Answer: The arc gives off ultraviolet light as a result of the high temperature of the arc which can be as high as above 3000°c.
Explanation: arc welding employs or uses electric current to generate heat for the purpose of joining metals. Metals being joined usually have high melting points above 3000°c, This high melting points of metals means the welding arc needs to attain a higher temperature to be able to join the metals. In the process of attaining that temperature needed to join the metals the arc gives of ultraviolet light (UV).
In an interference pattern, the intensity is Group of answer choices smaller in regions of constructive interference than in regions of destructive interference. unchanged in regions of destructive interference but smaller in regions of constructive interference. the same in both the regions of constructive interference and the regions of destructive interference. unchanged in regions of destructive interference but greater in regions of constructive interference. greater in regions of constructive interference than in regions of destructive interference.
Answer:
The same in both the regions of constructive interference and the regions of destructive interference.
Explanation:
Interference is a phenomenon which occurs when two waves meet while moving along the same medium . The amplitude formed as a result of the interference could be greater, lower, or the same amplitude.
Constructive and destructive interference result from the interaction of waves that are correlated or coherent with each other. This is because arose from the same source or they have the same or nearly the same frequency.
The waves being coherent, arising from the same source and having the same frequency explains why it’s the same in both the regions of constructive interference and the regions of destructive interference.
Answer:
Option E (greater in regions of constructive interference than in regions of destructive interference)Explanation:
Constructive interference occurs when the maxima of two waves add together (the two waves are in phase), so that the amplitude of the resulting wave is equal to the sum of the individual amplitudes. Whereas in destructive interference opposite is true.
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if two objects each have a mass of 10 kg , then the force of gravity between them...
A) is constant
B) is 100 kg
C)is greater when they are closer together
D) depends only on their masses
if two objects each have a mass of 10 kg, then the force of gravity between them C) is greater when they are closer together.
We have two objects, each with a mass of 10 kg. We can calculate the force of gravity (F) between them using Newton's law of universal gravitation.
[tex]F = G \frac{m_1m_2}{r^{2} }[/tex]
where,
G: gravitational constantm₁, m₂: masses of both objectsr: distance between the objectsAs we can see from this expression. the gravitational force between the objects depends on their masses and the distance between them. The closer they are, the stronger the gravity force.
if two objects each have a mass of 10 kg, then the force of gravity between them...
A) is constant. No. It varies with the distance.
B) is 100 kg. No. kg is not a unit of force.
C) is greater when they are closer together. Yes.
D) depends only on their masses. No. It also depends on the distance between them.
if two objects each have a mass of 10 kg, then the force of gravity between them C) is greater when they are closer together.
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Derivation of Eq. (3): o Basic physics principles: Justify equations (1) and (2) in your own words. . Doing the algebra: From equations (1) and (2), show that equation (3) holds. mu? =eu B This would be nice if we knew the velocity. Fortunately, we know the voltage through which the electrons are accelerated. Setting the change in electrical potential energy equal to the final kinetic energy of the electrons, we find: eV = From this point, some algebra lets us eliminate the velocity variable entirely, giving us: e 2V m (B_T)
Answer:
About eq (1)
[tex]mv^2/r = eVB[/tex]
when a charged particle (electron) enters into the magnetic field which is perpendicular to direction of motion than there will be magnetic force on particle and particle will travel in circular path in with constant speed.
So using force balance on charged particle:
[tex]F_{net} = Fc - Fm[/tex]
Since particle is traveling at constant speed, So acceleration is zero, and
[tex]F_{net} = 0[/tex]
[tex]Fc - Fm=0[/tex]
[tex]Fc = Fm[/tex]
Fc = centripetal force on particle [tex]= m*v^2/r[/tex]
Fm = magnetic force on electron = [tex]q*VxB = q*V*B*sin \theta[/tex]
q = charge on electron = e
since magnetic field is perpendicular to the velocity of particle, So theta = 90 deg
sin 90 deg = 1
So,
[tex]m*v^2/r = e*v*B[/tex]
About equation 2:
When this charged particle is released from rest in a potential difference V, and then it enters into above magnetic field, then using energy conservation on charge particle
[tex]KEi + PEi = KEf + PEf[/tex]
KEi = 0, since charged particle started from rest
[tex]PEi - PEf = q*dV[/tex]
[tex]PEi - PEf = eV[/tex]
KEf = final kinetic energy of particle when it leaves [tex]= (1/2)*m*v^2[/tex]
So,
[tex]0 + eV = (1/2)*m*v^2[/tex]
[tex]eV = (1/2)*m*v^2[/tex]
From above equation (1) and (2)
[tex]m*v^2/r = evB[/tex]
[tex]e/m = v/(r*B)[/tex]
Now
[tex]eV = (1/2)*m*v^2[/tex]
[tex]v = \sqrt{(2*e*V/m)}[/tex]
[tex]e/m = \sqrt{ (2*e*V/m)/(r*B)}[/tex]
[tex]\frac{e^2}{m^2} = \frac{2*e*V}{(m*r^2*B^2)}[/tex]
[tex]e/m = 2*V/(r^2*B^2)[/tex]
[tex]e/m = 2V/(Br)^2[/tex]
A block of plastic in the shape of a rectangular solid that has height 8.00 cm and area A for its top and bottom surfaces is floating in water. You place coins on the top surface of the block (at the center, so the top surface of the block remains horizontal). By measuring the height of the block above the surface of the water, you can determine the height hh below the surface. You measure hh for various values of the total mass mm of the coins that you have placed on the block. You plot h versus m and find that your data lie close to a straight line that has slope 0.0890 m/kg and y-intercept 0.0312 mm.
