha =\dfrac{\omega_i^2}{2\theta}\\\\\alpha =\dfrac{(65.73)^2}{2\times 251.32}\\\\\alpha =-8.59\ rad/s^2[/tex]
(c) Let the car move a distance d during the braking. So,
[tex]d=\theta r\\\\d=251.32\times 0.3\\\\d=75.39\ m[/tex]
Therefore, the above is the required explanation.
a. The angular speed is 65.73 rad/s.
b. The magnitude of the angular acceleration is -8.59 rad/s².
c. The distance should be 75.39m.
Calculation of the angular speed, magnitude, and the distance:Since
Initial speed of an automobile, u = 71 km/h = 19.72 m/s
Diameter of the tie, d = 60 cm
Radius, r = 30 cm
a. Now the angular speed should be
= v/r
= 19.72/0.3
= 65.73 rad/s
b. Now the magnitude is
= 65.73^2/2*251.32
= -8.59 rad/s^2
c. The distance should be
= 251.32*0.3
= 75.39 m
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Describe how the phase angle changes as you move from below resonance to above resonance Based on your results here and other textbook resources, what will happen if your frequency is much lower than resonance? What about much larger?
Answer:
Answer
Explanation:
In a LCR Circuit, the phase difference between voltage and current is usually summarized as;
tan∅ = XL- XC
R
=WL - 1/WC
R
When w=0 ( i.e when it is very low) tan∅ = ∞
or ∅ = -90
When w=0 ( i.e when it is very large) tan∅ =+ ∞
or ∅ = +90
20 POINTS! TRUE OR FALSE:
To increase the acceleration of an object, you would reduce its mass or increase the applied force.
Answer: TRUE
Explanation:
A circus act involves a trapeze artist and a baby elephant. They are going to balance on a teeter-totter that is 10 meters long and pivoted at the middle. The lady has a weight of 500 newtons and is standing on one end. The elephant has a weight of 2500 N and is walking toward her on the beam. How far does the elephant have to walk from the end toward the middle to balance the beam
Answer:
Explanation:
Balance point will be achieved as soon as the weight of the baby elephant creates torque equal to torque created by weight of woman about the pivot. torque by weight of woman
weight x distance from pivot
= 500x 5
= 2500 Nm
torque by weight of baby woman , d be distance of baby elephant from pivot at the time of balance
= 2500x d
for equilibrium
2500 d = 2500
d = 1 m
So elephant will have to walk up to 1 m close to pivot or middle point.
An electron in an electron gun is accelerated from rest by a potential of 25 kV applied over a distance of 1 cm.
The final velocity of the electrons is _____.
The mass of the electron is 9.1x10^(-31) kg and its charge is 1.6x10^(-19) C.
Answer:
[tex]9.38\times 10^7 m/s[/tex]
Explanation:
We are given that
Potential ,V=25 kV=[tex]25\times 10^3 V[/tex]
Distance,r =1 cm=[tex]\frac{1}{100}=0.01 m[/tex]
1 m=100 cm
Mass of electron, m=[tex]9.1\times 10^{-31} kg[/tex]
Charge, q=[tex]1.6\times 10^{-19} C[/tex]
We have to find the final velocity of the electron.
Speed of electron,[tex]v=\sqrt{\frac{2qV}{m}}[/tex]
Using the formula
[tex]v=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 25\times 10^3}{9.1\times 10^{-31}}[/tex]
v=[tex]9.38\times 10^7 m/s[/tex]
Hence, the final velocity of the electron=[tex]9.38\times 10^7 m/s[/tex]
Sally places a jar with some pennies into a pool of water and exactly half of the jar is submerged. The volume of the jar is 200 cm3 and the density of the water is 1 g/cm3.
a. What is the buoyant force acting on the jar?
b. What is the weight of the jar and pennies?
c. What is the density of the combined jar and pennies?
Answer:
Bouyant Force = 0.9N
The density of the jar and Penny was not given so we can't determine it's weigh and combined density
Explanation:
Bouyant force Fb = Volume of solid*density of liquid*acceleration
Fb= (200/2)*1*980.665
Fb= 98066.5/100000
Fb= 0.9N
A traditional set of cycling rollers has two identical, parallel cylinders in the rear of the device that the rear tire of the bicycle rests on. Assume that the rear tire is rotating at ω = 32k rad/s. What are the angular velocities of the two cylinders? Consider r1 = 460 mm and r2 = 46 mm.
Answer:
ω2 = 216.47 rad/s
Explanation:
given data
radius r1 = 460 mm
radius r2 = 46 mm
ω = 32k rad/s
solution
we know here that power generated by roller that is
power = T. ω ..............1
power = F × r × ω
and this force of roller on cylinder is equal and opposite force apply by roller
so power transfer equal in every cylinder so
( F × r1 × ω1) ÷ 2 = ( F × r2 × ω2 ) ÷ 2 ................2
so
ω2 = [tex]\frac{460\times 32}{34\times 2}[/tex]
ω2 = 216.47
A plate of glass with parallel faces having a refractive index of 1.57 is resting on the surface of water in a tank. A ray of light coming from above in air makes an angle of incidence 37.5 ∘ with the normal to the top surface of the glass.
θ₃ is 27.12°
Explanation:
Using Snell's law:
n₁ sin θ₁ = n₂ sin θ₂
where,
n₁ = refractive index of material with incident light
n₂ = refractive index of material with refracted light
(a)
1 → air
2 → glas
3 → water
n(water) = n₃ = 1.333
From air to glass,
n₁ sin θ₁ = n₂sin θ₂
The angle of refraction 2 is the angle of incidence when the light comes from glass to water.
From glass to water:
n₁ sin θ₁ = n₃ sin θ₃
1 X sin 37.5° = 1.333 sin θ₃
sin θ₃ = sin (37.5)/ 1.333
sin θ₃ = 0.456
θ₃ = 27.12°
Therefore, θ₃ is 27.12°
"An uncharged 30.0-µF capacitor is connected in series with a 25.0-Ω resistor, a DC battery, and an open switch. The battery has an internal resistance of 10.0 Ω and the open-circuit voltage across its terminals is 50.0 V. The leads have no appreciable resistance. At time t = 0, the switch is suddenly closed." "When does the maximum current occur?"
