Answer:
Assuming that the length of the magnet is much smaller than the separation between it and the charge. As a result of magnetic interaction (i.e., ignore pure Coulomb forces) between the charge and the bar magnet, the magnet will not experience any torque at all - option A
Explanation:
Assuming that the length of the magnet is much smaller than the separation between it and the charge. As a result of magnetic interaction (i.e., ignore pure Coulomb forces) between the charge and the bar magnet, the magnet will not experience any torque at all; the reason being that: no magnetic field is being produced by a charge that is static. Only a moving charge can produce a magnetic effect. And the magnet can not have any torque due to its own magnetic lines of force.
The magnet will experience No torque at all.
Ignoring of pure Coulomb forces:Here the length of the magnet should be less than the separation between it and the charge. Due to this there is magnetic interaction (i.e., ignore pure Coulomb forces) that lies between the charge and the bar magnet, the magnet should not experience any torque at all; the reason being that: no magnetic field should be generated via a charge i.e. here a moving charge can generate a magnetic effect. And the magnet can not have any torque because of its own magnetic lines of force.
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A dockworker applies a constant horizontal force of 90.0Nto a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves a distance 12.0min a time of 5.20s.If the worker stops pushing after 5.20s, how far does the block move in the next 5.30s
Answer:
Distance traveled in the next 5.30 seconds is 24.454 meters.
Explanation:
Considering the block moves with a constant acceleration when the force is applied, the average speed would be half of the speed at the end of 5.20 seconds of acceleration.
This means:
Average Speed = Total distance / Time
Average Speed = 12 / 5.20 = 2.307 m/s
Speed at end of 5.20 seconds = 2 * Average Speed
Speed at end of 5.20 seconds = 2 * 2.307
Speed at end of 5.20 seconds = 4.614 m/s
Now that we have the speed at which the block is traveling, and there is no friction to reduce or change the speed, we can solve for the distance covered in the next 5.30 seconds in the following way:
Distance traveled in the next 5.30 seconds:
Distance = Speed * Time
Distance = 4.614 * 5.30
Distance = 24.454 meters
An Atwood machine is constructed using a hoop with spokes of negligible mass. The 2.3 kg mass of the pulley is concentrated on its rim which is a distance 23.5 cm from the axle. The mass on the right is 1 kg and on the left is 1.65 kg.
What is the magnitude of the linear acceleration a of the hanging masses?
The magnitude of the linear acceleration of the hanging masses in the given Atwood machine is approximately 2.69 m/s².
Explanation:An Atwood machine is a system comprised of two different masses connected by a string passing over a pulley. In this case, we have a pulley with a mass of 2.3 kg. The radius of the pulley is given as 23.5 cm.
To find the linear acceleration of the hanging masses, we use the equation:
a = (m1 - m2) * g / (m1 + m2 + m_pulley)
Substituting the given values, we have:
m1 = 1 kg m2 = 1.65 kg m_pulley = 2.3 kg g = 9.8 m/s²Calculating the value of a, we get:
a = (1 - 1.65) * 9.8 / (1 + 1.65 + 2.3)
a ≈ -2.69 m/s²
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A block of mass m begins at rest at the top of a ramp at elevation h with whatever PE is associated with that height. The block slides down the ramp over a distance d until it reaches the bottom of the ramp. How much of its original total energy (in J) survives as KE when it reaches the ground? (In other words, the acceleration is not zero like it was in lab and friction does not remove 100% of the original PE. How much of that original energy is left over after the friction does work to remove some?) m = 2.9 kg h = 2.2 m d = 5 m μ = 0.3 θ = 36.87°
Answer:
The original energy that is left over after the friction does work to remove some is 33.724 J
Explanation:
The original energy that is left in the system can be obtained by removing the energy loss in the system.
Given the mass m = 2.9 kg
the height h = 2.2 m
the distance d = 5 m
coefficient of friction μ = 0.3
θ = 36.87°
g = 9.8 m/[tex]s^{2}[/tex]
Since the block is at rest the initial energy can be expressed as;
[tex]E_{i} = mgh[/tex]
= 2.9 kg x 9.8 m/[tex]s^{2}[/tex] x 2.2 m
= 62.524 J
The energy loss in the system can be obtained with the expression below;
[tex]E_{loss}[/tex] = (μmgcosθ) x d
The parameters have listed above;
[tex]E_{loss}[/tex] = 0.3 x 2.9 kg x 9.8 m/[tex]s^{2}[/tex] x cos 36.87° x 5 m
[tex]E_{loss}[/tex] = 28.8 J
The original energy that is left over after the friction does work to remove some can be express as;
[tex]E = E_{i} -E_{loss}[/tex]
E = 62.524 J - 28.8 J
E = 33.724 J
Therefore the original energy that is left over after the friction does work to remove some is 33.724 J
To calculate the remaining kinetic energy of a block sliding down an incline, subtract the work done by friction from the initial potential energy, considering the mass of the block, the height of the incline, the distance slid, the coefficient of friction, and the angle of the incline.
Explanation:The question concerns the calculation of the kinetic energy (KE) of a block sliding down an incline, taking into account the work done by friction and the conservation of energy. The block starts with potential energy (PE) due to its elevation h and ends with kinetic energy at the bottom of the incline. The force of friction, which depends on the coefficient of friction μ, the gravitational acceleration, and the normal force, does work along the distance d that removes some of this energy.
Initially, the block's total mechanical energy is all potential: PE = mgh. As it slides down the ramp, work done by friction (which is a non-conservative force) is given by Wfriction = μmgcos(θ)d. The final kinetic energy of the block when it reaches the bottom of the incline is calculated by subtracting the work done by friction from the initial potential energy: KE = PE - Wfriction. Substituting the given values and doing the math will provide us with the amount of energy that remains as kinetic when the block reaches the ground.