What is the mass of the block?
Answer:
m_{p} = 0.3506 kg
Explanation:
For this exercise we use Newton's equilibrium equation
B - Wc-Wp = 0
where B is the thrust of the water, Wc is the weight of the coins and Wp is the weight of the plastic block
B = Wc + Wp
the state push for the Archimeas equation
B = ρ_water g V
the volume of the water is the area of the block times the submerged height h, which is
h´ = 8 - h
where h is the height out of the water
ρ_water g A h´ = [tex]m_{c}[/tex] g + [tex]m_{p}[/tex] g
ρ_water A h´ = m_{c} + m_{p}
write this equation to make the graph
h´= 1 /ρ_water A (m_{c} +m_{p})
h´ = 1 /ρ_waterA (m_{c} + m_{p})
if we graph this expression, we get an equation of the line
y = m x + b
where
y = h´
m = 1 /ρ_water A
b = mp /ρ_water A
whereby
m_{p} = b ρ_water A
ρ_water = 1000 kg / m³
b = 0.0312 m
m = 0.0890 m / kg
we substitute the slope equation
b = m_{p} / m
calculate
m_{p}= 0.0312 / 0.0890
m_{p} = 0.3506 kg
A sphere of radius 5.00 cm c m carries charge 3.00 nC n C . Calculate the electric-field magnitude at a distance 4.00 cm c m from the center of the sphere and at a distance 6.00 cm c m from the center of the sphere if the sphere is a solid insulator with the charge spread uniformly throughout its volume.
To calculate the electric-field magnitude at points inside and outside a uniformly charged insulating sphere, we apply Gauss's Law, considering the proportion of enclosed charge within a Gaussian surface for points inside, and treating the sphere as a point charge for points outside.
Explanation:We apply Gauss's Law to understand the electric-field magnitude around a uniformly charged insulating sphere, which states that the electric flux through a closed surface is proportional to the enclosed charge. We'll calculate the electric field (E) at two distances from the centre of the sphere: 4.00 cm, which is inside the sphere, and 6.00 cm, which is outside the sphere.
For the inside point (r = 4.00 cm), we only consider the charge enclosed by a Gaussian surface of radius 4.00 cm. Using the fact that charge is distributed uniformly throughout the volume of the sphere, we calculate the enclosed charge (q_enclosed) by using the ratio of volumes: q_enclosed = (Q * (4/3 * π * r^3)) / (4/3 * π * R^3), where Q is the total charge of the sphere. R is the radius of the sphere.
The electric field inside the sphere at 4.00 cm is then E = (1 / (4 * π * ε_0)) * (q_enclosed / r^2). We substitute the values using the given radius and charge and solve for E.
For the outside point (r = 6.00 cm), we use the formula for the electric field outside a spherically symmetric charge distribution, which is E = (1 / (4 * π * ε_0)) * (Q / r^2), and solve for E using the sphere's total charge and the distance of 6.00 cm.
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A bullet with a mass m b = 11.9 mb=11.9 g is fired into a block of wood at velocity v b = 261 m/s. vb=261 m/s. The block is attached to a spring that has a spring constant k k of 205 N/m. 205 N/m. The block and bullet continue to move, compressing the spring by 35.0 cm 35.0 cm before the whole system momentarily comes to a stop. Assuming that the surface on which the block is resting is frictionless, determine the mass m w mw of the wooden block.
Answer:
0.372 kg
Explanation:
The collision between the bullet and the block is inelastic, so only the total momentum of the system is conserved. So we can write:
[tex]mu=(M+m)v[/tex] (1)
where
[tex]m=11.9 g = 11.9\cdot 10^{-3}kg[/tex] is the mass of the bullet
[tex]u=261 m/s[/tex] is the initial velocity of the bullet
[tex]M[/tex] is the mass of the block
[tex]v[/tex] is the velocity at which the bullet and the block travels after the collision
We also know that the block is attached to a spring, and that the surface over which the block slides after the collision is frictionless. This means that the energy is conserved: so, the total kinetic energy of the block+bullet system just after the collision will entirely convert into elastic potential energy of the spring when the system comes to rest. So we can write
[tex]\frac{1}{2}(M+m)v^2 = \frac{1}{2}kx^2[/tex] (2)
where
k = 205 N/m is the spring constant
x = 35.0 cm = 0.35 m is the compression of the spring
From eq(1) we get
[tex]v=\frac{mu}{M+m}[/tex]
And substituting into eq(2), we can solve to find the mass of the block:
[tex](M+m) \frac{(mu)^2}{(M+m)^2}=kx^2\\\frac{(mu)^2}{M+m}=kx^2\\M+m=\frac{(mu)^2}{kx^2}\\M=\frac{(mu)^2}{kx^2}-m=\frac{(11.9\cdot 10^{-3}\cdot 261)^2}{(205)(0.35)^2}-11.9\cdot 10^{-3}=0.372 kg[/tex]
The mass of the wooden block is approximately 0.404 kg.