Answer:
1.04x[tex]10^{-3}[/tex] s
Explanation:
->The maximum current through the resistor is
[tex]I_{max}[/tex] = V/R = V/[tex]Re^{-t/RC}[/tex]= V/R×[tex]e^{0}[/tex] = V/R
Voltage 'V'=50V
Effective resistance 'R'= 25.0-Ω+ 10.0 Ω= 35.0 Ω
Therefore, [tex]I_{max}[/tex]=50/35=> 1.43 A
->The maximum charge can be determined by
Q = CV
where,
Capacitance of the capacitor 'C' = 30.0µF = 30×10-⁶F
Therefore,
Q=30×10-⁶ x 50=>1.5 x [tex]10^{-3}[/tex]
In order to find that when does the maximum current occur, the time taken given the quantity of charge and the electric current is:
t= Q / I=> 1.5 x [tex]10^{-3}[/tex]/ 1.43
t= 1.04x[tex]10^{-3}[/tex] s
A comet has a period of 324 years; in other words, it orbits the Sun in 324 years. Most likely, this comet came from...(Hint: 324 years define a long period comet)
A. Somewhere outside the solar system
B. Oort's cloud
C. The ecliptic plane
D. The asteroid belt
E. The Kuiper belt
Comets are divided into two types. Long-period comets are the comets that take more than two hundred years to finish an orbit throughout the Sun originate from the Oort Cloud.
Explanation:
The Oort Cloud has sufficient distance apart of the Sun than the Kuiper Belt, it seems that the Oort Cloud objects were made closer to the Sun than the Kuiper Belt things.Long-period comets have highly eccentric orbits.The Oort cloud is considered to own a large range commencing from among 2,000 and 5,000 AU to 50,000 AU.The chemical makeup of long-period and short-period comets is quite alike.
Final answer:
A comet with a 324-year orbit is likely from the Oort Cloud, the distant spherical region of icy bodies surrounding our solar system, known as the source of long-period comets. The Kuiper Belt, on the other hand, sources short-period comets.
Explanation:
A comet with a period of 324 years is considered a long-period comet. These comets originate from a region far beyond the inner solar system. The Oort Cloud is a vast, spherical shell of icy bodies that extends up to about 50,000 astronomical units (AU) from the Sun and contains trillions of potential comets. This makes it the most likely source of long-period comets like the one mentioned. In contrast, the Kuiper Belt is a disk-shaped region beyond Neptune, extending to about 50 AU, which is known to source short-period comets, also known as Jupiter-family comets.
Long-period comets like the one with a 324-year orbit are thought to be influenced by gravitational perturbations from nearby passing stars or galactic tides, which can send them towards the inner solar system. Once these comets enter the inner solar system, they can have dramatic interactions with planets or the Sun, sometimes leading to their disintegration or alteration in orbit. An example of such an interaction was when Comet Shoemaker-Levy 9 broke apart and collided with Jupiter in 1994.
Electromagnetic radiation is more common than you think. Radio and TV stations emit radio waves when they broadcast their programs; microwaves cook your food in a microwave oven; dentists use X rays to check your teeth. Even though they have different names and different applications, these types of radiation are really all the same thing: electromagnetic (EM) waves, that is, energy that travels in the form of oscillating electric and magnetic fields.
Answer:
Explanation:
1. They have different wavelengths - Because These radiations form a spectra that differs by the size of the wavelength.
2. They have different Frequencies (f) that is frequency = 1/ wavelength
(f = 1/wavelength)
3. They propagate at different speed though a non vacuum media (non vacuum media affect the speed based on the wavelength)
Three different experiments are conducted that pertain to the oscillatory motion of a pendulum. For each experiment, the length of the pendulum and the mass of the pendulum are indicated in all experiments, the pendulum is released from the same angle with respect to the vertical. If the students collect data bout the kinetic energy of the pendulum as a function of time for each experiment, which of the following claims is true? a. The da ta collected from Experiment 1 will be the same as the data collected from Experiment 2. b. The data collected from Experiment 1 will be the same as the data collected from Experiment 3. C. The data collected from Experiment 2 will be the same as the data collected from Experiment 3. d. The data collected from each experiment will be different.
The data collected from all experiments by the students will be different. Option D is correct.
Kinetic energy:
The kinetic energy of the pendulum depends upon the length and mass and angle of the pendulum.
Since the angle is the same in all three experiments. Mass and length will decide the kinetic energy of the pendulum.
As students measure the kinetic energy against the function of time, the period of the pendulum will also vary.
The period of the pendulum can be calculated by the formula,
[tex]T = 2\pi \sqrt{ \dfrac lg}[/tex]
Where,
[tex]l[/tex] - length
[tex]g[/tex] - gravitational acceleration,
Therefore, the data collected from all experiments by the students will be different. Option D is correct.
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The motion of a simple pendulum is not affected by the mass of the bob; it is only influenced by the length of the pendulum and the acceleration due to gravity. If the pendulum lengths in the experiments are the same, then the kinetic energy data collected will be identical, even if the masses are different.
Explanation:If three different experiments are conducted that pertain to the oscillatory motion of a pendulum with varying lengths and masses, and all are released from the same angle, the kinetic energy as a function of time for each experiment will differ only if the lengths of the pendulums are different. The mass of the pendulum bob does not affect the motion of a simple pendulum; pendula are only affected by the period (which is related to the pendulum's length) and by the acceleration due to gravity. Therefore, if the lengths of the pendulums in the different experiments are the same, then claim a. The data collected from Experiment 1 will be the same as the data collected from Experiment 2 is true irrespective of the different masses of the pendulum bobs.
The movement of the pendula will not differ at all because the mass of the bob has no effect on the motion of a simple pendulum. The pendula are only affected by the period (which is related to the pendulum's length) and by the acceleration due to gravity.
A spherical gas-storage tank with an inside diameter of 8.0 m is being constructed to store gas under an internal pressure of 1.62 MPa. The tank will be constructed from steel that has a yield strength of 295 MPa. If a factor of safety of 3.0 with respect to the yield strength is required, determine the minimum wall thickness tmin required for the spherical tank.
Answer:
The minimum wall thickness Tmin required for the spherical tank is 65.90mm
Explanation:
Solution
Recall that,
Tmin = The minimum wall thickness =PD/2бp
where D = diameter of 8.0 m
Internal pressure = 1.62 MPa
Then
The yield strength = 295MPa/3.0 = 98.33
thus,
PD/2бp = 1.62 * 8000/ 2 *98.33
= 12960/196.66 = 65.90
Therefore the wall thickness Tmin required for the spherical tank is 65.90mm
The limit to the eye's acuity is actually related to diffraction by the pupil. What is the angle between two just‑resolvable points of light for a 3.75 mm 3.75 mm diameter pupil, assuming the average wavelength of 554 nm 554 nm ?
