Answer:
. The loop is pushed to the right, away from the magnetic field
Explanation
This decrease in magnetic strength causes an opposing force that pushes the loop away from the field
The advantage of a hydraulic lever is A : it transforms a small force acting over a large distance into a large force acting over a small distance. B : it transforms a small force acting over a small distance into a large force acting over a large distance. C : it allows you to exert a larger force with less work. D : it transforms a large force acting over a large distance into a small force acting over a small distance. E : it transforms a large force acting over a small distance into a small force acting over a large distance.
Answer:
A) it transforms a small force acting over a large distance into a large force acting over a small distance.
Explanation:
The hydraulic lever works based on Pascal's law of transmission of pressure through a fluid. In the hydraulic lever, the pressure transmitted is the same.
Pressure transmitted P = F/A
where F is the force applied
and A is the area over which the force is applied.
This pressure can be manipulated on the input end as a small force applied over a small area, and then be transmitted to the output end as a large force over a large area.
F/A = f/a
where the left side of the equation is for the output, and the right side is for the input.
The volume of the displaced fluid will be the same on both ends of the hydraulic lever. Since we know that
volume V = (area A) x (distance d)
this means that the the piston on the input smaller area of the hydraulic lever will travel a greater distance, while the piston on the larger output area of the lever will travel a small distance.
From all these, we can see that the advantage of a hydraulic lever is that it transforms a small force acting over a large distance into a large force acting over a small distance.
A Cannonball is shot at an angle of 35.0 degrees and is in flight for 11.0 seconds before hitting the ground at the same height from which it was shot.
A. What is the magnitude of the inital velocity?B. What was the maximum height reached by the cannonball?C. How far, horizontally, did it travel?
Answer:
Explanation:
According to Equations of Projectile motion :
[tex]Time\ of\ Flight = \frac{2vsin(x)}{g}[/tex]
vsin(x) = 11 * 9.8 / 2 = 53.9 m/sec
(A) v (Initial velocity) = 11 * 9.8 / 2 * sin(35) = 94.56 m/sec
[tex]Maximum Height = \frac{(vsinx)^{2} }{2g}[/tex]
(B) Maximum Height = 53.9 * 53.9 / 2 * 9.8 = 142.2 m
[tex]Horizontal Range = vcosx * t[/tex]
(C) Horizontal Range = 94.56 * 0.81 * 11 = 842.52 m
PLEASE HELP FAST Five-gram samples of brick and glass are at room temperature. Both samples receive equal amounts of energy due to heat flow. The specific heat capacity of brick is 0.22 cal/g°C and the specific heat capacity of glass is 0.22 cal/g°C. Which of the following statements is true? 1.The temperature of each sample will increase by the same amount. 2.The temperature of each sample will decrease by the same amount. 3.The brick will get hotter than the glass. 4.The glass will get hotter than the brick.
Answer:
1.The temperature of each sample will increase by the same amount
Explanation:
This is because, since their specific heat capacities are the same and we have the same mass of each substance, and the same amount of energy due to heat flow is supplied to both the glass and brick at room temperature, their temperatures would thereby increase by the same amount.
This is shown by the calculation below
Q = mcΔT
ΔT = Q/mc where ΔT = temperature change, Q = amount of heat, m = mass of substance and c = specific heat capacity of substance.
Since Q, m and c are the same for both substances, thus ΔT will be the same.
So, the temperature of each sample will increase by the same amount
A city of Punjab has a 15 percent chance of wet weather on any given day. What is the probability that it will take a week for it three wet weather on 3 separate days? Also find its Standard Deviation
Answer:
so the probability will be = 0.062
Standard deviation = 0.8925
Explanation:
The probability of rain = 15% = 15/100= 0.15
and the probability of no rain=q = 1-p= 1-0.15= 0.85
The number of trials = 7
so the probability will be
7C3 * ( 0.15)^3 (0.85)^4= 35* 0.003375 * 0.52200 =0.06166= 0.062
Taking this as binomial as the p and q are constant and also the trials are independent .