The initial kinetic energy of the bullet is given by:
KE_initial = 0.5 * mb * vb^2
where:
`m_b` is the mass of the bullet (11.9 g = 0.0119 kg)
`v_b` is the velocity of the bullet (261 m/s)
The potential energy stored in the compressed spring is given by:
PE_spring = 0.5 * k * x^2
where:
`k` is the spring constant (205 N/m)
`x` is the compression of the spring (0.35 m)
Since the system comes to a stop, the initial kinetic energy is converted to potential energy stored in the spring:
KE_initial = PE_spring
Solving for the mass of the wooden block, we get:
m_w = (2 * KE_initial) / (v_b^2 - (k * x) / m_b)
Plugging in the values, we get,
m_w = (2 * 0.5 * 0.0119 kg * (261 m/s)^2) / ((261 m/s)^2 - (205 N/m * 0.35 m) / 0.0119 kg)
m_w ≈ 0.404 kg
Therefore, the mass of the wooden block is approximately 0.404 kg.
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As a dilligent physics student, you carry out physics experiments at every opportunity. At this opportunity, you carry a 1.11-m-long rod as you jog at 3.07 m/s, holding the rod perpendicular to your direction of motion. What is the strength of the magnetic field that is perpendicular to both the rod and your direction of motion and that induces an EMF of 0.287 mV across the rod
Answer:
The strength of the magnetic field is 0.0842 mT
Explanation:
Given:
Velocity of rod [tex]v = 3.07[/tex] [tex]\frac{m}{s}[/tex]
Length of rod [tex]l = 1.11[/tex] m
Induced emf across the rod [tex]\epsilon = 0.287 \times 10^{-3}[/tex] V
According to the faraday's law
We have a special case for moving rod in magnetic field.
Induced emf in moving rod is given by,
[tex]\epsilon = Blv[/tex]
Where [tex]B =[/tex] strength of magnetic field
[tex]B = \frac{\epsilon}{lv}[/tex]
[tex]B = \frac{0.287 \times 10^{-3} }{3.07 \times 1.11}[/tex]
[tex]B = 0.0842 \times 10^{-3}[/tex] T
[tex]B = 0.0842[/tex] mT
Therefore, the strength of the magnetic field is 0.0842 mT
Block A weighs 12 N and has an apparent weight of 8 N when completely submerged in an ideal fluid. Block B weighs 20 N and has an apparent weight of 12 N when completely submerged in the same ideal fluid.
The ratio of the densities A/ B equals:
Final answer:
The calculated density ratio is 0.5.
The ratio of the densities of two blocks, A and B, is found by comparing the apparent weight reductions when they are submerged in a fluid. According to Archimedes' principle, this reduction is due to the buoyant force, which is equal to the weight of the fluid displaced.
Explanation:
The question revolves around calculating the ratio of the densities of two blocks, A and B, using the concept of buoyant force according to Archimedes' principle. The apparent weight of each block when submerged in a fluid gives us the buoyant force acting upon it, which is equal to the weight of the fluid displaced. Since both blocks are submerged in the same fluid, the difference in weights in air versus submerged gives us a measure of the volume of fluid displaced due to the respective densities of the blocks.
To determine the ratio of densities (A/B), we use the formula:
Ratio of densities = (Weight of A in air - Apparent weight of A) / (Weight of B in air - Apparent weight of B)
Substituting the given weights:
Ratio of densities = (12 N - 8 N) / (20 N - 12 N) = 4 N / 8 N = 0.5
Thus, the ratio of the densities of block A to block B is 0.5.
If the vertical displacement from crest to trough is 0.50 cm, what is the amplitude?
cm
0.50
0.25
1.0
Answer:
0.25
Explanation:
Since there is a vertical displacement of 0.5cm from the crest to the trough, there is half of that displacement to the midline, which is also known as the amplitude. Therefore, the amplitude is 0.5/2=0.25cm. Hope this helps!
A bat, flying at 5.1 m/s, pursues an insect that is flying at 1.1 m/s in the same direction. The bat emits a 47000-Hz sonar pulse. Take the speed of sound to be 343 m/s. show answer No Attempt At what frequency, in hertz, does the bat hear the pulse reflected back from the insect? f = | sin() cos() tan() cotan() asin() acos() atan() acotan() sinh() cosh() tanh() cotanh() Degrees Radians π ( ) 7 8 9 HOME E ↑^ ^↓ 4 5 6 ← / * 1 2 3 → + - 0 . END √() BACKSPACE DEL CLEAR Grade Summary Deductions 0% Potential 100% Submissions Attempts remaining: 3 (4% per attempt) detailed view Hints: 4% deduction per hint. Hints remaining: 2 Feedback: 5% deduction per feedback.