Answer: [tex]1.8(10)^{-4} rad[/tex]
Explanation:
This problem is related to the Rayleigh Criterion, which provides the following formula to find the acuity or limit of resolution of an optic system with circular aperture (the eye in this case):
[tex]\theta=1.22\frac{\lambda}{D}[/tex]
Where:
[tex]\theta[/tex] is the angle of resolution (related to the acuity)
[tex]\lambda=554 nm=554(10)^{-9} m[/tex] is the wavelength of the light
[tex]D=3.75 mm=3.75(10)^{-3} m[/tex] is the diameter of the pupil
Solving:
[tex]\theta=1.22 \frac{554(10)^{-9} m}{3.75(10)^{-3} m}[/tex]
[tex]\theta=1.8(10)^{-4} rad[/tex]
The human eye's acuity is limited by light's diffraction through the pupil, which determines the minimum angle, Δθ, between two resolvable points. This is calculated using the formula Δθ = 1.22 (λ / D), where λ is the light's wavelength and D the diameter of the pupil.
Explanation:The resolution of human visual acuity is related to the diffraction of light by the eye's pupil. This diffraction effect can be worked out using a formula that quantifies resolving power (Δθ), when the diameter of the aperture (in this case, the pupil size) and the wavelength of incident light are known.
In your example, a 3.75 mm diameter pupil encountering light of 554 nm wavelength can identify two points of light provided they subtend an angle (Δθ) at the pupil calculated by using the formula: Δθ = 1.22 (λ / D). For all light's range of visible wavelengths, the angle is extremely tiny - on the order of tens of thousands of a degree. This shows that the eye’s ability to distinguish two separate points of light is limited by the diffraction of light by the eye’s pupil.
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A wire carrying a 29.0 A current passes between the poles of a strong magnet such that the wire is perpendicular to the magnet's field, and there is a 2.25 N force on the 3.00 cm of wire in the field. What is the average field strength (in T) between the poles of the magnet
Answer:
2.59 T
Explanation:
Parameters given:
Current flowing through the wire, I = 29 A
Angle between the magnetic field and wire, θ = 90°
Magnetic force, F = 2.25 N
Length of wire, L = 3 cm = 0.03 m
The magnetic force, F, is related to the magnetic field, B, by the equation below:
F = I * L * B * sinθ
Inputting the given parameters:
2.25 = 29 * 0.03 * B * sin90
2.25 = 0.87 * B
=> B = 2.25/0.87
B = 2.59 T
The magnetic field strength between the poles is 2.59 T
A spring with a constant of 16 N/m has 98 J of energy stored in it when it is extended. How far is the spring extended?
The spring has been extended for 3.5 m
Explanation:
We have the formula,
PE =1/2 K X²
Rewrite the equation as
PE=1/2 K d²
multiply both the sides by 2/K to simplify the equation
2/k . PE= 1/2 K d² . 2/K
√d²=√2PE/K
Cancelling the root value and now we have,
d=√2PE/k
d=√2×98 J / 16N/m
d=√12.25
d=3.5 m
The spring has been extended for 3.5 m
Final answer:
To find the extension of a spring with a constant of 16 N/m and 98 J of energy stored, we use the formula for potential energy. The spring is extended by 3.5 meters.
Explanation:
The question involves calculating the extension of a spring based on the energy stored in it and the spring's constant. The formula for the potential energy stored in a spring is given by
U = 1/2 kx², where U is the potential energy, k is the spring constant, and x is the extension of the spring from its equilibrium position. In this case, the energy (U) is 98 J and the spring constant (k) is 16 N/m.
The spring in question has an energy of 98 J stored in it.
Using the formula for potential energy stored in a spring, we have:
[tex]PE = 1/2 k x^2[/tex]
By substituting the known values, we find that the spring is extended by 0.7 m (70 cm
We rearrange the formula to solve for x:
x = √(2U/k).
Substituting the given values,
x = √(2*98/16)
= √(196/16)
= √(12.25)
= 3.5 meters. Therefore, the spring is extended by 3.5 meters.
1. Consider a head-on collision between two identical billiard balls. Ball 1 is initially in motion toward ball 2, which is initially at rest. After the collision, ball 2 departs with the same velocity that ball 1 originally had. Disregard any friction between the balls and the surface. What happens to ball 1? What happens to ball 2?
Answer:
first ball is at rest after collision and the kinetic energy of the first ball is transferred to the second ball after collision.
Explanation:
mass of each ball = m
initial velocity of first ball = u
initial velocity of second ball = 0 m/s
final velocity of second ball = initial velocity of first ball = u
Use conservation of momentum
let v is the final velocity of the first ball after collision
m x u + m x 0 = m x v + m x u
So, v = 0
It means after the collision the second ball moves with the velocity which is equal to the initial velocity of the first ball and the first ball comes at rest.
As there is no friction between the balls during the collision, the collision of two balls is perfectly elastic and thus the kinetic energy of the system is conserved.
The entire kinetic energy of the first ball is transferred to the second ball after the collision.
The entire kinetic energy of ball 1 will be totally transferred to the second ball after the collision.
According to the law of conservation of momentum, the change in momentum of a body before the collision is equal to the change in momentum after the collision.
The formula for calculating momentum is expressed as:
[tex]\rho = mv \\[/tex]
According to the conservation of momentum
[tex]m_1u_1+m_2u_2=(m_1+m_2)v[/tex]
Since ball 2 is initially at rest [tex]u_2=0[/tex]
Also if after the collision, ball 2 departs with the same velocity that ball 1 originally had, hence [tex]v=u_1[/tex]
Substituting the given parameters into the formula:
[tex]m_1u_1+m_2(0)=(m_1+m_2)u_1\\m_1u_1=m_1u_1+m_2u_2\\m_2u_2=0[/tex]
This shows that the velocity of the second ball is also zero
Due to the absence of friction between the colliding object, the collision is elastic in nature showing that the kinetic energy of the system is conserved.
The entire kinetic energy of ball 1 will be totally transferred to the second ball after the collision.