For a binomial distribution
Standard deviation = npq= 0.15 *0.85 *7= 0.8925
Two wires carry current I1 = 73 A and I2 = 31 A in the opposite directions parallel to the x-axis at y1 = 3 cm and y2 = 13 cm. Where on the y-axis (in cm) is the magnetic field zero?
Answer:
The position on the y-axis where the magnetic field is zero is at y = 10 cm
Explanation:
The magnetic field B due to a long straight wire carrying a current, i at a distance R from the wire is given by
B = μ₀i/2πR
Now, let y be the point where the magnetic fields of both wires are equal.
So, the magnetic field due to wire 1 carrying current i₁ = 73 A is
B₁ = μ₀i₁/2π(y - 3) and
the magnetic field due to wire 2 carrying current i₂ = 31 A is
B₂ = μ₀i₂/2π(13 - y)
At the point where the magnetic field is zero, B₁ = B₂. So,
μ₀i₁/2π(y - 3) = μ₀i₂/2π(13 - y)
cancelling out μ₀ and 2π, we have
i₁/(x - y) = i₂/(13 - y)
cross-multiplying, we have
(13 - y)i₁ = (y - 3)i₂
Substituting the values of i₁ and i₂, we have
(13 - y)73 = (y - 3)31
949 - 73y = 31y - 93
Collecting like terms, we have
949 + 93 = 73y + 31y
1042 = 104y
dividing through by 104, we have
y = 1042/104
y = 10.02 cm
y ≅ 10 cm
So, the position on the y-axis where the magnetic field is zero is at y = 10 cm
A girl is sitting on the edge of a pier with her legs dangling over the water. Her soles are 80.0 cm above the surface of the water. A boy in the water looks up at her feet and wants to touch them with a reed. (nwater =1.333). He will see her soles as being:____
a. right at the water surface.
b. 53.3 cm above the water surface.
c. exactly 80.0 cm above the water surface.
d. 107 cm above the water surface.
e. an infinite distance above the water surface.
Answer:
d. 107 cm above the water surface.
Explanation:
The refractive index of water and air = 1.333
The real height of the girl's sole above water = 80.0 cm
From the water, the apparent height of the girl's sole will be higher than it really is in reality by a factor that is the refractive index.
The boy in the water will therefore see her feet as being
80.0 cm x 1.333 = 106.64 cm above the water
That is approximately 107 cm above the water
If you wanted to make your own lenses for a telescope, what features of a lens do you think would affect the images that you can see
Answer:
Therefore the characteristics to be found are:
* the focal length must be large and the focal length of the eyepiece must be small
* The diameter of the objective lens should be as large as possible, to be able to collect small without need from light
* The system must be configured to the far sight tip,
Explanation:
The length of the telescope is
L = f_ocular + f_objetive
the magnification of the telescope is
m = - f_objective / f_ocular
These are the two equations that describe the behavior of the telescope. Therefore the characteristics to be found are:
* the focal length must be large and the focal length of the eyepiece must be small
* The diameter of the objective lens should be as large as possible, to be able to collect small without need from light
* The system must be configured to the far sight tip,
You want the current amplitude through a 0.450 mH inductor (part of the circuitry for a radio receiver) to be 1.50 mA when a sinusoidal voltage with an amplitude of 13.0 V is applied across the inductor. What frequency is required?
Answer:
3.067MHzExplanation:
The formula for calculating the voltage across an inductor is expressed as
[tex]V_l = IX_l\\\\Since\ X_l = 2\pi fL\\V_l = I(2\pi fL)[/tex]
Given parameters
current amplitude I = 1.50mA = 1.5*10⁻³A
inductance L = 0.450mH = 0.450*10⁻³H
Voltage across the inductor [tex]V_l[/tex] = 13.0V
Required
frequency f
Substituting the given parametres into the formula, we have;
[tex]V_l = I(2\pi fL)\\\\13 = 1.50*10^{-3}(2*3.14*f*0.450*10^{-3})\\\\13 = 4.239*10^{-6}f\\\\f = \frac{13}{4.239*10^{-6}} \\\\f = 3,066,761 Hertz\\\\f = 3.067MHz[/tex]
Hence, the frequency required is 3.067MHz
Consider a wire of a circular cross-section with a radius of R = 3.17 mm. The magnitude of the current density is modeled as J = cr2 = 9.00 ✕ 106 A/m4 r2. What is the current (in A) through the inner section of the wire from the center to r = 0.5R?