Answer:
The frequency of the sona-pulse reflected back is [tex]f_b = 48109.22Hz[/tex]
Explanation:
From the question we are told that
The speed of the bat is [tex]v = 5.1 m/s[/tex]
The speed of the insect is [tex]v_i = 1.1 m/s[/tex]
The frequency emitted by the bat is [tex]f = 47000 \ Hz[/tex]
The speed of sound is [tex]v_s = 343 m/s[/tex]
Let look at this question in this manner
At the first instant the that the bat emits the sonar pulse
Let the bat be the source of sound
Let the insect be the observer
This implies that the frequency of sound the the insect would receive is mathematically represented as
[tex]f_a = [\frac{v_s - v_i}{v_s - v}] f[/tex]
Substituting values
[tex]f_a = [\frac{343 - 1.1}{345 -5.1} ] * 47000[/tex]
[tex]f_a = 47556.4 Hz[/tex]
Now at the instant the sonar pules reaches the insect
Let the bat be the observer
Let the insect be the source of the sound
Here the sound wave is reflected back to the bat
This implies that the frequency of sound the the bat would receive is mathematically represented as
[tex]f_b =[ \frac{x}{y} ] * f_a[/tex]
[tex]f_b =[ \frac{v_s + v}{v_s + v_i} ] * f_a[/tex]
[tex]f_b =[ \frac{343 + 5.1}{343 + 1.1} ] * 47556.4[/tex]
[tex]f_b = 48109.22Hz[/tex]
Final answer:
To calculate the frequency a bat hears from a sonar pulse reflected back by an insect, the Doppler effect formula is used. By substituting the given values into the formula, the result is approximately 47,544 Hz, which is the frequency at which the bat detects the echo from the insect.
Explanation:
To calculate the frequency at which a bat hears the sonar pulse reflected back from an insect, we can use the Doppler effect formula for sound in the same direction as the source of the sound is moving:
[tex]f' =\frac{f (v + v_d)}{(v + v_s)}[/tex]
Where:
f' is the frequency heard by the bat.
f is the emitted frequency of the bat's sonar pulse (47,000 Hz).
v is the speed of sound (343 m/s).
[tex]v_d[/tex] is the speed of the detector (bat), which is 5.1 m/s.
[tex]v_s[/tex] is the speed of the source (insect), which is 1.1 m/s.
Substituting the values into the equation:
[tex]f' = \frac{47000 Hz \times (343 m/s + 5.1 m/s)}{(343 m/s + 1.1 m/s)}[/tex]
[tex]f' = \frac{47000 Hz * 348.1 m/s}{344.1 m/s}[/tex]
[tex]f' = 47000 Hz \times 1.0116[/tex]
f' = 47544.2 Hz
Therefore, the bat hears the pulse reflected back from the insect at approximately 47,544 Hz.
A copper rod is sliding on two conducting rails that make an angle of 19o with respect to each other, as in the drawing. The rod is moving to the right with a constant speed of 0.60 m/s. A 0.63-T uniform magnetic field is perpendicular to the plane of the paper. Determine the magnitude of the average emf induced in the triangle ABC during the 7.5-s period after the rod has passed point A.
Answer:
0.2923 V
Explanation:
Given that
Angle between the rails, θ = 19°
Speed of the rod, v = 0.6 m/s
Magnetic field present, B = 0.63 T
Time used, t = 7.5 s
E = -ΔΦ/Δt
where, Φ = BA, so
E = -BΔA / Δt
To get the area, if we assume the rails are joined in a triangular fashion(see attachment)
E = -B(1/2 * AC * BC) / Δt
E = -B(vΔt * vΔt tanθ) / 2Δt
E = -(B * v² * Δt² * tanθ) / 2Δt
E = -Bv²Δt.tanθ/2
E = -(0.63 * 0.6² * 7.5 * tan 19) / 2
E = -0.5857 / 2
E = -0.2923
Thus, the magnitude of average emf induced if 0.2923 V
g 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 0.462 of the escape speed from Earth and (b) its initial kinetic energy is 0.462 of the kinetic energy required to escape Earth
Answer:
a.)r=4RE/4
b.)r=2RE
c.)ZERO
Explanation:
a.) given v= 0.462 which is V= 0.466Ve
Since the projectile is shot directly away from earth surface the speed it escape with;
v=√2GM/RE......................eqn(1)
M is the Earth's Mass
RE is the Radius
From the Law of conservation of Energy
K+U=0...............................eqn(2)
K₁+U₁=K₂+U₂....................eqn(3)
where K₁ and U₁ is initial kinetic and potential energy
K₂ and U₂ are final kinetic and potential energy
Kinetic Energy (K.E) decrease with time as the projectile moves up and there is decrease in Potential Energy (P.E) , it will let to a point where K.E will turn to zero i.e K₂=0
U₂=K₁U₂ ..........................eqn(3)
From Gravitational Law
U₁= -GMm/RE ..................(5)
U₂= -GMm/r .....................(6)
Where "r" is the distance
v= 0.462√2GM/RE
v= √GM/2RE
GM/4RE - GM/RE = -GM/r
r= 4RE/3
b.) "r" is calculated by this equation;
K₁=0.466
K₁= 1/4MVe².............................eqn(9)
substitute eqn(1) into eqn (9) then
1/4m2GM/RE=0.466GM/2Re
GM/2RE - GM/RE =-GM/r
r=2RE
c.)The potential energy and kinetic energy is the same in terms of their size both in different directions, while the potential energy face outward, the kinetic energy face inward therefore the least initial mechanical energy
required at launch if the projectle is to escape is ZERO
20 POINTS! TRUE OR FALSE
Friction is a force that speeds up motion between two surfaces that are in contact with each other.
A) TRUE
B) FALSE
Answer:
False
Explanation:
Friction is the resisting force of motion that converts kinetic energy into heat energy. Kinetic energy is the energy of movement. Removing kinetic energy makes the object slower.
Rank from largest to smallest the values of the magnetic field at the following distances from the axis of the conducting cylinder: Ra = 7.75 cm , Rb = 4.95 cm , r = 5.40 cm , and r>Ra.