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A hospital needs 0.100 g of 133 54Xe for a lung-imaging test. If it takes 10 days to receive the shipment, what is the minimal amount mXe of xenon that the hospital should order?
Answer:
the hospital needs to order for a minimum amount of 0.4 g of [tex]\left \ {{133} \atop {54} \right. Xe[/tex]
Explanation:
Given that:
A hospital needs 0.100 g of [tex]\left \ {{133} \atop {54} \right. Xe[/tex]
and it takes 10 days for the shipment to arrive:
the half life [tex]t_{1/2}[/tex] = 5 days
So, since the half life = 5 days ;
decay constant [tex]\lambda = \frac{In_2}{t_{1/2}}[/tex]
where:
[tex]N_o= ??? \\ \\ N = 0.1 00 \\ \\ t = 10 days \\ \\ N(t)= N_oe^{-\lambda t}\\ \\0.1 = N_oe^{\frac{-In_2}{5} *10} \\ \\0.1 = N_oe^{ - In \ 4} \\ \\ 0.1 = N_oe^{ In \frac{1}{ 4}} \\ \\ 0.1 = \frac{N_o}{4} \\ \\ N_o = 0.1*4 \\ \\ N_o = 0.4 \ g[/tex]
Therefore in order to get 0.100 g of [tex]\left \ {{133} \atop {54} \right. Xe[/tex], the hospital needs to order for a minimum amount of 0.4 g of
A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE gives the radial distance (from the Earth's center) the projectile reaches if (a) its initial speed is 0.785 of the escape speed from Earth and (b) its initial kinetic energy is 0.785 of the kinetic energy required to escape Earth
Answer:
Explanation:
Given that,
We have a projectile that is shot from the earth
Then, it's escape velocity is
Ve = √(2GM / R)
Ve² = 2GM / R
M is the mass of earth
R is the radius of Earth
Using conservation of energy
Ki + Ui = Kf + Uf
The final kinetic energy is zero since the body is finally at rest
Then,
Ki + Ui = Uf
Kinetic energy can be determine using
K = ½mv²
Potential energy can be determine using
U = -GmM / R
Where m is mass of the body
Then,
Ki + Ui = Uf
½mv² - GmM / R = -GmM / r
r is the maximum height reached
m cancel out
½ v² - GM / R = -GmM / r
A. We want to find the maximum height reached when initial speed (v) is equal to 0.785 escape velocity
So, v = 0.785•Ve
So,
½v² - GM / R = -GM / r
½(0.785•Ve)² - GM / R = -GM / r
0.308•Ve² - GM / R = -GM / r
0.308 × 2GM / R - GM / R = -GM / r
Divide through by GM
0.616 / R - 1 / R = -1 / r
-0.384 / R = -1 / r
Cross multiply
-0.384r = -R
r = -R / -0.384
r = 2.6 R
B. When it initial kinetic energy is is 0.785 of the kinetic energy required to escape Earth
Ki = 0.785 Ke
Ki = 0.785 ½mVe²
Ki = 0.785 × ½m× 2GM / R
Ki = 0.785•G•m•M / R
So,
Ki + Ui = Uf
0.785•G•m•M / R - GmM / R = -GmM / r
Let G•m•M cancels out
0.785 / R - 1 / R = -1 / r
-0.215 / R = -1 / r
Cross multiply.
-0.215r = -R
r = -R / -0.215
r = 4.65 R
The frequency of a sound wave is 300 Hz and the room temperature is 30 Celsius. What is the wavelength of this sound wave. Two children stretch a jump rope between them and send wave pulses back and forth. The rope is 4 meters long and its mass is .4 kg and the force exerted on it by the children is 50 N. What is the speed of the waves on the rope?
Answer:
Explanation:
Velocity of sound wave at 30 degree = 350 m /s
frequency of sound = 300 Hz .
wavelength of sound in air. = velocity / frequency
= 350 / 300
= 1.167 m
for wave formed in the rope :
velocity of wave in the rope
= [tex]\sqrt{\frac{T}{m} }[/tex]
T is tension in the rope and m is mass per unit length .
m = .4 / 4
= .1
Putting the given values in the equation above
v = [tex]\sqrt{\frac{50}{.1} }[/tex]
v = 22.36 m /s .
velocity of wave in the rope.
= 22.36 m /s .
Below is a sketch of the initial state of the situation described in this problem. Draw the most suitable set of coordinate axes for this problem. Note that even though you can choose the y=0y=0 level to be wherever you like, in most situations it is best to set the zero height to coincide with either the initial or final position, so that the calculations for the gravitational potential energy become easier. For this reason, in this particular problem place the origin of your coordinate axes on the black dot marking the performer's initial position. Draw only the positive portion of the coordinate axes. Draw the vectors starting at the black dot. The location and orientation of the vectors will be graded. The length of the vectors will not be graded.
Answer:
This problem is incomplete, it says like this: The Great Sandini is a 60 kg circus performer who is shot from a cannon (actually a spring gun). You dont find many men of his caliber, so you help him design a new gun. This new gun has a very large spring with a very small mass and a force constant of 1100 N/m that will compress with a force of 4400 N. The inside of the gun barrel is coated with teflon, so the average friction force will be only 40 N during the 4 m his moves in the barrel. At what speed will he emerge from the end of the barrel, 2.5 m above his initial rest position?
The speed is 15.5 m/s, and the image shows the vector drawing.
Explanation:
For the calculation of the force that is due to the spring is equal to:
F = kx
Where
F = 4400 N
k = 1100 N/m
x = F/k = 4400/1100 = 4 m
For the calculation of the total energy is equal to:
[tex]\frac{1}{2} kx^{2} =\frac{1}{2}mv^{2} +mgh+Fx\\v=\sqrt{\frac{2}{m} (\frac{1}{2}kx^{2} -mgh-Fx) }[/tex]
Where
m = 60 kg
h = 2.5 m
Replacing:
[tex]v=\sqrt{\frac{2}{60}(\frac{1}{2}*1100*4^{2}-60*9.8*2.5-(40*4)) } =15.5m/s[/tex]
The problem is about setting coordinate axes and vectors for a physics problem, with the origin placed at the performer's initial position. The vectors represent various physical quantities depending on their orientation and direction, always starting from the origin.