Determine the power required for a 1150-kg car to climb a 100-m-long uphill road with a slope of 308 (from horizontal) in 12 s (a) at a constant velocity, (b) from rest to a final velocity of 30 m/s, and (c) from 35 m/s to a final velocity of 5 m/s. Disregard friction, air drag, and rolling resistance.
Explanation:
Below is an attachment containing the solution.
Answer:
the question is incomplete, below is the complete question
"Determine the power required for a 1150-kg car to climb a 100-m-long uphill road with a slope of 30 (from horizontal) in 12 s (a) at a constant velocity, (b) from rest to a final velocity of 30 m/s, and (c) from 35 m/s to a final velocity of 5 m/s. Disregard friction, air drag, and rolling resistance."
a. [tex]47KW[/tex]
b. 90.1KW
c. -10.5KW
Explanation:
Data given
mass m=1150kg,
length,l=100m
angle, =30
time=12s
Note that the power is define as the rate change of the energy, note we consider the potiential and the
Hence
[tex]Power,P=\frac{energy}{time} \\P=\frac{mgh+1/2mv^{2}}{t}[/tex]
to determine the height, we draw the diagram of the hill as shown in the attached diagram
a. at constant speed, the kinetic energy is zero.Hence the power is calculated as
[tex]p=\frac{mgh}{t} \\p=\frac{1150*9.81*100sin30}{12}\\p=47*10^{3}W\\[/tex]
b.
for a change in velocity of 30m/s
we have the power to be
[tex]P=\frac{mgh}{t} +\frac{1/2mv^2}{t}\\ P=47Kw+\frac{ 0.5*1150*30^2}{12}\\ P=47kw +43.1kw\\P=90.1KW[/tex]
c. when i decelerate, we have the power to be
[tex]P=\frac{mgh}{t} +\frac{1/2mv^2}{t}\\ P=47Kw+\frac{ 0.5*1150*\\((5^2)-(35^2))}{12}\\ P=47kw -57.5kw\\P=-10.5KW[/tex]
Suppose you hit a 0.058-kg tennis ball so that the ball then moves with an acceleration of 10 m/s2. If you were to hit a basketball of mass 0.58 kg with the same force, what would the acceleration a of the basketball be?
Answer:
1 m/s²
Explanation:
Force = mass × acceleration
F = ma ............ Equation 1
Where F = force, m = mass, a = acceleration.
Given: m = 0.058 kg, a = 10 m/s²
Substitute into equation 1
F = 0.058(10)
F = 0.58 N.
If the same force was used to hit the baseball,
F = m'a
a = F/m'.............. Equation 2
Where M' = mass of the baseball.
Given: F = 0.58 N, m' = 0.58 kg.
Substitute into equation 2
a = 0.58/0.58
a = 1 m/s²
Answer:
1 m/s^2.
Explanation:
Note:
Force = mass x acceleration.
Given:
mass = 0.058 kg
acceleration = 10 m/s2
Therefore, force = 0.058 x 10
= 0.58 N.
Since the same force is to be used, this same value is used for the other condition.
mass = 0.58 kg
Force = 0.58 N
F = m × a
a = 0.58/0.58
= 1 m/s^2.
Which most simplified form of the law of conservation of energy describes the motion of the block as it slides on the floor from the bottom of the ramp to the moment it stops?
The principle of mechanical energy conservation describes the motion of the block as it slides on the floor from the bottom of the ramp to the moment it stops.
Explanation:The most simplified form of the law of conservation of energy that describes the motion of the block as it slides on the floor from the bottom of the ramp to the moment it stops is the principle of mechanical energy conservation. This principle states that the total mechanical energy of a system remains constant as long as no external forces do work on the system. In this case, as the block slides, the potential energy it loses due to its change in height is transformed into kinetic energy until the block stops and all of its energy is in the form of kinetic energy.
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A deuteron, with the same charge but twice the mass of a proton, moves with a speed of 6.00 × 105 m/s perpendicular to a uniform magnetic field of 0.0525 T. Which of the paths described below would it follow? (qp = 1.60 × 10−19 C and md = 3.34 × 10−27 kg)
Explanation:
The given data is as follows.
v = [tex]6.00 \times 10^{5} m/s[/tex]
B = 0.0525 T, q = [tex]1.60 \times 10^{-19}[/tex]
m = [tex]3.34 \times 10^{-27} kg[/tex]
It is known that relation between mass and magnetic field is as follows.
[tex]\frac{mv^{2}}{r} = Bvq[/tex]
or, r = [tex]\frac{mv^{2}}{Bvq}[/tex]
So, putting the given values into the above formula and we will calculate the radius as follows.
r = [tex]\frac{mv^{2}}{Bvq}[/tex]
= [tex]\frac{3.34 \times 10^{-27} kg \times 6.00 \times 10^{5} m/s}{0.0525 T \times 1.60 \times 10^{-19}}[/tex]
= [tex]\frac{20.04 \times 10^{-22}}{0.084 \times 10^{-19}}[/tex]
= 0.238 m
Thus, we can conclude that radius of the circular path is 0.238 m.
Answer:
The radius is [tex]238.57\times10^{-3}\ m[/tex]
Explanation:
Given that,
Speed [tex]v=6.00\times10^{5}\ m/s[/tex]
Magnetic field = 0.0525 T
We need to calculate the radius
Using relation of centripetal force and magnetic force
[tex]F=qvB[/tex]
[tex]\dfrac{mv^2}{r}=qvB[/tex]
[tex]r=\dfrac{mv^2}{qvB}[/tex]
[tex]r=\dfrac{mv}{qB}[/tex]
Put the value into the formula
[tex]r=\dfrac{3.34\times10^{-27}\times6.00\times10^{5}}{1.60\times10^{-19}\times0.0525}[/tex]
[tex]r=0.238\ m[/tex]
[tex]r=238.57\times10^{-3}\ m[/tex]
Hence, The radius is [tex]238.57\times10^{-3}\ m[/tex]
P7.16 A thin flat plate 55 by 110 cm is immersed in a 6-m/s stream of SAE 10 oil at 20C. Compute the total friction drag if the stream is parallel to (a) the long side and (b) the short side.