The ranking from largest to smallest the values of the magnetic field should be like
Rb = 4.95 cm
r= 5.40 cm
Ra =7.75 cm
r> Ra
What is the magnetic field?It is a vector field that explained the magnetic impact on moving electric charges, electric currents, and magnetic materials.
In this, the force should be perpendicular to the velocity and the magnetic field. Also, it should be inversely proportional with respect to the distance with the axis of the cylinder.
So,
The largest magnetic field =Rb =4.95 cm
And,
Smallest magnetic field = r>Ra
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The magnetic field inside a conducting cylinder is maximized at certain points before dropping off outside. The ranked values from largest to smallest are: r = 5.40 cm, Rb = 4.95 cm, Ra = 7.75 cm, and r > Ra.
To determine the magnetic field at different distances from the axis of a conducting cylinder, we need to understand how the magnetic field varies inside and outside a cylinder carrying a current.
Given the distances Ra = 7.75 cm, Rb = 4.95 cm, r = 5.40 cm, and r > Ra, we will rank the magnetic field intensities:
1.) r = 5.40 cm:
Since r is within the cylinder where current is distributed uniformly, the magnetic field at this point can be calculated using Ampère's Law.
2.) Rb = 4.95 cm:
Still within the cylinder but closer to the center, thus the field here will not yet be maximized.
3.) Ra = 7.75 cm:
Outside the cylinder where the current configuration influences how the field drops off with increasing distance.
4.) r > Ra:
This point is furthest from the axis, hence the magnetic field will be the smallest.
The magnetic field reaches a maximum inside the conducting cylinder before decreasing outside of it.
A step-up transformer is connected to a generator that is delivering 141 V and 110 A. The ratio of the turns on the secondary to the turns on the primary is 1110 to 5. What voltage is across the secondary
Final answer:
To find the voltage across the secondary winding of a step-up transformer with a primary voltage of 141 V and a turns ratio of 1110 to 5, the secondary voltage is calculated to be 31302 V.
Explanation:
Calculating the Voltage Across the Secondary in a Step-Up Transformer
To calculate the voltage across the secondary winding of a step-up transformer, we use the transformer equation that relates the primary and secondary voltages with the number of turns on the primary (Np) and secondary (Ns) windings:
Vs / Vp = Ns / Np
Given that the primary voltage (Vp) is 141 V, and the ratio of the number of turns is 1110 to 5 (Ns / Np), we can rearrange the equation to solve for the secondary voltage (Vs) as follows:
Vs = Vp × (Ns / Np)
Vs = 141 V × (1110 / 5)
Vs = 141 V × 222
Vs = 31302 V
Thus, the voltage across the secondary winding is 31302 V.
The symbol, "I" represents-
a. voltage
b. power
c. current
d. resistance
Answer:
The answer is C.
Explanation:
Current can be represented in the symbol of I.
The symbol for voltage is V.
The symbol for power is W.
The symbol for resistance is R.
Most days in the year are weekdays (as opposed to weekends). There will come a day when a huge asteroid strikes the earth. Thus, the day a huge asteroid strikes the earth will likely be a weekday. – The previous argument is Group of answer choices
Answer:
True.
The day a huge asteroid strikes the earth will likely be a weekday.
Explanation:
Mathematically, there are 5 weekdays in the week and there are only 2 weekend days.
Total number of days in a week = 7
Since, the asteroids can hit on any day,
The probability of the asteroid hitting on a weekend day = (2/7) = 0.2857
The probability that an asteroid hitting on a weekday = (5/7) = 0.7143.
So, by the virtue of the probability of the asteroid hitting on a weekday or on a weekend day, the asteroid is more likely to hit on a weekday as
P(weekday) > P(weekend)
0.7143 > 0.2857.
Hope this Helps!!!
Interactive LearningWare 22.2 reviews the fundamental approach in problems such as this. A constant magnetic field passes through a single rectangular loop whose dimensions are 0.46 m x 0.68 m. The magnetic field has a magnitude of 3.0 T and is inclined at an angle of 67o with respect to the normal to the plane of the loop. (a) If the magnetic field decreases to zero in a time of 0.49 s, what is the magnitude of the average emf induced in the loop
Explanation:
The dimension of a single rectangular loop is 0.46 m x 0.68 m.
Magnetic field, B = 3 T
The loop is inclined at an angle of 67 degrees with respect to the normal to the plane of the loop.
It is required to find the magnitude of the average emf induced in the loop if the magnetic field decreases to zero in a time of 0.49 s.
Te induced emf in the loop is given by :
[tex]\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=\dfrac{-d(BA)}{dt}\\\\\epsilon=A\dfrac{dB}{dt}\cos\theta\\\\\epsilon=0.46\times 0.68\times \dfrac{3}{0.49}\times \cos(67)\\\\\epsilon=0.74\ V[/tex]
So, the magnitude of the average emf induced in the loop is 0.74 V.
The rotor of an electric motor has rotational inertia Im = 4.36 x 10-3 kg·m2 about its central axis. The motor is used to change the orientation of the space probe in which it is mounted. The motor axis is mounted along the central axis of the probe; the probe has rotational inertia Ip = 6.07 kg·m2 about this axis. Calculate the number of revolutions of the rotor required to turn the probe through 30.6° about its central axis.