Explanation:This is a typical problem in physics dealing with vectors, coordinate axes, and natural phenomena like gravitational potential energy. Setting up the appropriate coordinate system is crucial. We set the origin, or the (0,0) point, at the performer's initial position as mentioned in the problem.
The y-axis should be vertical, aligned with the direction of gravity. This makes computations involving gravity much easier. It's important to note that increasing height (direction opposite to that of gravity) is typically represented as positive y.
The vectors originating from the black dot (our origin) represent quantities such as displacement, velocity, force etc. The direction of the vector indicates the direction of the quantity.
For example, if there's a vector pointing directly upwards from the origin, it could represent an upward force or movement. Keep in mind, the orientation and location of the vectors are determined by the physical reality the problem is describing, and they should always start from the origin in the problem's context.
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A 6.0 g marble is fired vertically upward using a spring gun. The spring must be compressed 9.4 cm if the marble is to just reach a target 22 m above the marble's position on the compressed spring. (a) What is the change ΔUg in the gravitational potential energy of the marble-Earth system during the 22 m ascent? (b) What is the change ΔUs in the elastic potential energy of the spring during its launch of the marble? (c) What is the spring constant of the spring?
Answer:
a) [tex]\Delta U_{g} = 12.945\,J[/tex], b) [tex]\Delta U_{k} = 12.945\,J[/tex], c) [tex]k = 2930.059\,\frac{N}{m}[/tex]
Explanation:
a) The change in the gravitational potential energy of the marble-Earth system is:
[tex]\Delta U_{g} = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (22\,m)[/tex]
[tex]\Delta U_{g} = 12.945\,J[/tex]
b) The change in the elastic potential energy of the spring is equal to the change in the gravitational potential energy, then:
[tex]\Delta U_{k} = 12.945\,J[/tex]
c) The spring constant of the gun is:
[tex]\Delta U_{k} = \frac{1}{2} \cdot k \cdot x^{2}[/tex]
[tex]k = \frac{2\cdot \Delta U_{k}}{x^{2}}[/tex]
[tex]k = \frac{2\cdot (12.945\,J)}{(0.094\,m)^{2}}[/tex]
[tex]k = 2930.059\,\frac{N}{m}[/tex]
A loop of wire lies flat on a horizontal surface. A bar magnet is held above the center of a loop with south pole pointing down. Magnet is released.
As magnet approaches the loop of wire, the induced current in this wire is:
a) Clockwise
b) Counterclockwise
c) Zero
Answer:
A
Explanation:
Current, magnetic field and motion are mutually dependent and perpendicular to one another
Answer:
b) clockwise
Explanation:
The right hand rule states that to determine the direction of the magnetic force on a positive moving charge, point the thumb of the right hand in the direction of the potential v, the fingers in the direction of the magnetic field, B, and a perpendicular to the palm points in the direction of the force, F.
Since the south pole of the bar magnet is pointing downwards, the induced current developed in the wire repels it and moves in the clockwise direction.
A toroidal coil has a mean radius of 16 cm and a cross-sectional area of 0.25 cm2; it is wound uniformly with 1000 turns. A second toroidal coil of 750 turns is wound uniformly over the first coil. Ignoring the variation of the magnetic field within a toroid, determine the mutual inductance of the two coils.
Answer:
Explanation:
Mutual inductance is equal to magnetic flux induced in the secondary coli due to unit current in the primary coil .
magnetic field in a torroid B = μ₀ n I , n is number of turns per unit length and I is current .
B = 4π x 10⁻⁷ x (1000 / 2π x .16 )x 1 ( current = 1 A)
flux in the secondary coil
= B x area of face of coil x no of turns of secondary
= 4π x 10⁻⁷ x (1000 /2π x .16 ) .25 x 10⁻⁴ x 750
= 2 x 1000 x .25 x( 750 /.16) x 10⁻¹¹
2343.75 x 10⁻⁸
= 23.43 x 0⁻⁶ H.
.
Answer:
2.5 x 10^-5 henry
Explanation:
The mutual inductance between the toroids is same.
mean radius of the toroid, r = 16 cm = 0.16 m
Area of crossection, A = 0.25 cm²
Number of turns in the first toroid, N1 = 1000
Number of turns in the second toroid, N2 = 750
The formula for the mutual inductance is given by
[tex]M =\frac{\mu_{0}N_{1}N_{2}A}{l}[/tex]
Where, l is the length
l = 2 x 3.14 x r = 2 x 3.14 x 0.16 = 1.0048 m
[tex]M =\frac{4\pi\times 10^{-7}\times 1000\times 750\times 0.25\times 10^{-4}}{1.0048}[/tex]
M = 2.5 x 10^-5 henry
Thus, the mutual inductance between the two toroid is 2.5 x 10^-5 henry.