Answer:
a
The total friction drag for the long side of the plate is 107 N
b
The total friction drag for the long side of the plate is 151.4 N
Explanation:
The first question is to obtain the friction drag when the fluid i parallel to the long side of the plate
The block representation of the this problem is shown on the first uploaded image
Where the U is the initial velocity = 6 m/s
So the equation we will be working with is
[tex]F = \frac{1}{2} \rho C_fAU^2[/tex]
Where [tex]\rho[/tex] is the density of SAE 10W = [tex]870\ kg/m^3[/tex] This is obtained from the table of density at 20° C
[tex]C_f[/tex] is the friction drag coefficient
This coefficient is dependent on the Reynolds number if the Reynolds number is less than [tex]5*10^5[/tex] then the flow is of laminar type and
[tex]C_f[/tex] = [tex]\frac{1.328}{\sqrt{Re} }[/tex]
But if the Reynolds number is greater than [tex]5*10^5[/tex] the flow would be of Turbulent type and
[tex]C_f = \frac{0.074}{Re_E^{0.2}}[/tex]
Where Re is the Reynolds number
To obtain the Reynolds number
[tex]Re = \frac{\rho UL}{\mu}[/tex]
where L is the length of the long side = 110 cm = 1.1 m
and [tex]\mu[/tex] is the Dynamic viscosity of SAE 10W oil [tex]= 1.04*10^{-1} kg /m.s[/tex]
This is gotten from the table of Dynamic viscosity of oil
So
[tex]Re = \frac{870 *6*1.1}{1.04*10^{-1}}[/tex]
[tex]= 55211.54[/tex]
Since 55211.54 < [tex]5.0*10^5[/tex]
Hence
[tex]C_f = \frac{1.328}{\sqrt{55211.54} }[/tex]
[tex]= 0.00565[/tex]
[tex]A[/tex] is the area of the plate = [tex]\frac{ (110cm)(55cm)}{10000}[/tex] =[tex]0.55m^2[/tex]
Since the area is immersed totally it should be multiplied by 2 i.e the bottom face and the top face are both immersed in the fluid
[tex]F = \frac{1}{2} \rho C_f(2A)U^2[/tex]
[tex]F =\frac{1}{2} *870 *0.00565*(2*0.55)*6^2[/tex]
[tex]F = 107N[/tex]
Considering the short side
To obtain the Reynolds number
[tex]Re = \frac{\rho U b}{\mu}[/tex]
Here b is the short side
[tex]Re =\frac{870*6*0,55}{1.04*10^{-1}}[/tex]
[tex]=27606[/tex]
Since the value obtained is not greater than [tex]5*10^5[/tex] then the flow is laminar
And
[tex]C_f = \frac{1.328}{\sqrt{Re} }[/tex]
[tex]= \frac{1.328}{\sqrt{27606} }[/tex]
[tex]= 0.00799[/tex]
The next thing to do is to obtain the total friction drag
[tex]F = \frac{1}{2} \rho C_f(2A)U^2[/tex]
Substituting values
[tex]F = \frac{1}{2} * 870 * 0.00799 * 2( 0.55) * 6^2[/tex]
[tex]= 151.4 N[/tex]
Final answer:
To calculate the total friction drag on the plate immersed in the given fluid stream, use the drag force formula. Compute the drag for both the long and short sides by multiplying the relevant dimensions with the velocity and viscosity.
Explanation:
The drag force per unit area on the plate is given by Fdrag = μSA, where μ is the viscosity of the fluid and A is the area of the plate. Here, the total friction drag can be calculated by multiplying the drag force per unit area by the total area of the plate.
For the long side: Total friction drag = μ(6)(0.55)(1.1)
For the short side: Total friction drag = μ(6)(0.11)(1.1)
To apply the law of conservation of energy to an object launched upward in Earth's gravitational field.
In the absence of nonconservative forces such as friction and air resistance, the total mechanical energy in a closed system is conserved. This is one particular case of the law of conservation of energy.
In this problem, you will apply the law of conservation of energy to different objects launched from Earth. The energy transformations that take place involve the object's kinetic energy K=(1/2)mv^2 and its gravitational potential energy U=mgh. The law of conservation of energy for such cases implies that the sum of the object's kinetic energy and potential energy does not change with time. This idea can be expressed by the equation
K_{\rm i}+U_{\rm i}=K_{\rm f}+U_{\rm f}\;\;\;\;,
where "i" denotes the "initial" moment and "f" denotes the "final" moment. Since any two moments will work, the choice of the moments to consider is, technically, up to you. That choice, though, is usually suggested by the question posed in the problem.
Using conservation of energy, find the maximum height h_max to which the object will rise
Answer:
The maximum height is [tex]h_{max} = \frac{v^2}{2g}[/tex]
Explanation:
From the question we are given that
[tex]K_{\rm i}+U_{\rm i}=K_{\rm f}+U_{\rm f}\;\;\;\;,[/tex]
and [tex]K=(1/2)mv^2[/tex]
while [tex]U=mgh[/tex]
Now at the minimum height the kinetic energy is maximum and the potential energy is 0
[tex]K_i \ is \ max[/tex]
[tex]U_i = 0[/tex]
At maximum height [tex]h_{max}[/tex]
The kinetic energy is 0 and kinetic energy is
Hence the above equation
[tex]K_i = U_f[/tex]
[tex]\frac{1}{2}mv^2 = mgh_{max}[/tex]
Making [tex]h_{max}[/tex] the subject of the formula we have
[tex]h_{max} = \frac{v^2}{2g}[/tex]
A relatively large plate of a glass is subjected to a tensile stress of 50 MPa. If the specific surface energy and modulus of elasticity for this glass are 0.5 J/m2 and 80 GPa, respectively, determine the maximum length of a surface flaw that is possible without fracture.