Answer:
The number of revolutions of the rotor required to turn the probe is 118 revolutions
Explanation:
Given;
rotational inertia of the electric motor, Im = 4.36 x 10⁻³ kg·m²
rotational inertia of the probe, Ip = 6.07 kg·m²
the angular position of the probe, θ = 30.6°
From the principle of conservation of angular momentum;
[tex]I_m \omega _m = I_p \omega _p \\\\Also;\\\\I_m \theta _m = I_p \theta _p[/tex]
where;
[tex]\omega _m[/tex] is the angular velocity of the electric motor
[tex]\omega _p[/tex] is the angular velocity of the probe
[tex]\theta _m[/tex] is the angular position of the electric motor
[tex]\theta _p[/tex] is the angular position of the probe
[tex]\theta _m = \frac{I_p \theta_p}{I_m} \\\\\theta _m = \frac{6.07* 30.6^o}{4.36*10^{-3}} = 42601.4^o[/tex]
360° = One revolution
42601.4° = ?
Divide 42601.4° by 360°
= 118 revolutions
Therefore, the number of revolutions of the rotor required to turn the probe is 118 revolutions
A vessel at rest at the origin of an xy coordinate system explodes into three pieces. Just after the explosion, one piece, of mass m, moves with velocity (-47 m/s) and a second piece, also of mass m, moves with velocity (-47 m/s) . The third piece has mass 3m. Just after the explosion, what are the (a) magnitude and (b) direction (as an angle relative to the +x axis) of the velocity of the third piece?
Answer:
a) 22.06m/s
b)45°
Explanation:
Let 'm' be mass of 1st and 2nd pieces
mass of 3rd piece=3m
[tex]v_{1}[/tex]=velocity of 1st piece = -47iˆ m/s
[tex]v_{2}[/tex]=velocity of 2nd piece = -47jˆ m/s
[tex]v_{3}[/tex]=velocity of 3rd piece=?
By considering conservation of linear momentum, we have
m[tex]v_{1}[/tex] + m[tex]v_{2}[/tex] + 3m[tex]v_{3}[/tex]=0
[tex]v_{1}[/tex] + [tex]v_{2}[/tex] + 3[tex]v_{3}[/tex]=0
3[tex]v_{3}[/tex]= - ([tex]v_{1}[/tex] + [tex]v_{2}[/tex] )
[tex]v_{3}[/tex] = -[tex]\frac{1}{3}[/tex] ([tex]v_{1}[/tex] + [tex]v_{2}[/tex] )
Substituting the values of [tex]v_{1}[/tex] and [tex]v_{2}[/tex] in above equation
[tex]v_{3}[/tex] = -[tex]\frac{1}{3}[/tex] (-47iˆ -47jˆ ) => 15.6iˆ + 15.6jˆ
(a)Magnitude of the velocity of the third piece is given by
|[tex]v_{3}[/tex]| =√15.6²+15.6² => 22.06m/s
(b) its direction (as an angle relative to the +x axis) 'θ'
θ= [tex]tan^{-1} (\frac{15.6}{15.6} )[/tex] => [tex]tan^{-1} (1 )[/tex] =>45°
In astronomy, distances are often expressed in light-years. One light-year is the distance traveled by light in one year. If the distance to a star is 3.6 light-years, what is this distance in meters? (There are 365.25 days in one year.)
Answer:
The distance of the star is [tex]3.40x10^{16}[/tex] meters
Explanation:
It is known that the speed of light has a value of [tex]3x10^{8}m/s[/tex] in vacuum. That is, it travels [tex]3x10^{8]m[/tex] in one second, according with the following equation:
[tex]v = \frac{x}{t}[/tex]
Where v is the speed, x is the distance and t is the time.
[tex]x = v\cdot t[/tex] (1)
Equation 1 can be used to determine the distance that the light travels in 1 year:
It is necessary to find how many seconds are in 1 year (365.25 days).
[tex]365.25 days \cdot \frac{86400s}{1 day}[/tex] ⇒ [tex]31557600s[/tex]
[tex]x = (3x10^{8}m/s)(31557600s)[/tex]
[tex]x = 9.46x10^{15}m[/tex]
Therefore, in 1 year, light travels [tex]9.46x10^{15}[/tex] meters.
If the distance to a star is 3.6 light-years, what is this distance in meters?
A simple conversion between units can be used to get the distance in meters
[tex]x_{star} = 3.6ly \cdot \frac{9.46x10^{15}m}{1ly}[/tex] ⇒ [tex]3.40x10^{16}m[/tex]
Hence, the distance of the star is [tex]3.40x10^{16}[/tex] meters.
A light-year, a distance unit used in astronomy, is approximately 9.46 x [tex]10^{15[/tex] meters. Hence, a distance of 3.6 light years would convert to approximately 3.4 x [tex]10^{16[/tex] meters.
Explanation:A light-year is a unit of distance used in astronomy, which represents the distance light travels in one year. Light travels at a speed of 186,000 miles per second. Over the course of a year, this distance adds up significantly.
To calculate a light-year in meters, we would carry out the following calculation - light travels at 299,792 kilometers per second. This converts to exactly 299,792,000 meters per second. If we multiply this by the number of seconds in a year (60 seconds/minute, 60 minutes/hour, 24 hours/day, 365.25 days/year) we find that one light-year is approximately 9.46 x [tex]10^{15[/tex] meters.
Therefore, if the distance to a star is 3.6 light-years, then this would convert to about 3.4 x [tex]10^{16[/tex] meters.