A 6200 line/cm diffraction grating is 3.14 cm wide. If light with wavelengths near 624 nm falls on the grating, how close can two wavelengths be if they are to be resolved in any order
Answer:
[tex]1.6026299569\times 10^{-11}\ m[/tex]
Explanation:
Grating constant
[tex]d=\dfrac{1}{6200}=0.000161\ cm=0.000161\times 10^{-2}\ m[/tex]
Number of slits
[tex]N=3.14\times 6200=19468[/tex]
Order
[tex]m=\dfrac{d}{\lambda}\\\Rightarrow m=\dfrac{0.000161\times 10^{-2}}{624\times 10^{-9}}\\\Rightarrow m\approx 2[/tex]
At m = 1
[tex]\Delta\lambda=\dfrac{\lambda}{mN}\\\Rightarrow \Delta\lambda=\dfrac{624\times 10^{-9}}{1\times 19468}\\\Rightarrow \Delta\lambda=3.2052599137\times 10^{-11}\ m[/tex]
At m = 2
[tex]\Delta\lambda=\dfrac{\lambda}{mN}\\\Rightarrow \Delta\lambda=\dfrac{624\times 10^{-9}}{2\times 19468}\\\Rightarrow \Delta\lambda=1.6026299569\times 10^{-11}\ m[/tex]
The wavelengths can be close by [tex]1.6026299569\times 10^{-11}\ m[/tex]
(a) For what frequencies does a 23.0-μF capacitor have a reactance below 160 Ω? f Correct: Your answer is correct. 6.95 Incorrect: Your answer is incorrect. Hz (b)What is the reactance of a 41.0-μF capacitor over this same frequency range? Xc Correct: Your answer is correct. Ω
Answer:
(a) So range of frequency [tex]f > 43.27[/tex] Hz
(b) the reactance is 89.75 Ω
Explanation:
Given:
(a)
Capacitance of a capacitor [tex]C= 23 \times 10^{-6}[/tex] F
Reactance of capacitive circuit [tex]X_{C} =[/tex] 160 Ω
From the formula of reactance,
[tex]X_{C} = \frac{1}{\omega C}[/tex]
[tex]X_{C} = \frac{1}{2\pi fC}[/tex]
[tex]f = \frac{1}{2\pi X_{C} C }[/tex]
[tex]f = \frac{1}{6.28 \times 160 \times 23 \times 10^{-6} }[/tex]
[tex]f = 43.27[/tex] Hz
So range of frequency [tex]f > 43.27[/tex] Hz
(b)
Capacitance [tex]C = 41 \times 10^{-6}[/tex] F
Frequency [tex]f = 43.27[/tex] Hz
From the formula of reactance,
[tex]X_{C} = \frac{1}{2\pi fC}[/tex]
[tex]X_{C} = \frac{1}{6.28 \times 43.27 \times 41 \times 10^{-6} }[/tex]
[tex]X_{C} =[/tex] 89.75 Ω
Therefore, the reactance is 89.75 Ω
A block of mass 500 g is attached to a spring of spring constant 80 N/m. The other end of the spring is attached to a support while the mass rests on a rough surface with a coefficient of friction of 0.20 that is inclined at angle of 30°. The block is pushed along the surface till the spring compresses by 10 cm and is then released from rest. (a) How much potential energy was stored in the block-spring-support system when the block was just released? (b) Determine the speed of the block when it crosses the point when the spring is neither compressed nor stretched. (c) Determine the position of the block where it just comes to rest on its way up the incline.
Answer:
a) Vki = 0.4 J
b) v = 0.98 m/s
c) final position = 0.11745 m from initial position
Explanation:
Given:-
- The mass of block, m = 500 g
- The spring constant, k = 80 N/m
- The coefficient of the surface, u = 0.20
- The inclination of the block, θ = 30°
- The block is initially compressed, xi = 10 cm
- The block is initially at rest, vi = 0 m/s
Find:-
(a) How much potential energy was stored in the block-spring-support system when the block was just released?
(b) Determine the speed of the block when it crosses the point when the spring is neither compressed nor stretched.
(c) Determine the position of the block where it just comes to rest on its way up the incline.
Solution:-
- The potential energy initially stored in a spring "Vki" with constant "k" which has undergone a displacement of "x" is given by:
Vki = 0.5*k*xi^2
Vki = 0.5*80*0.1^2
Vki = 0.4 J
- The block is released from rest when the energy stored in the spring is dispensed in three forms of energy.
- There would be an increase in the potential energy " Ep " of the block as it moves up.
- There would be an increase in kinetic energy for some time " Ek "
- There would be loss of total energy due to fictitious force ( Ff) the work done against friction will dissipate energy.
- The increase in potential energy "Ep" at displacement xo = 0 from mean position is given by:
ΔEp = m*g*Δh ... change in vertical height h
ΔEp = m*g*( xi - xo )*sin ( θ )
ΔEp = 0.5*9.81*(0.1)*sin ( 30 )
ΔEp = 0.24525 J
- The increase in kinetic energy "Ek" at displacement xi = 10 cm, vi = 0 m/s from initial position to mean position at xo = 0 ,its velocity is "vf" is given by:
ΔEk = 0.5*m*( vf^2 - vi^2 ) ... change in velocity
ΔEk = 0.5*m*( vf )^2
- There would be loss of total energy due to fictitious force ( Ff ) the work done against friction will dissipate energy. First apply the equilibrium conditions on the block normal to slope and determine the contact force ( Nc ):
Nc - m*g*cos ( θ ) = 0
Nc = m*g*cos ( θ )
- The friction force ( Ff ) is given by:
Ff = Nc*u
Ff = u*m*g*cos ( θ )
- The work done by block against friction is given by:
Wf = -Ff*( xi - xo )
Wf = -u*m*g*xi*cos ( θ )
Wf = -0.2*0.5*9.81*0.1*cos ( 30 )
Wf = -0.08495 J
- We can now express the work done principle for the block:
Vki = ΔEp + ΔEk + Wf
ΔEk = - ΔEp - Wf + Vki
0.5*m*( vf )^2 = -0.24525 + 0.08495 +0.4
vf^2 = 4*(-0.24525 + 0.08495 +0.4 )
vf = √0.9588
vf = 0.98 m/s
- We will denote the extension of spring at top most position from mean position as "x".
- From mean position xo = 0 m. The block will further move up the slope and expense all its kinetic energy "Ek" in the form of gain in potential energy and gain in elastic potential energy "Vk" and work is done against friction.
Vki = Ep2 + Wf + Vk2
0.4 = mg*x*sin ( 30 ) + 0.24525 + 0.08495 + u*m*g*x*cos ( θ ) + 0.5*k*x^2
0.5*80*x^2 + x*(0.5*9.81*sin(30) + 0.2*0.5*9.81*cos ( 30 ) ) + 0.3302-0.4 =0
40x^2 + 3.30207x - 0.0698 = 0
Solve the quadratic equation:
x = -0.1 m (10 cm) - initial compression at rest ;
x = 0.01745 m
- So the extension in spring at the rest position is x = 0.01745 m. The position of the next resting point is:
final position = xi + x
final position = 0.1 + 0.01745
final position = 0.11745 m from initial position.
x =
(a) The stored PE will be "0.40 J".
(b) The block's speed will be "0.48 m/s".
(c) The block's position will be at "12 cm".
Potential energy (P.E)According to the question,
Mass, m = 500 h
Spring constant = 80 N/m
Coefficient of friction = 0.20
Inclined angle = 30°
(a) We know the formula,
Spring P.E = [tex]\frac{1}{2}[/tex] kx²
By substituting the values,
= [tex]\frac{1}{2}[/tex] × 80 × (0.1)²
= 0.40 J
(b) We know that,
→ [tex]P.E_s[/tex] = ΔKE + Work done by friction + Δ[tex]PE_g[/tex]
or,
= [tex]\frac{1}{2}[/tex] mv² + [tex]\mu_R[/tex] mgx Cosθ + mgx Sinθ
By substituting the values,
0.40 = [tex]\frac{1}{2}[/tex] × 0.5 × v² + 0.2 × 0.5 × 9.8 × Cos30° × 0.1 + 0.5 × 9.8 × 0.1 × Sin30°
= 0.48 m/s
(c) By using the above formula,
→ 0.40 = [tex]\mu_R[/tex] mgl Cosθ + mgl Sinθ
= 0.2 × 0.5 × 9.8 × Cos30° × l + 0.5 × 9.8 × l × Sin30°
l = 0.12 m or,
= 12 cm
Thus the above responses are correct.