Answer:
The value of the maximum length of a surface flaw that is possible without fracture a = 1.02 × [tex]10^{-11}[/tex] mm
Explanation:
Given data
Tensile stress [tex]\sigma[/tex] = 50 [tex]\frac{N}{mm^{2} }[/tex]
Specific surface energy [tex]\gamma_{s}[/tex] = 0.5 [tex]\frac{J}{m^{2} }[/tex] = 0.5 × [tex]10^{-6}[/tex] [tex]\frac{J}{mm^{2} }[/tex]
Modulus of elasticity E = 80 × [tex]10^{3}[/tex] [tex]\frac{N}{mm^{2} }[/tex]
The critical stress is given by [tex]\sigma_{c}^{2}[/tex] = [tex]\frac{2 E \gamma_{s} }{\pi a}[/tex] ----- (1)
In the limiting case [tex]\sigma[/tex] = [tex]\sigma_{c}[/tex]
⇒ [tex]\sigma^{2}[/tex] = [tex]\frac{2 E \gamma_{s} }{\pi a}[/tex] ------ (2)
Put all the values in above formula we get,
⇒ [tex]50^{2}[/tex] = 2 × 80 × [tex]10^{3}[/tex] × 0.5 × [tex]10^{-6}[/tex] × [tex]\frac{1}{3.14}[/tex] × [tex]\frac{1}{a}[/tex]
⇒ a = 1.02 × [tex]10^{-11}[/tex] mm
This is the value of the maximum length of a surface flaw that is possible without fracture.
A steam stream of 400 m3 /s, standard (standard cubic meter per second) is cooled by adding 7 m3 /s cold water. The fluid leaving the system is liquid water. The crosssectional area of the outlet port is 0.5 m2 . What is the velocity of the leaving water stream? (Hint: steam is a gas. Its specific gravity is about 18/29.)
Answer: 510 m/s
Explanation: specific gravity of steam is 18/29 = 0.620
It is the ratio of the density of steam over density of water
400m3/s of steam =
400m3ms * 0.620 of water
= 248m3/s of water
Total flow rate Q = 248 + 7 = 255m3/s
Using Q = AV
Where A is area and V is velocity
V = Q/A
V = 255/0.5 = 510m/s
three charged particles lie on a straight line and are separated by distances d. Charges q1 and q2 are held fixed. Charge q3 is free to move but happens to be in equilibrium (no net electrostatic force acts on it)
Answer:
[tex]\boxed {q_1=-4q_2}[/tex]
Explanation:
Using the attached figure
Considering that the distance of separation is 2d then
[tex]F_1=\frac {q_1q_3}{4\pi\epsilon_o(2d)^{2}}[/tex]
Also, considering that distance of separation between and is d then
[tex]F_2=\frac {q_1q_3}{4\pi\epsilon_o(d)^{2}}[/tex]
The net force acting on is
[tex]F=F_1+F_2=0\\F=\frac {q_1q_3}{4\pi\epsilon_o(2d)^{2}}+ \frac {q_1q_3}{4\pi\epsilon_o(d)^{2}}=0\\F=\frac {q_3}{4\pi \epsilon_o d^{2}}(q_2+0.25q_1)=0\\F=0.25q_1+q_2=0[/tex]
Therefore
[tex]\boxed {q_1=-4q_2}[/tex]
The question pertains to a classical physics problem dealing with the equilibrium of three charged particles along a line. It is solved using Coulomb's law to balance the forces acting on the 3rd charge, leading to a condition that determines the values of the charges.
Explanation:This situation falls under the domain of Physics, specifically the study of electromagnetism. When the charges are in equilibrium, it means the net electrostatic force acting on the third charge, q3, is zero. This equilibrium condition allows us to create an equation. The electrostatic force F between two charges q1 and q2 separated by distance d is described by Coulomb's law: F = k*q1*q2/d^2, where k is Coulomb's constant. It follows then that for q3 to be in equilibrium, the forces from q1 and q2 must balance out. That is, the force of attraction or repulsion between q1 and q3 must equal the force between q2 and q3.
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A farmer uses a tractor to pull a 150 kg bale of hay up a 15∘ incline to the barn at a steady 5.0 km/h. The coefficient of kinetic friction between the bale and the ramp is 0.40.Part A What is the tractor's power output?
Answer:
the tractor's power output = 1318.47 watts
Explanation:
Weight of bale = (mg) = (150 kg * 9.81 m/s²) = 1471.5 newtons
Resolve that weight force into its components at right angles to the ramp and parallel to the slope.
Down the slope we have 1471.5sin(15°) = 380.85 N
At right angles to the ramp we have 1471.5cos(15°) = 1421.35 N
OK, now we need the relationship between the coefficient of friction (μ), the friction force (Ff) and the normal force (Fn) {The normal force is the force at right angles to the friction surface. In this case Fn is equal in magnitude to the component of weight force at right angles to the surface.}
Ff = μ * Fn = (0.40 * 1421.35) = 568.54 N
The bale is not accelerating, so the "pull force" up the incline = component of weight + friction force down
= (380 N + 568.54 N)
= 948.54N
We need the speed in m/s not km/h
5.0 km/h = 5000 m/h
= (5000/3600)
= 1.39 m/s
Power = (948.54 N * 1.39m/s)
= 1318.47 N.m/s
= 1318.47 watts
The power output of the tractor is 1318.47 W
Power:According to the question the weight of bale
mg = 150 × 9.81 = 1471.5 N
Resolving the weight we get:
The normal reaction perpendicular to the ramp is given by:
N = 1471.5cos(15°) = 1421.35 N
The force of friction is given by:
[tex]f = \mu N = 0.40 \times 1421.35\\\\f = 568.54 N[/tex]
The bale is not accelerating, so the force up the incline = component of weight + friction force down
F = mgsin(15°) + f
F = 380 N + 568.54 N
F = 948.54N
Now, the engine speed is:
v = 5.0 km/h = 5000 m/h
v = (5000/3600)
v = 1.39 m/s
The power is defined as the product of force and speed, so:
Power P = Fv
P = 948.54 × 1.39
P = 1318.47 W
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There are two important isotopes of uranium, 235U
and 238U; these isotopes are nearly identical chemically
but have different atomic masses. Only 235U is very
useful in nuclear reactors. Separating the isotopes is called
uranium enrichment (and is often in the news as of this
writing, because of concerns that some countries are
enriching uranium with the goal of making nuclear
weapons.) One of the techniques for enrichment, gas
diffusion, is based on the different molecular speeds of
uranium hexafluoride gas, UF6 . (a) The molar masses of
235U and 238UF6 are 349.0 g/mol and 352.0 g/mol,
respectively. What is the ratio of their typical speeds vrms ?