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An open-delta three-phase transformer system has one transformer center-tapped to provide a neutral for single-phase voltages. If the voltage from line to center tap is 277 V, what is the high-leg voltage?
Answer:
The high-leg voltage is [tex]V_h= 497.76V[/tex]
Explanation:
A diagram showing the arrangement of this three phase transformer is shown on the first uploaded image
The high - leg terminal is in the diagram is from [tex]Z_1 - n[/tex]
Generally the high-leg terminal is 1.732 times of the the voltage from line to center tap which is given as 277 V
The high -voltage can be computed as follows
[tex]V_h = 1.732 *277[/tex]
[tex]V_h= 497.76V[/tex]
A battery with emf e m f is connected in series with an inductance L and a resistance R. (a) Assuming the current has reached steady state when it is at 99% of its maximum value, how long does it take to reach steady state, assuming the initial current is zero? (Use any variable or symbol stated above as necessary. To represent e m f, use E.) t99% = (b) If an emergency power circuit needs to reach steady state within 1.2 ms of turning on and the circuit has a total resistance of 72 Ω, what values of the total inductance of the circuit are needed to satisfy the requirement? (Give the maximum value.) H
Answer:
a) [tex]t = 4.6\tau[/tex]
b) [tex]L = 0.0187 \: H[/tex]
Explanation:
The current flowing in a R-L series circuit is given by
[tex]I = I_{0} (1 - e^{\frac{-t}{\tau} })[/tex]
Where τ is the time constant and is given by
[tex]\tau = \frac{L}{R}[/tex]
Where L is the inductance and R is the resistance
Assuming the current has reached steady state when it is at 99% of its maximum value,
[tex]0.99I_{0} = I_{0} (1 - e^{\frac{-t}{\tau} })\\0.99 = (1 - e^{\frac{-t}{\tau} })\\1 - 0.99 = e^{\frac{-t}{\tau}}\\ln(0.01) = ln(e^{\frac{-t}{\tau}})\\-4.6 = \frac{-t}{\tau}\\t = 4.6\tau\\[/tex]
Therefore, it would take t = 4.6τ to reach the steady state.
(b) If an emergency power circuit needs to reach steady state within 1.2 ms of turning on and the circuit has a total resistance of 72 Ω, what values of the total inductance of the circuit are needed to satisfy the requirement?
[tex]t = 4.6\frac{L}{R}\\t = 4.6\frac{L}{R}\\0.0012 = 4.6\frac{L}{72}\\0.0864 = 4.6 L\\L = 0.0864/4.6\\L = 0.0187 \: H[/tex]
Therefore, an inductance of 0.0187 H is needed to satisfy the requirement.
An MRI technician moves his hand from a region of very low magnetic field strength into an MRI scanner's 1.80 T field with his fingers pointing in the direction of the field. His wedding ring has a diameter of 2.23 cm, and it takes 0.320 s to move it into the field.
1. What average current is induced in the ring in A if its resistance is 0.0100 Ω?
2. What average power is dissipated in mW?
3. What magnetic field is induced at the center of the ring in T?
Given that,
Magnetic field strength is
B = 1.8T
The wedding ring has a diameter of
d = 2.23 cm = 0.023m
Time take t = 0.32 secs
A. Current induced
From ohms law
V= iR, given that R = 0.01Ω
So, we need to get the induced emf
Using
ε = -NdΦ / dt
Where Φ = BA
ε = -A ∆B / ∆t
ε = -¼πd²(B2-B1) / (t2-t1)
ε = -¼ × π × 0.023² × -1.8 / 0.32
ε = 0.0023371 V
Then
I = ε / R
I = 0.002337 / 0.01
I =0.2337 A
B. Power discippated?
Power is given as
P = iV
P = 0.2337 × 0.002337
P = 0.0005462 W
P = 0.56 mW
C. The magnetic field at the centre of the ring.
The electric field at the centre of the ring is zero because each part of the ring will cause a symmetrical opposite magnitude at every point,
Then, B = 0 T
Jet fighter planes are launched from aircraft carriers with the aid of their own engines and a catapult. If in the process of being launched from rest, the thrust of a jet's engines is 2.30 105 N and after moving through a distance of 90 m the plane lifts off with a kinetic energy of 3.60 107 J, what is the work done (in J) on the jet by the catapult?
Answer:
The work done on the jet by the catapult is 1.53 x 10⁷ J
Explanation:
Given;
force of thrust engines, F = 2.3 x 10⁵ N
distance moved by the thrust engines, d = 90 m
kinetic energy of the plane, K.E = 3.60 x 10⁷ J
Based on work-energy theorem;
the net work done on the plane by catapult and the thrust engine is given as;
[tex]W_n = K.E_f = 3.6 *10^7 \ J[/tex]
work done by the thrust engine on the jet;
[tex]W_T_._e_n = fd\\\\W_T_._e_n = 2.3*10^5 * 90\\\\W_T_._e_n = 2.07*10^7 \ J[/tex]
Work done on the jet by the catapult;
[tex]W_c_p= W_n - W_T_._e_g\\\\W_c_p= (3.6*10^7 - 2.07*10^7) J\\\\W_c_p= 1.53*10^7 \ J[/tex]
Therefore, the work done on the jet by the catapult is 1.53 x 10⁷ J
A gas is enclosed in a cylinder fitted with a light frictionless piston and maintained at atmospheric pressure. When 254 kcal of heat is added to the gas, the volume is observed to increase slowly from 12.0 m3 to 16.2 m3 . Part A Calculate the work done by the gas. Express your answer with the appropriate units. W = nothing nothing Request Answer Part B Calculate the change in internal energy of the gas. Express your answer with the appropriate units. ΔU = nothing nothing
The work done by the gas when the volume increases from 12.0 m³ to 16.2 m³ at atmospheric pressure is 425.565 kJ. The change in internal energy of the gas when 254 kcal of heat is added is 636.931 kJ.