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A laser beam is incident on a plate of glass that is 2.8 cm thick. The glass has an index of refraction of 1.6 and the angle of incidence is 36°. The top and bottom surfaces of the glass are parallel. What is the distance b between the beam formed by reflection off the top surface of the glass and the beam reflected off the bottom surface of the glass?
Final answer:
The distance between the reflected light beams can be found using reflection and refraction laws, applying Snell's law to calculate the angle of refraction, and subsequently using trigonometry to determine the light path within the glass and the shift caused by both reflections.
Explanation:
To determine the distance b between the light beam reflected from the top and the bottom surfaces of a glass plate, we need to use the laws of reflection and Snell's law for the refraction of light. The given information includes the thickness of the glass (2.8 cm), the index of refraction of the glass (1.6), and the angle of incidence (36°). When the laser beam strikes the top surface at an angle of incidence of 36°, it reflects at the same angle (law of reflection).
The angle of refraction can be found using Snell's law: n1 * sin(θ1) = n2 * sin(θ2), where n1 and n2 are the indices of refraction of the air and glass, respectively, and θ1 and θ2 are the angles of incidence and refraction, respectively. Once the angle of refraction is determined, the actual path of the beam within the glass can be calculated using trigonometry, and subsequently the shift b caused by the refraction at the two interfaces (again using trigonometry).
Space scientists have a large test chamber from which all the air can be evacuated and in which they can create a horizontal uniform electric field. The electric field exerts a constant hor-izontal force on a charged object. A 15 g charged projectile is launched with a speed of 6.0 m/s at an angle 35° above the hori-zontal. It lands 2.9 m in front of the launcher. What is the magni-tude of the electric force on the projectile?
Answer:
the magnitude of the electric force on the projectile is 0.0335N
Explanation:
time of flight t = 2·V·sinθ/g
= (2 * 6.0m/s * sin35º) / 9.8m/s²
= 0.702 s
The body travels for this much time and cover horizontal displacement x from the point of lunch
So, use kinematic equation for horizontal motion
horizontal displacement
x = Vcosθ*t + ½at²
2.9 m = 6.0m/s * cos35º * 0.702s + ½a * (0.702s)²
a = -2.23 m/s²
This is the horizontal acceleration of the object.
Since the object is subject to only electric force in horizontal direction, this acceleration is due to electric force only
Therefore,the magnitude of the electric force on the projectile will be
F = m*|a|
= 0.015kg * 2.23m/s²
= 0.0335 N
Thus, the magnitude of the electric force on the projectile is 0.0335N
Answer:
Magnitude of electric force = 0.03345 N
Explanation:
We are given;
Mass; m = 15g = 0.015kg
Angle above horizontal; θ = 35°
Speed; v = 6 m/s
Horizontal displacement; d = 2.9m
Now formula for time of flight is given as;
time of flight; t = (2Vsinθ)/g
Thus, plugging in values, we have
t = (2 x 6.0 x sin35)/9.8
t = (12 x 0.5736)/9.8
t = 0.7024 s
Now, let's find the acceleration
The formula for horizontal displacement is given by;
d = (Vcosθ)t + (1/2)at²
Plugging in the relevant values ;
2.9 = [6(cos35) x 0.7024] + (1/2)a(0.7024)²
2.9 = (4.2144 x 0.8192) + (0.2467)a
2.9 = 3.45 + (0.2467)a
(0.2467)a = 2.9 - 3.45
a = -0.55/0.2467
a = -2.23 m/s²
Since we are looking for the magnitude of the electric force, we will take the absolute value of a. Thus, a = 2.23 m/s²
We know that F = ma
Thus,Force = 0.015kg x 2.23m/s² =
= 0.03345 N
A 3.0 kg object is moving along the x-axis in a region where its potential energy as a function of x is given as U(x) = 4.0x2 , where U is in joules and x is in meters. When the object passes the point x = -0.50 m, its velocity is +2.0 m/s. All forces acting on the object are conservative. Calculate the total mechanical energy of the object Calculate the x-coordinate of any points at which the object has zero kinetic energy. Calculate the magnitude of the momentum of the object at x = 0.60 m. Calculate the magnitude of the acceleration of the object as it passes x = 0.60 m. On the axes below, sketch graphs of the object’s position x versus time t and kinetic energy K versus time t. Assume that x = 0 at time t = 0 . The two graphs should cover the same time interval and use the same scale on the horizontal axes.
Answer:
a) [tex]E_{tot} = 7\,J[/tex], b) [tex]x = \pm\sqrt{\frac{7}{4} }[/tex]c) [tex]p = 5.775\,\frac{kg\cdot m}{s}[/tex]
Explanation:
a) The total energy of the object is:
[tex]E_{tot} = U + K[/tex]
[tex]E_{tot} = 4\cdot x^{2} + \frac{1}{2}\cdot m \cdot v^{2}[/tex]
[tex]E_{tot} = 4\cdot (-0.50\,m)^{2} + \frac{1}{2}\cdot (3\,kg)\cdot (2\,\frac{m}{s} )^{2}[/tex]
[tex]E_{tot} = 7\,J[/tex]
b) The total energy of the object is:
[tex]E_{tot} = U[/tex]
[tex]7\,J = 4\cdot x^{2}[/tex]
[tex]x = \pm\sqrt{\frac{7}{4} }[/tex]
c) The speed of the object is clear in the total energy expression:
[tex]E_{total} = U + K[/tex]
[tex]K = E_{total}-U[/tex]
[tex]\frac{1}{2}\cdot m \cdot v^{2} = E_{total} - 4\cdot x^{2}[/tex]
[tex]v^{2} = \frac{2\cdot (E_{total}-4\cdot x^{2})}{m}[/tex]
[tex]v = \sqrt{\frac{2\cdot (E_{total}-4\cdot x^{2})}{m} }[/tex]
[tex]v = \sqrt{\frac{2\cdot [7\,J- 4\cdot (0.6\,m)^{2}]}{3\,kg} }[/tex]
[tex]v \approx 1.925\,\frac{m}{s}[/tex]
The magnitude of the momentum is:
[tex]p = (3\,kg)\cdot (1.925\,\frac{m}{s} )[/tex]
[tex]p = 5.775\,\frac{kg\cdot m}{s}[/tex]
d) Before calculating the acceleration experimented by the object, it is required to determine the net force exerted on it. There is a relationship between potential energy and net force:
[tex]F = -\frac{dU}{dx}[/tex]
[tex]F = -8\cdot x[/tex]
Acceleration experimented by the object is:
[tex]a = -\frac{8\cdot x}{m}[/tex]
[tex]a = -\frac{8\cdot (0.6\,m)}{3\,kg}[/tex]
[tex]a = -1.6\,\frac{m}{s^{2}}[/tex]
e) The position of the object versus time is found by solving the following differential equation:
[tex]\frac{d^{2}x}{dt} +\frac{8\cdot x}{m} = 0[/tex]
[tex]s^{2}\cdot X(s)- s\cdot v(0) - x(0) + \frac{8}{m}\cdot X(s) = 0[/tex]
[tex](s^{2} + \frac{8}{m})\cdot X(s) = s\cdot v(0)+x(0)[/tex]
[tex]X(s) = \frac{s\cdot v(0)+x(0)}{(s^{2}+\frac{8}{m} )}[/tex]
[tex]X(s) = v(0)\cdot \frac{s}{s^{2}+\frac{8}{m} } +\frac{m\cdot x(0)}{8} \cdot \frac{\frac{8}{m}}{s^{2}+\frac{8}{m}}[/tex]
[tex]x(t) = v(0) \cdot \cos \left(\frac{8}{m}\cdot t \right)+\frac{m\cdot x(0)}{8}\cdot \sin \left(\frac{8}{m}\cdot t \right)[/tex]
The velocity is obtained by deriving the previous expression:
[tex]v(t) = -\frac{8\cdot v(0)}{m}\cdot \sin \left(\frac{8}{m}\cdot t \right)+x(0)\cdot \cos \left(\frac{8}{m}\cdot t \right)[/tex]
Speed of the object at [tex]x = 0[/tex] is:
[tex]v = \sqrt{\frac{2\cdot (E_{total}-4\cdot x^{2})}{m} }[/tex]
[tex]v = \sqrt{\frac{2\cdot [7\,J- 4\cdot (0\,m)^{2}]}{3\,kg} }[/tex]
[tex]v \approx 2.160\,\frac{m}{s}[/tex]
The equation of the motion are:
[tex]x(t) = 2.160\cdot \cos \left(2.667\cdot t \right)[/tex]
[tex]v(t) = -5.76\cdot \sin (2.667\cdot t)[/tex]
The expression for the kinetic energy of the object is:
[tex]K = \frac{1}{2}\cdot m \cdot v^{2}[/tex]
[tex]K = \frac{1}{2}\cdot (3\,kg)\cdot [33.178\cdot \sin^{2}(2.667\cdot t)][/tex]
[tex]K = 49.767\cdot \sin^{2}(2.667\cdot t)[/tex]
Graphics are included below as attachments. (Position versus time, kinetic energy vs time).
Following are the responses to the given points:
Given:
object mass [tex](m)= 3.0\ kg \\\\[/tex]
Potential energy [tex]U(x)=4.0\ x^2\\\\[/tex]
[tex]\to x=-0.5 \ m\\\\ \to velocity \ (v)=2.0 \ \frac{m}{s}\\\\[/tex]
Solution:
For point (a)
Total Energy [tex](TE) = PE + KE \\\\[/tex]
[tex]\to PE = 4.0 (0.5 m)^2 = 1\ J\\\\ \to KE = \frac{1}{2} mv^2 = 0.5 \times 3.0\ kg (2.0 \frac{m}{s})^2 = 6\ J \\\\\to TE = 7\ J\\\\[/tex]
For point (b)
If an object's potential energy is 7 J, it has 0 kinetic energy.
[tex]\to U(x) = 4x^2 = 7\ J \\\\\to x=+1.2 \ m, -1.2 \ m[/tex]
For point (c)
[tex]\to x=0.60\ m \\\\ \to TE=4x^2 + \frac{1}{2}mv^2 = 7\ J\\\\ \to 4(0.60)^2 +0.5 \times 3.0 \ kg \times v^2 = 7\ J \\\\ \to v=1.92 \ \frac{m}{s}\\\\ \to Momentum (p) = mv \\\\\to p= 3.0\ kg \times 1.92 \ \frac{m}{s}\\\\ \to p= 5.76 kg \ \frac{m}{s}\\\\[/tex]
For point (d)
[tex]\to x_1= 0.6 \ m\\\\ \to v_1 = 1.92 \frac{m}{s}\\\\ \to x_2 = 1.2\ m[/tex] the velocity is found by
[tex]\to 4x^2 + \frac{1}{2} mv^2= 7\ J \\\\ \to 4(1.2)^2 +0.5 \times 3.0\ kg \times v^2 = 7\ J \\\\ \to v_2 = 0.9 \ \frac{m}{s}\\\\[/tex]
Calculating the acceleration:
[tex]\to V^2_{2}-v^2_{1}= 2a(x_2-x_1) \\\\\to (0.9 \ \frac{m}{s})^2 - (1.92\ \frac{m}{s})^2 = 2 a(1.2\ m - 0.6\ m) \\\\\to a= -2.4 \frac{m}{s^2}\\\\[/tex]
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An electron is placed in a magnetic field that is directed along a z axis. The energy difference between parallel and antiparallel alignments of the z component of the electron's spin magnetic moment with is 7.50 × 10-25 J. What is the magnitude of ?
Answer:
B = 0.0404 т
Explanation:
Given
ΔE = 7.50 x 10⁻²⁵ J
μb = 927.4 x 10⁻²⁶ J/т
ΔE = 2 * μb * B
B = ΔE / 2 μb
B = 7.50 x 10⁻²⁵ / 2 * 927.4 x 10⁻²⁶
B = 7.50 / 1854.8 x 10⁻²⁵⁻⁽⁻²⁶⁾
B = 0.00404 x 10¹
B = 0.0404 т