(b) At what temperature would their typical speeds differ by
1.00 m/s? (c) Do your answers in this problem imply that
this technique may be difficult?
Answer:
a) (vᵣₘₛ₁/vᵣₘₛ₂) = 1.00429
where vᵣₘₛ₁ represents the vᵣₘₛ for 235-UF6 and vᵣₘₛ₂ represents the vᵣₘₛ for 238-UF6.
b) T = 767.34 K
c) The answers point to a difficult seperation technique, as the two compounds of the different isotopes have very close rms speeds and to create a difference of only 1 m/s In their rms speeds would require a high temperature of up to 767.34 K.
Explanation:
The vᵣₘₛ for an atom or molecule is given by
vᵣₘₛ = √(3RT/M)
where R = molar gas constant = J/mol.K
T = absolute temperature in Kelvin
M = Molar mass of the molecules.
₁₂
Let the vᵣₘₛ of 235-UF6 be vᵣₘₛ₁
And its molar mass = M₁ = 349.0 g/mol
vᵣₘₛ₁ = √(3RT/M₁)
√(3RT) = vᵣₘₛ₁ × √M₁
For the 238-UF6
Let its vᵣₘₛ be vᵣₘₛ₂
And its molar mass = M₂ = 352.0 g/mol
√(3RT) = vᵣₘₛ₂ × √M₂
Since √(3RT) = √(3RT)
vᵣₘₛ₁ × √M₁ = vᵣₘₛ₂ × √M₂
(vᵣₘₛ₁/vᵣₘₛ₂) = (√M₂/√M₁) = [√(352)/√(349)]
(vᵣₘₛ₁/vᵣₘₛ₂) = 1.00429
b) Recall
vᵣₘₛ₁ = √(3RT/M₁)
vᵣₘₛ₂ = √(3RT/M₂)
(vᵣₘₛ₁ - vᵣₘₛ₂) = 1 m/s
[√(3RT/M₁)] - [√(3RT/M₂)] = 1
R = 8.314 J/mol.K, M₁ = 349.0 g/mol = 0.349 kg/mol, M₂ = 352.0 g/mol = 0.352 kg/mol, T = ?
√T [√(3 × 8.314/0.349) - √(3 × 8.314/0.352) = 1
√T (8.4538 - 8.4177) = 1
√T = 1/0.0361
√T = 27.7
T = 27.7²
T = 767.34 K
c) The answers point to a difficult seperation technique, as the two compounds of the different isotopes have very close rms speeds and to create a difference of only 1 m/s In their rms speeds would require a high temperature of up to 767.34 K.
4.77 Augment the rectifier circuit of Problem 4.70 with a capacitor chosen to provide a peak-to-peak ripple voltage of (i) 10% of the peak output and (ii) 1% of the peak output. In each case: (a) What average output voltage results? (b) What fraction of the cycle does the diode conduct? (c) What is the average diode current? (d) What is the peak diode current?
To augment the rectifier circuit with a capacitor, connect it in parallel with the load resistor. Calculate the capacitor value to achieve the desired ripple voltage. Determine the average output voltage, fraction of cycle during diode conduction, average diode current, and peak diode current using the ripple voltage and peak output voltage.
Explanation:To augment the rectifier circuit with a capacitor, we need to connect the capacitor in parallel with the load resistor. The peak-to-peak ripple voltage depends on the value of the capacitor. To calculate the average output voltage, we need to consider the ripple voltage and the peak output voltage. The fraction of the cycle during which the diode conducts, average diode current, and peak diode current can also be determined using the ripple voltage and peak output voltage.
(a) To achieve a peak-to-peak ripple voltage of 10% of the peak output, we can calculate the capacitor value using the equation Vr = (1/2πfc) * (Id / Ic) * Vp, where Vr is the ripple voltage, Id is the diode current, Ic is the capacitor current, Vp is the peak output voltage, f is the frequency, and c is the capacitor value. With the calculated capacitor value, we can find the average output voltage.
(b) The fraction of the cycle during which the diode conducts can be calculated by dividing the time during which the diode conducts by the total time of the cycle.
(c) The average diode current can be calculated by dividing the total charge passing through the diode by the total time period of the cycle.
(d) The peak diode current can be calculated by multiplying the average current by the diode conduction time.
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Consider a laser pointer that emits red light with wavelength 650 nm. This light is used for a photoelectric effect experiment where the anode in the evacuated glass tube is made up of a material that has work function equal to 1 eV. 1. What is the energy of an individual photon that comes out of the laser pointer?
Answer:
The energy of an individual photon that comes out of the laser pointer is 1.91 eV
Explanation:
The energy of a photon can be obtained using the expression below
E = hc/λ
where E is the energy;
h is the Planck's constant = 6.626 x 10-34 Js;
c is the speed of light = 3.00 x [tex]10^{8}[/tex] m/s (speed of light);
λ is the wavelength = 650 nm =650 x [tex]10^{-9}[/tex] m.
E = (6.626 x 10-34 Js) x (3.00 x [tex]10^{8}[/tex] m/s) /650 x [tex]10^{-9}[/tex] m
E = 3.058 x [tex]10^{-19}[/tex] J
1 joule = 6.242 x [tex]10^{18}[/tex] eV
3.058 x [tex]10^{-19}[/tex] J = 3.058 x [tex]10^{-19}[/tex] J x 6.242 x [tex]10^{18}[/tex] eV = 1.91 eV
Therefore the energy of an individual photon that comes out of the laser pointer is 1.91 eV
You give a crate a push to set it in motion. After the push is over, the crate skids to a stop on a rough, level floor. Describe the energy changes after the push (with the crate as the system).
Answer: potential to kinetic and then potential energies
Explanation:
Answer:
KE₁ = PE₂
Explanation:
From the principle of conservation of energy we know that
KE₁ + PE₁ = KE₂ + PE₂
where KE is the kinetic energy and PE is potential energy
KE = ½mv²
PE = mgd
Initially, you gave a push to crate and it started moving with some velocity
KE₁ = ½mv₁²
Initially, the potential energy is zero since it didnt cover any distance
PE₁ = mgd₁ = 0
At the final state, the crate stopped and its finally velocity become zero, therefore, the kinetic energy is zero.
KE₂ = ½mv₂² = 0
At the final state, the crate covered some distance and the final potential energy is
PE₂ = mgd₂
Therefore, the energy relation becomes
KE₁ + PE₁ = KE₂ + PE₂
KE₁ + 0 = 0 + PE₂
KE₁ = PE₂
Hence, the initial kinetic energy of the system is converted to final potential energy of the system.
A spaceship flies from Earth to a distant star at a constant speed. Upon arrival, a clock on board the spaceship shows a total elapsed time of 8 years for the trip. An identical clock on Earth shows that the total elapsed time for the trip was 10 years. What was the speed of the spaceship relative to the Earth?
Answer:
35 288 mile/sec
Explanation:
This is a problem of special relativity. The clocks start when the spaceship passes Earth with a velocity v, relative to the earth. So, out and back from the earth it will take:
[tex]10 years = \frac{2d}{v}[/tex]
If we use the Lorentz factor, then, as observed by the crew of the ship, the arrival time will be:
[tex]0.8 = \sqrt{1-\frac{v^{2} }{c^{2} } }[/tex]
Then the amount of time wil expressed as a reciprocal of the Lorentz factor. Thus:
[tex]0.8 = \sqrt{1 - \frac{v^{2} }{c^{2} } }[/tex]
[tex]0.64 = 1-\frac{v^{2} }{186282^{2} }[/tex]
solving for v, gives = 35 288 miles/s
Compute VO when Vin = 0.5 V, in two different ways: a) using the equation VO = G (V+-V-) with G = 106; and b) using the Golden Rules. In both cases you can assume that the currents into the input terminals of the op-amp are negligible.
Final answer:
To compute Vout with Vin = 0.5 V: a) Using the equation VO = G (V+-V-) with G = 10⁶ would theoretically give 500,000 V, but this is practically limited by the op-amp's supply voltage. b) Using the Golden Rules, assuming an ideal op-amp, Vout would be equal to Vin, thus Vout = 0.5 V.
Explanation:
To compute Vout when Vin = 0.5 V, we can use two different methods:
a) Using the equation VO = G (V+-V-) with G = 10⁶, we assume that for an ideal operational amplifier (op-amp), the voltage at the non-inverting input (V+) is equal to the voltage at the inverting input (V-). Since the question suggests that V- is at circuit common and therefore is 0 V, we have VO = G * (V+ - V-). Substituting values gives VO = 10⁶ * (0.5 V - 0 V) = 500,000 V. However, because real op-amps cannot output such high voltages, Vout would typically be limited by the supply voltage of the op-amp.
b) Using the Golden Rules of op-amps, which state that no current flows into the input terminals of an op-amp and the voltage at the inverting terminal is equal to the voltage at the non-inverting terminal, we can infer that because Vin = 0.5 V is applied to the non-inverting terminal and no current flows into the op-amp, then the voltage difference across the inputs is zero. Thus, Vout must also be 0.5 V to maintain this condition. Therefore, Vout = Vin = 0.5 V under the ideal op-amp approximation.
Consider the following systems: I) water behind a dam; II) a swinging pendulum; III) an apple on an apple tree; IV) the space shuttle in orbit. In which of the systems is potential energy present?
Final answer:
Potential energy is present in water behind a dam and in a swinging pendulum.
Explanation:
In the given systems, potential energy is present in water behind a dam and in the swinging pendulum.
Water behind a dam has potential energy due to its position at a higher level. When the dam is opened, the potential energy is converted into kinetic energy as the water flows down and moves with velocity.
A swinging pendulum also exhibits potential energy. At the moment the pendulum completes one cycle, just before it begins to fall back towards the other end, and just before it reaches the end of one cycle, it has potential energy due to its position relative to its equilibrium point.
When the temperature of 2.35 m^3 of a liquid is increased by 48.5 degrees Celsius, it expands by 0.0920 m^3. What is its coefficient of volume expansion?
Answer:
a liquid is increased by 48.5 degrees Celsius, it expands by 0.0920 m^3. Explanation:
The coefficient of volume expansion is [tex]0.000810 \, \text{per} \, ^\circ\text{C}.[/tex]
The coefficient of volume expansion (\(\beta\)) can be calculated using the formula:
[tex]\[\Delta V = V_0 \beta \Delta T\][/tex]
where:
- [tex]\(\Delta V\)[/tex] is the change in volume,
- [tex]\(V_0\)[/tex] is the initial volume,
- [tex]\(\beta\)[/tex] is the coefficient of volume expansion,
- [tex]\(\Delta T\)[/tex] is the change in temperature.
We need to solve for [tex]\(\beta\):[/tex]
[tex]\[\beta = \frac{\Delta V}{V_0 \Delta T}\][/tex]
Substituting the given values:
[tex]\[\beta = \frac{0.0920 \, \text{m}^3}{2.35 \, \text{m}^3 \times 48.5 \, ^\circ\text{C}}\][/tex]
First, calculate the denominator:
[tex]\[2.35 \, \text{m}^3 \times 48.5 \, ^\circ\text{C} = 113.575 \, \text{m}^3 \cdot ^\circ\text{C}\][/tex]
Now, calculate [tex]\(\beta\)[/tex]:
[tex]\[\beta = \frac{0.0920 \, \text{m}^3}{113.575 \, \text{m}^3 \cdot ^\circ\text{C}}\][/tex]
[tex]\[\beta \approx 0.000810 \, \text{per} \, ^\circ\text{C}\][/tex]
Therefore, the coefficient of volume expansion is approximately [tex]\[\beta \approx 8.10 \times 10^{-4} \, ^\circ\text{C}^{-1}\][/tex].
As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff springs arranged side by side so that they are parallel to each other. When you push the platform, you compress the springs. You do an amount of work of 79.0 J when you compress the springs a distance of 0.190 m from their uncompressed length.(a) What magnitude of force must you apply to hold the platform in this position?
Explanation:
The given data is as follows.
Work done by the force, (W) = 79.0 J
Compression in length (x) = 0.190 m
So, formula for parallel combination of springs equivalent is as follows.
[tex]k_{eq} = K_{1} + K_{2}[/tex]
= 2k
Hence, work done is as follows.
W = [tex]\frac{1}{2}k_{eq} \times x^{2}[/tex]
[tex]k_{eq} = \frac{2W}{x^{2}}[/tex]
= [tex]\frac{2 \times 79}{(0.19)^{2}}[/tex]
= [tex]\frac{158}{0.0361}[/tex]
= 4376.73 N/m
Hence, magnitude of force required to hold the platform is as follows.
F = [tex]k_{eq}x[/tex]
= [tex]4376.73 N/m \times 0.19 m[/tex]
= 831.58 N
Thus, we can conclude that magnitude of force you must apply to hold the platform in this position is 831.58 N.
In an insulated container, liquid water is mixed with ice. What can you conclude about the phases present in the container when equilibrium is established?
a. There would be ice only.
b. There would be liquid water only.
c. There would be both ice and liquid water.
d. There is no way of knowing the phase composition without more information.
Answer:
d. There is no way of knowing the phase composition without more information.
Explanation:
There is no way to ascertain the phases present in the container when equilibrium is established because we are not furnished with enough information.
The most important information needed to have a deeper understanding and provide solution is temperature. Liquid water and ice can both exist at the same temperature. So we cannot conclude based on the information at hand.In this example, if the emf of the 4 V battery is increased to 15 V and the rest of the circuit remains the same, what is the potential difference Vab?
Answer:
In this example, if the emf of the 4 V battery is increased to 15 V and the rest of the circuit remains the same, what is the potential difference Vab?
The image of the circuit has been attached
At 12 emf Vab = 9.5 V
At 15 emf Vab = 12.94 V
Explanation:
Kirchhoff loop rule states that the sum of the currents coming into a junction equals the sum of the currents going out of a junction. That is to say that the sum is equal to zero.
Calculations
The voltage in a circuit can be calculated using the expression;
V= IR .............1
since the Applying Kirchhoff rule to the circuit we have;
Calculating Vab (the voltage across ab) when the emf is 12 v
let us obtain the value of the current flowing across the circuit using equation 1 and Kirchhoff loop rule
+12 - (I x 2) -(I x 3)-(I x 4)-(4)-(I x 7) = 0
I = 8/16 = 0.5 A
calculating the voltage across Vab we have;
Vab = 4V + (I x 7) + (I x 4)
Vab = 4V + (0.5 x 7) + (0.5 x 4)
Vab = 4 +3.5+2
Vab = 9.5 V
at 12v emf Vab is 9.5V
calculating Vab at 15 emf value using equation and also Kirchhoff's loop rule we have;
+15 - (I x 4) -(I x 3)-(I x 2)-(12)-(I x 7) = 0
I = 3/16
I =0.1875 A
Vab = 15 V -(I x 7) - (I x 4)
Vab = 15 - ( 0.1875 x 4)-(0.1875 x 7)
Vab = 15 - 0.75-1.3125
Vab = 12.94 V
The 10-lb block has a speed of 4 ft>s when the force of F = (8t2) lb is applied. Determine the velocity of the block when it moves s = 30 ft. The coefficient of kinetic friction at the surface is ms = 0.2.
v=6ft/sec
To determine the velocity of the block when it moves a distance of 30 ft, we can use the equations of motion and consider the force applied and the friction acting on the block. First, let's find the acceleration of the block using the force applied. Next, we need to consider the frictional force acting on the block. Finally, we can calculate the net force and use the equation of motion to determine the velocity of the block.
Explanation:To determine the velocity of the block when it moves a distance of 30 ft, we can use the equations of motion and consider the force applied and the friction acting on the block.
First, let's find the acceleration of the block using the force applied. The force applied is given by F = 8t^2 lb, where t is the time in seconds. So, at t = 0, the force is F = 0 lb, and at t = 1, the force is F = 8 lb.
We can now use the equations of motion to find the acceleration. Since the mass of the block is not given, we can assume it to be 10 lb (as mentioned in the question). From the second law of motion, F = ma, where F is the net force, m is the mass, and a is the acceleration. Therefore, 8 = 10a. Solving for a, we get a = 0.8 ft/s^2.
Next, we need to consider the frictional force acting on the block. The coefficient of kinetic friction is given as μs = 0.2. The frictional force, Ff, can be calculated using Ff = μs * N, where N is the normal force. The normal force is equal to the weight of the block, N = mg, where g is the acceleration due to gravity, approximately 32 ft/s^2. So, N = 10 * 32 = 320 lb. Substituting the values, we get Ff = 0.2 * 320 = 64 lb.
Now, we can calculate the net force acting on the block. The net force, Fn, is given as Fn = F - Ff, where F is the force applied. Substituting the values, Fn = 8t^2 - 64 lb.
Finally, we can use the equation of motion, v^2 = u^2 + 2as, to determine the velocity of the block when it moves a distance of 30 ft. Here, v is the final velocity, u is the initial velocity (which is 0 ft/s because the block starts from rest), a is the acceleration, and s is the distance traveled. Substituting the values, v^2 = 0 + 2 * 0.8 * 30. Solving for v, we find
v = 6 ft/s.
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Suppose that you wish to construct a simple ac generator having 64 turns and an angular velocity of 377 radians/second (this is the frequency point of 60 Hz). A uniform magnetic field of 0.050 T is available. If the area of the rotating coil is 0.01 m 2, what is the maximum output voltage?
Answer:
[tex]\epsilon_{max} =12.064\ V[/tex]
Explanation:
Given,
Number of turns, N = 64
angular velocity, ω = 377 rad/s
Magnetic field, B = 0.050 T
Area, a = 0.01 m²
maximum output voltage = ?
We know,
[tex]\epsilon_{max} = NBA\omega[/tex]
[tex]\epsilon_{max} = 64\times 0.05\times 0.01\times 377[/tex]
[tex]\epsilon_{max} =12.064\ V[/tex]
The maximum output voltage is equal to 12.064 V.
An ambulance is traveling north at 60.3 m/s, approaching a car that is also traveling north at 33.4 m/s. The ambulance driver hears his siren at a frequency of 696 Hz. Ambulance 60.3 m/s 33.4 m/s Car What is the wavelength at any position in front of the ambulance for the sound from the ambulance’s siren? The velocity of sound in air is 343 m/s. Answer in units of m.
Answer:
0.44999 m
Explanation:
f = Actual wavelength = 696 Hz
v = Speed of sound in air = 343 m/s
[tex]v_o[/tex] = Velocity of observer = 33.4 m/s
[tex]v_s[/tex] = Velocity of source = 60.3 m/s
From Doppler's effect we have
[tex]f_o=f\left(\dfrac{v-v_o}{v-v_s}\right)\\\Rightarrow f_o=696\left(\dfrac{343-33.4}{343-60.3}\right)\\\Rightarrow f_o=762.22709\ Hz[/tex]
Wavelength is given by
[tex]\lambda=\dfrac{v}{f}\\\Rightarrow \lambda=\dfrac{343}{762.22709}\\\Rightarrow \lambda=0.44999\ m[/tex]
The wavelength at any position in front of the ambulance for the sound from the ambulance’s siren is 0.44999 m.
As an object moves from point A to point B only two forces act onit: one force is nonconservative and does -30 J of work, the otherforce is conservative and does +50 J of work. Between A andB,
a. the kineticenergy of object increases, mechanical energydecreases.
b. the kineticenergy of object decreases, mechanical energydecreases.
c. the kineticenergy of object decreases, mechanical energyincreases.
d. the kineticenergy of object increases, mechanical energyincreases.
e. None of theabove.
To solve this problem we will apply the principles of energy conservation. On the one hand we have that the work done by the non-conservative force is equivalent to -30J while the work done by the conservative force is 50J.
This leads to the direct conclusion that the resulting energy is 20J.
The conservative force is linked to the movement caused by the sum of the two energies, therefore there is an increase in kinetic energy. The decrease in the mechanical energy of the system is directly due to the loss given by the non-conservative force, therefore there is a decrease in mechanical energy.
Therefore the correct answer is A. Kintetic energy increases and mechanical energy decreases.
Based on the given question, we can infer that a. the kinetic energy of object increases, mechanical energy decreases.
What is Kinetic Energy?This refers to the form of energy which makes use of the motion of an object to move.
Hence, we can see that based on the principles of energy conservation, we can see that the work done by the non-conservative force is equivalent to -30J while the work done by the conservative force is 50J.
With this in mind, we can see that the resulting energy is 20J.
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You need to purchase a motor to supply 400 joules [J] in 10 seconds [s]. All of the motors you can choose from are 80% efficient. What is the minimum wattage [W] on the motor you need to choose
Answer:
32 W.
Explanation:
Power: This can be defined as the ratio of energy to time. The S.I unit of power is watt(W). The formula for power is given as,
P = W/t.................... Equation 1
Where P = power, W = work done, t = time.
Given: W = 400 J, t = 10 s.
Substitute into equation 1
P = 400/10
P = 40 W.
If the motors that is choose is 80% efficient,
P' = P(0.8)
Where P' = minimum power
P' = 40(0.8)
P' = 32 W.
A flat coil of wire has an area A, N turns, and a resistance R. It is situated in a magnetic field, such that the normal to the coil is parallel to the magnetic field. The coil is then rotated through an angle of 90°, so that the normal becomes perpendicular to the magnetic field. The coil has an area of 1.5 10-3 m2, 50 turns, and a resistance of 166 Ω. During the time while it is rotating, a charge of 7.3 10-5 C flows in the coil. What is the magnitude of the magnetic field?
Explanation:
Expression for magnitude of the induced emf is as follows.
[tex]\epsilon = N \frac{BA}{t}[/tex]
[tex]\frac{Q}{t}R = \frac{NBA}{t}[/tex]
So, magnitude of the magnetic field is as follows.
B = [tex]\frac{RQ}{A \times N}[/tex]
It is given that,
A = [tex]1.5 \times 10^{-3} m^{2}[/tex]
Q =[tex]7.3 \times 10^{-5} C[/tex]
N = 50
R = 166 [tex]\ohm[/tex]
Putting the given values into the above formula as follows.
B = [tex]\frac{RQ}{A \times N}[/tex]
= [tex]\frac{166 \times 7.3 \times 10^{-5}}{1.5 \times10^{-3} \times 50}[/tex]
= [tex]\frac{1211.8 \times 10^{-5}}{75 \times 10^{-3}}[/tex]
= [tex]16.157 \times 10^{-2}[/tex]
= 0.1615 T
Thus, we can conclude that magnitude of the magnetic field is 0.1615 T.