Answer:
The current is [tex]I = 8.9 *10^{-5} \ A[/tex]
Explanation:
From the question we are told that
The radius is [tex]r = 3.17 \ mm = 3.17 *10^{-3} \ m[/tex]
The current density is [tex]J = c\cdot r^2 = 9.00*10^{6} \ A/m^4 \cdot r^2[/tex]
The distance we are considering is [tex]r = 0.5 R = 0.001585[/tex]
Generally current density is mathematically represented as
[tex]J = \frac{I}{A }[/tex]
Where A is the cross-sectional area represented as
[tex]A = \pi r^2[/tex]
=> [tex]J = \frac{I}{\pi r^2 }[/tex]
=> [tex]I = J * (\pi r^2 )[/tex]
Now the change in current per unit length is mathematically evaluated as
[tex]dI = 2 J * \pi r dr[/tex]
Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows
[tex]I = 2\pi \int\limits^{0.5 R}_{0} {( 9.0*10^6A/m^4) * r^2 * r} \, dr[/tex]
[tex]I = 2\pi * 9.0*10^{6} \int\limits^{0.001585}_{0} {r^3} \, dr[/tex]
[tex]I = 2\pi *(9.0*10^{6}) [\frac{r^4}{4} ] | \left 0.001585} \atop 0}} \right.[/tex]
[tex]I = 2\pi *(9.0*10^{6}) [ \frac{0.001585^4}{4} ][/tex]
substituting values
[tex]I = 2 * 3.142 * 9.00 *10^6 * [ \frac{0.001585^4}{4} ][/tex]
[tex]I = 8.9 *10^{-5} \ A[/tex]
a car moves for 10 minutes and travels 5,280 meters .What is the average speed of the car?
Answer:use the formular distance over time i.e distance/time. Make sure to convert the distance from metres to kilometers and time from minutes to hours .
Explanation:
The average speed of the car is 31,680 meters per hour.
To calculate the average speed of the car, you need to divide the total distance traveled by the time it took to travel that distance.
Given:
Time taken (t) = 10 minutes = 10 minutes × (1 hour / 60 minutes) = 10/60 hours = 1/6 hours
Distance traveled (d) = 5,280 meters
Average Speed (v) = Distance (d) / Time (t)
Average Speed (v) = 5280 meters / (1/6) hours
To simplify, when you divide by a fraction, it's equivalent to multiplying by its reciprocal:
Average Speed (v) = 5280 meters × (6/1) hours
Average Speed (v) = 31,680 meters per hour
Hence, the average speed of the car is 31,680 meters per hour.
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A car moving at 36 m/s passes a stationary police car whose siren has a frequency of 500 Hz. What is the change in the frequency (in Hz) heard by an observer in the moving car as he passes the police car? (The speed of sound in air is 343 m/s.)
Answer:
Change in the frequency (in Hz) = 104.96 Hz
Explanation:
Given:
Speed of sound in air (v) = 343 m/s
Speed of car (v1) 36 m/s
Frequency(f) = 500 Hz
Find:
Change in the frequency (in Hz)
Computation:
Frequency hear by the observer(before)(f1) = [f(v+v1)] / v
Frequency hear by the observer(f1) = [500(343+36)] / 343
Frequency hear by the observer(f1) = 552.48 Hz
Frequency hear by the observer(after)(f2) = [f(v-v1)] / v
Frequency hear by the observer(f2) = [500(343-36)] / 343
Frequency hear by the observer(f2) = 447.52 Hz
Change in the frequency (in Hz) = f1 - f2
Change in the frequency (in Hz) = 552.48 Hz - 447.52 Hz
Change in the frequency (in Hz) = 104.96 Hz
You perform a double‑slit experiment in order to measure the wavelength of the new laser that you received for your birthday. You set your slit spacing at 1.11 mm and place your screen 8.63 m from the slits. Then, you illuminate the slits with your new toy and find on the screen that the tenth bright fringe is 4.71 cm away from the central bright fringe (counted as the zeroth bright fringe). What is your laser's wavelength lambda expressed in nanometers?
Answer:
λ = 605.80 nm
Explanation:
These double-slit experiments the equation for constructive interference is
d sin θ = m λ
where d is the distance between the slits, λ the wavelength of light and m an integer that determines the order of interference.
In this case, the distance between the slits is d = 1.11 mm = 1.11 10⁻³ m, the distance to the screen is L = 8.63 m, the range number m = 10 and ay = 4.71 cm
Let's use trigonometry to find the angle
tan θ = y / L
as the angles are very small
tan θ = sin θ / cos θ = sin θ
we substitute
sin θ = y / L
we substitute in the first equation
d y / L = m λ
λ = d y / m L
let's calculate
λ = 1.11 10⁻³ 4.71 10⁻²/ (10 8.63)
λ = 6.05805 10⁻⁷ m
let's reduce to nm
λ = 6.05805 10⁻⁷ m (10⁹ nm / 1m)
λ = 605.80 nm
Two hoops, staring from rest, roll down identical incline planes. The work done by nonconservative forces is zero. The hoops have the same mass, but the larger hoop has twice the radius. Which hoop will have the greater total kinetic energy at the bottom
Answer:
They both have the same total K.E at the bottom
Explanation:
This Is because If assuming no work is done by non conservative forces, total mechanical energy must be conserved
So
K1 + U1 = K2 + U2
But If both hoops start from rest, and and at the bottom of the incline the level for gravitational potential energy is zero for reference
thus
K1 = 0 , U2 = 0
ΔK = ΔU = m g. h
But if the two inclines have the same height, and both hoops have the same mass m,
So difference in kinetic energy, must be the same for both hoops.
Electrons are accelerated through a voltage difference of 270 kV inside a high voltage accelerator tube. What is the final kinetic energy of the electrons?
Each electron winds up with kinetic energy of
(270 keV)
plus
(whatever KE it had when it started accelerating).
Suppose you observed the equation for a traveling wave to be y(x, t) = A cos(kx − ????t), where its amplitude of oscillations was 0.15 m, its wavelength was two meters, and the period was 2/15 s. If a point on the wave at a specific time has a displacement of 0.12 m, what is the transverse speed of that point?
Answer:
15m/sExplanation:
The equation for a traveling wave as expressed as y(x, t) = A cos(kx − [tex]\omega[/tex]t) where An is the amplitude f oscillation, [tex]\omega[/tex] is the angular velocity and x is the horizontal displacement and y is the vertical displacement.
From the formula; [tex]k =\frac{2\pi x}{\lambda} \ and \ \omega = 2 \pi f[/tex] where;
[tex]\lambda \ is\ the \ wavelength \ and\ f \ is\ the\ frequency[/tex]
Before we can get the transverse speed, we need to get the frequency and the wavelength.
frequency = 1/period
Given period = 2/15 s
Frequency = [tex]\frac{1}{(2/15)}[/tex]
frequency = 1 * 15/2
frequency f = 15/2 Hertz
Given wavelength [tex]\lambda[/tex] = 2m
Transverse speed [tex]v = f \lambda[/tex]
[tex]v = 15/2 * 2\\\\v = 30/2\\\\v = 15m/s[/tex]
Hence, the transverse speed at that point is 15m/s
A small helium-neon laser emits red visible light with a power of 5.40 mW in a beam of diameter 2.30 mm.
Required:
a. What is the amplitude of the electric field of the light? Express your answer with the appropriate units.
b. What is the amplitude of the magnetic field of the light?
c. What is the average energy density associated with the electric field? Express your answer with the appropriate units.
d. What is the average energy density associated with the magnetic field? Express your answer with the appropriate units.
E) What is the total energy contained in a 1.00-m length of the beam? Express your answer with the appropriate units.
Answer:
A. 990v/m
B.330x10^-8T
C.2.19x10^-6J/m³
D.1.45x10^-11J
Explanation:
See attached file
Question 18(Multiple Choice Worth 2 polnis)
When riding your skateboard you crash into a curb, the skateboard stops, and you continue moving forward. Which law of
motion is being described in this scenario?
O Law of Universal Gravitation
o Newton's Second Law of Motion
o Law of Conservation of Energy
o Newton's First Law of Motion
A single-slit diffraction pattern is formed on a distant screen. Assuming the angles involved are small, by what factor will the width of the central bright spot on the screen change if the slit width is doubled
Answer:
y ’= y / 2
thus when the slit width is doubled the pattern width is halved
Explanation:
The diffraction of a slit is given by the expressions
a sin θ = m λ
where a is the width of the slit, λ is the wavelength and m is an integer that determines the order of diffraction.
sin θ = m λ / a
If this equation
a ’= 2 a
we substitute
2 a sin θ'= m λ
sin θ'= (m λ / a) 1/2
sin θ ’= sin θ / 2
We can use trigonometry to find the width
tan θ = y / L
as the angle is small
tan θ = sin θ / cos θ = sin θ
sin θ = y / L
we substitute
y ’/ L = y/L 1/2
y ’= y / 2
thus when the slit width is doubled the pattern width is halved
Two blocks A and B have a weight of 11 lb and 5 lb , respectively. They are resting on the incline for which the coefficients of static friction are μA = 0.16 and μB = 0.23. Determine the incline angle θ for which both blocks begin to slide. Also find the required stretch or compression in the connecting spring for this to occur. The spring has a stiffness of k = 2.1 lb/ft .
Answer:
[tex]\theta=10.20^{\circ}[/tex]
[tex]\Delta l=0.10 ft[/tex]
Explanation:
First of all, we analyze the system of blocks before starting to move.
[tex]\Sum F_{x}=P_{A}sin(\theta)+P_{B}sin(\theta)-F_{fA}-F_{fB}=0[/tex]
[tex]\Sum F_{x}=11sin(\theta)+5sin(\theta)-0.16N_{A}-0.23N_{B}=0[/tex]
[tex]11sin(\theta)+5sin(\theta)-0.16P_{A}cos(\theta)-0.23P_{B}cos(\theta)=0[/tex]
[tex]11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0[/tex]
[tex]11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0[/tex]
[tex]16sin(\theta)-2.91cos(\theta)=0[/tex]
[tex]tan(\theta)=0.18[/tex]
[tex]\theta=arctan(0.18)[/tex]
[tex]\theta=10.20^{\circ}[/tex]
Hence, the incline angle θ for which both blocks begin to slide is 10.20°.
Now, if we do a free body diagram of block A we have that after the block moves, the spring force must be taken into account.
[tex]P_{A}sin(\theta)-F_{fA}-F_{spring}=0[/tex]
Where:
[tex]F_{spring} = k\Delta l=2.1\Delta l[/tex]
[tex]P_{A}sin(\theta)-0.16*11cos(\theta)-2.1\Delta l=0[/tex]
[tex]\Delta l=\frac{11sin(\theta)-0.16*11cos(\theta)}{2.1}[/tex]
[tex]\Delta l=0.10 ft[/tex]
Therefore, the required stretch or compression in the connecting spring is 0.10 ft.
I hope it helps you!
(a) The inclined angle for which both blocks begin to slide is 10.3⁰.
(b) The compression of the spring is 0.22 ft.
The given parameters;
mass of block A, = 11 lbmass of block B, = 5 lbcoefficient of static friction for A, = 0.16coefficient of static friction for B, = 0.23 spring constant, k = 2.1 lb/ftThe normal force on block A and B:
[tex]F_n_A = m_Agcos \ \theta\\\\F_n_B = m_Bgcos \ \theta[/tex]
The frictional force on block A and B:
[tex]F_f_A = \mu_s_AF_n_A \\\\F_f_B = \mu_s_BF_n_A[/tex]
The net force on the blocks when they starts sliding;
[tex](m_Ag sin \theta+ m_Bgsin\theta) - (F_f_A + F_f_B) = 0\\\\m_Ag sin \theta+ m_Bgsin\theta = F_f_A + F_f_B\\\\m_Ag sin \theta+ m_Bgsin\theta = \mu_Am_Agcos\theta \ + \ \mu_Bm_Bgcos\theta\\\\gsin\theta(m_A + m_B) = gcos\theta (\mu_Am_A + \mu_Bm_B)\\\\\frac{sin\theta}{cos \theta} = \frac{\mu_Am_A\ + \ \mu_Bm_B}{m_A\ + \ m_B} \\\\tan\theta = \frac{(0.16\times 11) \ + \ (0.23 \times 5)}{11 + 5} \\\\tan\theta = 0.1819\\\\\theta = tan^{-1}(0.1819)\\\\\theta = 10.3 \ ^0[/tex]
The change in the energy of the blocks is the work done in compressing the spring;
[tex]\Delta E = W\\\\F_A (sin \theta )d- \mu F_n d= \frac{1}{2} kd^2\\\\F_A sin\theta \ - \ \mu F_A cos\theta = \frac{1}{2} kd\\\\d = \frac{2F_A(sin\theta - \mu cos \theta) }{k} \\\\d = \frac{2\times 11(sin \ 10.3\ - \ 0.16\times cos \ 10.3) }{2.1} \\\\d = 0.22 \ ft[/tex]
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Equipotential lines are lines with equal electric potential (for example, all the points with an electric potential of 5.0 V). Using the plot tool that comes with voltmeter (pencil icon) make two equipotential lines at r = 0.5 m and r = 1.5 m. Enable electric field vectors in the simulation. Put an electric field sensor at different points on the equipotential line and note the direction of the electric field vector. What can you conclude about the direction of the electric field vector in relation to the equipotential lines?
The direction for each field vector is perpendicular to equipotential lines.
Take a snapshot of the simulation showing equipotential lines and paste to a word document.
....................
At what speed (in m/s) will a proton move in a circular path of the same radius as an electron that travels at 7.45 ✕ 106 m/s perpendicular to the Earth's magnetic field at an altitude where the field strength is 1.10 ✕ 10−5 T
Answer:
The speed of the proton is 4059.39 m/s
Explanation:
The centripetal force on the particle is given by;
[tex]F = \frac{mv^2}{r}[/tex]
The magnetic force on the particle is given by;
[tex]F = qvB[/tex]
The centripetal force on the particle must equal the magnetic force on the particle, for the particle to remain in the circular path.
[tex]\frac{mv^2}{r} = qvB\\\\r = \frac{mv^2}{qvB} \\\\r = \frac{mv}{qB}[/tex]
where;
r is the radius of the circular path moved by both electron and proton;
⇒For electron;
[tex]r = \frac{(9.1*10^{-31})(7.45*10^6)}{(1.602*10^{-19})(1.1*10^{-5})}\\\\r = 3.847 \ m[/tex]
⇒For proton
The speed of the proton is given by;
[tex]r = \frac{mv}{qB}\\\\mv = qBr\\\\v = \frac{qBr}{m} \\\\v = \frac{(1.602*10^{-19})(1.1*10^{-5})(3.847)}{1.67*10^{-27}} \\\\v = 4059.39 \ m/s[/tex]
Therefore, the speed of the proton is 4059.39 m/s
An electron moving in the direction of the +x-axis enters a magnetic field. If the electron experiences a magnetic deflection in the -y direction, the direction of the magnetic field in this region points in the direction of the:______
Answer:
-z axis
Explanation:
According to the left hand rule for an electron in a magnetic field, hold the thumb of the left hand at a right angle to the rest of the fingers, and the rest of the fingers parallel to one another. If the thumb represents the motion of the electron, and the other fingers represent the direction of the field, then the palm will push in the direction of the force on the electron. In this case, the left hand will be held out with the thumb pointing to the right (+x axis), and the palm facing your body (-y axis). The magnetic field indicated by the other fingers will point down in the the -z axis.
The difference between a DC and an AC generator is that
a. the DC generator has one unbroken slip ring.
b. the AC generator has one unbroken slip ring
c. the DC generator has one slip ring splitin two halves.
d. the AC generator has one slip ring split in two halves.
e The DC generator has twounbroken sip rings
Answer:
The AC generator has one unbroken slip ring
Explanation:
In physics, the application of electromagnetic induction can be seen in generators and dynamos. Electromagnetic induction is the process of generating electricity using magnets. It found applications in generators and the types of generator they found application is in AC and DC generator.
An AC generator is also called a Dynamo. A DC generator contains what is called a SPLIT RING fixed to the end of the coil which can be separated and coupled back according to the name "split". An AC generator also called a Dynamo makes use of a SLIP ring which cannot be divided into two. It comes as an entity. The presence of this rings is what differentiates a DC generator from an AC generator.
We can replace split rings with slip rings when converting a DC generator to an AC generator and vice versa.
It can therefore be concluded that the difference between a DC and an AC generator is that the AC generator has one unbroken slip ring.
3. Which of the following accurately describes circuits?
O A. In a parallel circuit, the same amount of current flows through each part of the circuit
O B. In a series circuit, the amount of current passing through each part of the circuit may vary
O C. In a series circuit, the current can flow through only one path from start to finish
O D. In a parallel circuit, there's only one path for the current to travel.
Answer:
Option (c)
Explanation:
In a Series circuit, as the components are connected end-to-end ,the current can flow through only one path from start to finish.
(C.) is the only correct statement in the list of choices.
In a series circuit, the current can flow through only one path from start to finish.
With the same block-spring system from above, imagine doubling the displacement of the block to start the motion. By what factor would the following change?
A. Kinetic energy when passing through the equilibrium position.
B. Speed when passing through the equilibrium position.
Answer:
A) K / K₀ = 4 b) v / v₀ = 4
Explanation:
A) For this exercise we can use the conservation of mechanical energy
in the problem it indicates that the displacement was doubled (x = 2xo)
starting point. At the position of maximum displacement
Em₀ = Ke = ½ k (2x₀)²
final point. In the equilibrium position
[tex]Em_{f}[/tex] = K = ½ m v²
Em₀ = Em_{f}
½ k 4 x₀² = K
(½ K x₀²) = K₀
K = 4 K₀
K / K₀ = 4
B) the speed value
½ k 4 x₀² = ½ m v²
v = 4 (k / m) x₀
if we call
v₀ = k / m x₀
v = 4 v₀
v / v₀ = 4
An electrostatic paint sprayer contains a metal sphere at an electric potential of 25.0 kV with respect to an electrically grounded object. Positively charged paint droplets are repelled away from the paint sprayer's positively charged sphere and towards the grounded object. What charge must a 0.168-mg drop of paint have so that it will arrive at the object with a speed of 18.8 m/s
Answer:
The charge is [tex]Q = 2.177 *10^{-9} \ C[/tex]
Explanation:
From the question we are told that
The electric potential is [tex]V = 25.0 \ kV = 25.0 *10^{3}\ V[/tex]
The mass of the drop is [tex]m = 0.168 \ m g = 0.168 *10^{-3} \ g = 0.168 *10^{-6}\ kg[/tex]
The speed is [tex]v = 18.8 \ m/s[/tex]
Generally the charge on the paint drop due to the electric potential which will give it the speed stated in the question is mathematically represented as
[tex]Q = \frac{m v^2 }{ 2 * V }[/tex]
Substituting values
[tex]Q = \frac{0.168 *10^{-6} (18)^2 }{ 2 * 25*10^3 }[/tex]
[tex]Q = 2.177 *10^{-9} \ C[/tex]
A toroidal solenoid with 400 turns of wire and a mean radius of 6.0 cm carries a current of 0.25 A. The relative permeability of the core is 80.
(a) What is the magnetic field in the core?
(b) What part of the magnetic field is due to atomic currents?
Answer:
A) 0.0267 T
B) 0.0263 T
Explanation:
Given that
The number of turns, N = 400
Radius of the wire, r = 6 cm = 0.06 m
Current in the wire, I = 0.25 A
Relative permeability, K(m) = 80
See the attached picture for the calculation
A semi-circular loop consisting of one turn of wire is place in the x-y plane. A constant magnetic field B=1.7T points along the negative z-axis(into the page), and a current I=0.7A flows counterclockwisefrom the positive z-axis. The net magnetic force on the circular section of the loop points in what direction? What is the net magnetice force on the circular section of the loop?
Answer:
The direction of net magnetic force on the circular section of the loop is in the positive y-axis
The net magnetic force on the circular section of the loop is 3.74 N
Explanation:
The magnetic field strength [tex]B[/tex] = 1.7 T
the current [tex]I[/tex] = 0.7 A
The diameter of the loop = 2 m
the length of the circular section of the semi-circular loop [tex]l[/tex] = πd/2
==> [tex]l[/tex] = (3.142 x 2)/2 = 3.142 m
The force on the semi-circular is given as
F = [tex]BIl[/tex] sin ∅
but the loop is perpendicular to the field, therefore
sin ∅ = sin 90° = 1
F = 1.7 x 0.7 x 3.142 x 1 = 3.74 N
The right hand rule states that "if the fingers of the right hand are held parallel to each other in the direction of the magnetic field, and the thumb is held at right angle to the other fingers in the direction of the flow of current. The palm will push in the direction of the magnetic force on the conductor".
According to the right hand rule, the direction of net magnetic force on the circular section of the loop is in the positive y-axis
Two football teams, the Raiders and the 49ers are engaged in a tug-of-war. The Raiders are pulling with a force of 5000N. Which of the following is an accurate statement?
A. The tension in the rope depends on whether or not the teams are in equilibrium.
B. The 49ers are pulling with a force of more than 5000N because of course they’d be winning.
C. The 49ers are pulling with a force of 5000N.
D. The tension in the rope is 10,000N.
E. None of these statements are true.
Answer:
E. None of these statements are true.
Explanation:
We can't say the exact or approximate amount of tension on the rope, since we do know for sure from the statement who is winning.
for A, the tension on the rope does not depend on if both teams pull are in equilibrium.
for B, the 49ers would be pulling with a force more than 5000 N, if they were winning. The problem is that we can't say with all confidence that they'd be winning.
for C, we don't know how much tension exists on the rope, and its direction, so we can't work out how much tension the 49ers are pulling the rope with.
for D, just as for C above, we can't work out how much tension there is on the rope, since we do not know how much force the 49ers are pulling with.
we go with option E.
A merry-go-round spins freely when Diego moves quickly to the center along a radius of the merry-go-round. As he does this, it is true to say that
Answer:
A) the moment of inertia of the system decreases and the angular speed increases.
Explanation:
The complete question is
A merry-go-round spins freely when Diego moves quickly to the center along a radius of the merry-go-round. As he does this, It is true to say that
A) the moment of inertia of the system decreases and the angular speed increases.
B) the moment of inertia of the system decreases and the angular speed decreases.
C) the moment of inertia of the system decreases and the angular speed remains the same.
D) the moment of inertia of the system increases and the angular speed increases.
E) the moment of inertia of the system increases and the angular speed decreases
In angular momentum conservation, the initial angular momentum of the system is conserved, and is equal to the final angular momentum of the system. The equation of this angular momentum conservation is given as
[tex]I_{1} w_{1} = I_{2} w_{2}[/tex] ....1
where [tex]I_{1}[/tex] and [tex]I_{2}[/tex] are the initial and final moment of inertia respectively.
and [tex]w_{1}[/tex] and [tex]w_{2}[/tex] are the initial and final angular speed respectively.
Also, we know that the moment of inertia of a rotating body is given as
[tex]I = mr^{2}[/tex] ....2
where [tex]m[/tex] is the mass of the rotating body,
and [tex]r[/tex] is the radius of the rotating body from its center.
We can see from equation 2 that decreasing the radius of rotation of the body will decrease the moment of inertia of the body.
From equation 1, we see that in order for the angular momentum to be conserved, the decrease from [tex]I_{1}[/tex] to [tex]I_{2}[/tex] will cause the angular speed of the system to increase from [tex]w_{1}[/tex] to [tex]w_{2}[/tex] .
From this we can clearly see that reducing the radius of rotation will decrease the moment of inertia, and increase the angular speed.