Explanation:To calculate the work done by the gas during a quasi-static expansion, we can use the formula W = P ΔV, where W is work, P is pressure, and ΔV is the change in volume. Given the atmospheric pressure is 1 atm or 101,325 Pa, and the volume change is from 12.0 m³ to 16.2 m³, the work done by the gas can be calculated as:
W = P ΔV = 101,325 Pa × (16.2 m³ - 12.0 m³)
W = 101,325 Pa × 4.2 m³
W = 425,565 J or 425.565 kJ
To calculate the change in internal energy (ΔU) of the gas, we can use the first law of thermodynamics, which states that ΔU = Q - W, where Q is the heat added to the system. The heat is given as 254 kcal, which needs to be converted to joules (1 kcal = 4.184 kJ).
ΔU = Q - W = (254 kcal × 4.184 kJ/kcal) - 425.565 kJ
ΔU = 1062.496 kJ - 425.565 kJ
ΔU = 636.931 kJ
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A 7.7 kg sphere makes a perfectly inelastic collision with a second sphere initially at rest. The composite system moves with a speed equal to one third the original speed of the 7.7 kg sphere. What is the mass of the second sphere
Answer:
15.4 kg.
Explanation:
From the law of conservation of momentum,
Total momentum before collision = Total momentum after collision
mu+m'u' = V(m+m').................... Equation 1
Where m = mass of the first sphere, m' = mass of the second sphere, u = initial velocity of the first sphere, u' = initial velocity of the second sphere, V = common velocity of both sphere.
Given: m = 7.7 kg, u' = 0 m/s (at rest)
Let: u = x m/s, and V = 1/3x m/s
Substitute into equation 1
7.7(x)+m'(0) = 1/3x(7.7+m')
7.7x = 1/3x(7.7+m')
7.7 = 1/3(7.7+m')
23.1 = 7.7+m'
m' = 23.1-7.7
m' = 15.4 kg.
Hence the mass of the second sphere = 15.4 kg
Answer:
The mass of the second sphere is 15.4 kg
Explanation:
Given;
mass of the first sphere, m₁ = 7.7 kg
initial velocity of the second sphere, u₂ = 0
let mass of the second sphere = m₂
let the initial velocity of the first sphere = u₁
final velocity of the composite system, v = ¹/₃ x u₁ = [tex]\frac{u_1}{3}[/tex]
From the principle of conservation of linear momentum;
Total momentum before collision = Total momentum after collision
m₁u₁ + m₂u₂ = v(m₁ + m₂)
Substitute the given values;
[tex]7.7u_1 + 0=\frac{u_1}{3} (7.7+m_2)[/tex]
Divide through by u₁
7.7 = ¹/₃(7.7 + m₂)
multiply both sides by 3
23.1 = 7.7 + m₂
m₂ = 23.1 - 7.7
m₂ = 15.4 kg
Therefore, the mass of the second sphere is 15.4 kg
A wire of length L is wound into a square coil with 117 turns and used in a generator that operates at 60.0 Hz and 120 V rms value in a 0.023-T magnetic field. What is the length L of the wire used to construct the coil?
Answer: The length L of the wire used to construct the coil is 191.4m
Explanation: Please see the attachments below
The eyes of amphibians such as frogs have a much flatter cornea but a more strongly curved (almost spherical) lens than do the eyes of air-dwelling mammals. In mammalian eyes, the shape (and therefore the focal length) of the lens changes to enable the eye to focus at different distances. In amphibian eyes, the shape of the lens doesn't change. Amphibians focus on objects at different distances by using specialized muscles to move the lens closer to or farther from the retina, like the focusing mechanism of a camera. In air, most frogs are nearsighted; correcting the distance vision of a typical frog in air would require contact lenses with a power of about −6.0D .A frog can see an insect clearly at a distance of 10cm . At that point the effective distance from the lens to the retina is 8mm .If the insect moves 5cm farther from the frog, by how much and in which direction does the lens of the frog's eye have to move to keep the insect in focus?0.02cm, toward the retina.0.02cm, away from the retina.0.06cm, toward the retina.0.06cm, away from the retina.
Answer:
0.2cm towards the retina.
Explanation:
the focal length of the frog eye is
(1/f) = (1/10) + (1/0.8)
f = 0.74cm
Since the distance of the object is 15cm Hence
(1/0.74) = (1/15) + (1/V)
V = 0.78cm
Therefore the distance the retina is to move is
0.78cm - 0.8cm = 0.02cm towards the retina.
A man pushes a shopping cart at a constant speed. What happens if the man increases the force on the cart?
Answer:
The cart would speed up.
Explanation:
According to Newton's 1st law, object subjected to no force, or net force 0, would have a constant speed. In our case the cart is initially at constant speed, meaning the man exerts a force that is equal to friction force. If he increases the force on the cart, the net force would no longer be 0. The cart would gain an acceleration and increases its speed.
Answer:
A. The cart accelerates
Explanation: