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
Alpha particles (charge = +2e, mass = 6.68 × 10-27 kg) are accelerated in a cyclotron to a final orbit radius of 0.30 m. The magnetic field in the cyclotron is 0.80 T. The period of the circular motion of the alpha particles is closest to: A. 0.25 μs B. 0.16 μs C. 0.49 μs D. 0.40 μs E. 0.33 μs
Answer:
Option B: T ≈ 0.16 μs
Explanation:
We are given;
Mass; m = 6.68 × 10^(-27) kg
Magnetic field;B = 0.80 T
Charge;q = 2e
Now, e is the charge on an electron and it has a value of 1.6 × 10^(-19) C
So, q = 2 × 1.6 × 10^(-19)
q = 3.2 × 10^(-19) C
The period of the circular motion of the alpha particles moving along a in the presence of the magnetic field is given by;
T = 2πm/qB
Where ;
m, q and B are as stated earlier.
Plugging in the relevant values, we have;
T = (2π × 6.68 × 10^(-27))/(3.2 × 10^(-19) × 0.8)
T = 0.16395 × 10^(-6) s
This can also be written as;
T ≈ 0.16 μs
How are electricity and magnets connected
Answer: The properties of magnets are used to make electricity. Moving magnetic fields pull and push electrons. Moving a magnet around a coil of wire, or moving a coil of wire around a magnet, pushes the electrons in the wire and creates an electrical current.
A lamp has the shape of a parabola when viewed from the side. The lamp is centimeters wide and centimeters deep. How far is the light source from the bottom of the lamp if the light source is placed at the focus
The question is not complete so i have attached it.
Answer:
The light source is 2 cm from the bottom of the lamp
Explanation:
From the attached image, we can see that the parabola opens up with its vertex at the origin.
Now, the standard form of equation for a parabola is:
x² = 4ay
Looking at the parabola in the attachment, the top right edge of the lamp has a coordinate of (12,18)
Thus, we have;
12² = 4a(18)
144 = 72a
a = 144/72
a = 2
Looking at the parabola again, the line of symmetry is at x = 0
Thus, axis of symmetry is at x = 0.
Thus, focus is at (0, 2)
So, if the light source is placed at the focus, the distance of the light source from the bottom of the lamp is 2 cm
The distance of the light source from the bottom of the lamp is 2 cm.
The given parameters;
the top right edge of the lamp has a coordinate of (12,18)Apply standard parabola equation to determine the distance of the light source from the bottom of the lamp;
[tex]x^2 = 4ay\\\\12^2 = 4a(18)\\\\144 = 72 a\\\\a = \frac{144}{72} \\\\a = 2 \ cm[/tex]
Thus, the distance of the light source from the bottom of the lamp is 2 cm.
"Your question is not complete, it seems to be missing the following information";
the top right edge of the lamp has a coordinate of (12,18)
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If the rods with diameters and lengths listed below are made of the same material, which will undergo the largest percentage length change given the same applied force along its length?a. d, 3L b. 3d, L c. 2d, 2L d. 4d, L
Answer:
The highest percentage of change corresponds to the thinnest rod, the correct answer is a
Explanation:
For this exercise we are asked to change the length of the bar by the action of a force applied along its length, in this case we focus on the expression of longitudinal elasticity
F / A = Y ΔL/L
where F / A is the force per unit length, ΔL / L is the fraction of the change in length, and Y is Young's modulus.
In this case the bars are made of the same material by which Young's modulus is the same for all
ΔL / L = (F / A) / Y
the area of the bar is the area of a circle
A = π r² = π d² / 4
A = π / 4 d²
we substitute
ΔL / L = (F / Y) 4 /πd²
changing length
ΔL = (F / Y 4 /π) L / d²
The amount between paracentesis are all constant in this exercise, let's look for the longitudinal change
a) values given d and 3L
ΔL = cte 3L / d²
ΔL = cte L /d² 3
To find the percentage, we must divide the change in magnitude by its value and multiply by 100.
ΔL/L % = [(F /Y 4/π 1/d²) 3L ] / 3L 100
ΔL/L % = cte 100%
b) 3d and L value, we repeat the same process as in part a
ΔL = cte L / 9d²
ΔL = cte L / d² 1/9
ΔL / L% = cte 100/9
ΔL / L% = cte 11%
c) 2d and 2L value
ΔL = (cte L / d ½ )/ 2L
ΔL/L% = cte 100/4
ΔL/L% = cte 25%
d) value 4d and L
ΔL = cte L / d² 1/16
ΔL/L % = cte 100/16
ΔL/L % = cte 6.25%
The highest percentage of change corresponds to the thinnest rod, the correct answer is a
What is the power P of the eye when viewing an object 61.0 cm away? Assume the lens-to-retina distance is 2.00 cm , and express the answer in diopters.
Answer:
The power of the eye is 51.64 diopters
Explanation:
The power of the eye is given by;
[tex]P = \frac{1}{f} = \frac{1}{d_o} +\frac{1}{d_i}[/tex]
where;
P is the power of the eye in diopter
f is the focal length of the eye
[tex]d_o[/tex] is the distance between the eye and the object
[tex]d_i[/tex] is the distance between the eye and the image
Given;
[tex]d_o[/tex] = 61.0 cm = 0.61 m
[tex]d_i[/tex] = 2.0 cm = 0.02 m
[tex]P = \frac{1}{d_o} +\frac{1}{d_i} \\\\P = \frac{1}{0.61} + \frac{1}{0.02} \\\\P = 51.64 \ D[/tex]
Therefore, the power of the eye is 51.64 diopters.
The power P of the eye when viewing an object 61.0 cm away is 51.639D
The power of a lens is a reciprocal of its focal length and it is expressed as:
[tex]P=\frac{1}{f}[/tex]
According to the mirror formula
[tex]\frac{1}{f} =\frac{1}{d_i} +\frac{1}{d_0}[/tex]
where
[tex]d_i[/tex] is the distance from the lens to the image = 61.0cm = 0.61m
[tex]d_0[/tex] is the distance from the lens to the object = 2.00cm = 0.02m
[tex]P=\frac{1}{f} =\frac{1}{0.02} +\frac{1}{0.61}\\P=50+1.639\\P=51.639D[/tex]
Hence the power P of the eye when viewing an object 61.0 cm away is 51.639D
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An LR circuit consists of a 35-mH inductor, a resistance of 12 ohms, an 18-V battery, and a switch. What is the current 5.0 ms after the switch is closed
Answer:
Current, I = 1.23 A
Explanation:
Given that,
Inductance, L = 35 mH
Resistance, R = 12 ohms
Potential difference, V = 18 V
We need to find current 5 ms after the switch is closed. Current in LR circuit is given by :
[tex]I=I_o(1-e^{-t/\tau })[/tex] ....(1)
Here,
[tex]I_o[/tex] is final current
[tex]I_o=\dfrac{V}{R}\\\\I_o=\dfrac{18}{12}=1.5\ A[/tex]
[tex]\tau[/tex] is time constant
[tex]\tau=\dfrac{L}{R}\\\\\tau=\dfrac{35\times 10^{-3}}{12}\\\\\tau=0.00291\ s[/tex]
So, equation (1) becomes :
[tex]I=1.5\times (1-e^{-5\times 10^{-3}/0.00291})\\\\I=1.23\ A[/tex]
So, after 5 ms the current in the circuit is 1.23 A.
2. The nuclear model of the atom held that
a. electrons were randomly spread through "a sphere of uniform positive
electrification."
b. matter was made of tiny electrically charged particles that were smaller than the
atom
C. matter was made of tiny, indivisible particles.
d. the atom had a dense, positively charged nucleus.
Answer:
the atom had a dense, positively charged nucleus.
Explanation:
Ernest Rutherford, based on the experiment carried out by two of his graduate students, established the authenticity of the nuclear model of the atom.
According to the nuclear model, an atom is made up of a dense positive core called the nucleus. Electrons are found to move round this nucleus in orbits. This is akin to the movement of the planets round the sun in the solar system.
One day, after pulling down your window shade, you notice that sunlight is passing through a pinhole in the shade and making a small patch of light on the far wall. Having recently studied optics in your physics class, you're not too surprised to see that the patch of light seems to be a circular diffraction pattern. It appears that the central maximum is about 1 cm across, and you estimate that the distance from the window shade to the wall is about 4 m.
Estimate:
a. The average wavelength of the sunlight (in nm)
b. The diameter of the pinhole (in mm).
Given that,
Central maximum = 1 cm
Distance from the window shade to the wall =4 m
We know that,
The visible range of the sun light is 400 nm to 700 nm.
(a). We need to calculate the average wavelength
Using formula of average wavelength
[tex]\lambda_{avg}=\dfrac{\lambda_{1}+\lambda_{2}}{2}[/tex]
Put the value into the formula
[tex]\lambda_{avg}=\dfrac{400+700}{2}[/tex]
[tex]\lambda_{avg}=550\ nm[/tex]
(b). We need to calculate the diameter of the pinhole
Using formula for diameter
[tex]w=\dfrac{2.44\lambda L}{D}[/tex]
[tex]D=\dfrac{2.44\lambda L}{w}[/tex]
Put the value into the formula
[tex]D=\dfrac{2.44\times550\times10^{-9}\times4}{1\times10^{-2}}[/tex]
[tex]D=0.537\ mm[/tex]
Hence, (a). The average wavelength 550 nm.
(b). The diameter of the pinhole is 0.537 mm.
Rank the following types of electromagnetic waves by the wavelength of the wave.
a. Microwaves
b. X-rays
c. Radio waves
d. Visible light
Explanation:
In order of Increasing Wavelength of the Electromagnetic Spectrum :
B) X rays
D) Visible light
A) Microwave
C) Radio Waves
Electromagnetic waves in order of decreasing wavelength is X-rays,visible light,microwaves and radio waves.
What are electromagnetic waves?The electromagnetic radiation consists of waves made up of electromagnetic field which are capable of propogating through space and carry the radiant electromagnetic energy.
The radiation are composed of electromagnetic waves which are synchronized oscillations of electric and magnetic fields . They are created due to change which is periodic in electric as well as magnetic fields.
In vacuum ,all the electromagnetic waves travel at the same speed that is with the speed of air.The position of an electromagnetic wave in an electromagnetic spectrum is characterized by it's frequency or wavelength.They are emitted by electrically charged particles which undergo acceleration and subsequently interact with other charged particles.
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A horizontal circular platform rotates counterclockwise about its axis at the rate of 0.945 rad/s. You, with a mass of 69.7 kg, walk clockwise around the platform along its edge at the speed of 1.01 m/s with respect to the platform. Your 20.7 kg poodle also walks clockwise around the platform, but along a circle at half the platform's radius and at half your linear speed with respect to the platform. Your 17.7 kg mutt, on the other hand, sits still on the platform at a position that is 3/4 of the platform's radius from the center. Model the platform as a uniform disk with mass 93.1 kg and radius 1.93 m. Calculate the total angular momentum of the system.
Answer:
317.22
Explanation:
Given
Circular platform rotates ccw 93.1kg, radius 1.93 m, 0.945 rad/s
You 69.7kg, cw 1.01m/s, at r
Poodle 20.2 kg, cw 1.01/2 m/s, at r/2
Mutt 17.7 kg, 3r/4
You
Relative
ω = v/r
= 1.01/1.93
= 0.522
Actual
ω = 0.945 - 0.522
= 0.42
I = mr^2
= 69.7*1.93^2
= 259.6
L = Iω
= 259.6*0.42
= 109.4
Poodle
Relative
ω = (1.01/2)/(1.93/2)
= 0.5233
Actual
ω = 0.945- 0.5233
= 0.4217
I = m(r/2)^2
= 20.2*(1.93/2)^2
= 18.81
L = Iω
= 18.81*0.4217
= 7.93
Mutt
Actual
ω = 0.945
I = m(3r/4)^2
= 17.7(3*1.93/4)^2
= 37.08
L = Iω
= 37.08*0.945
= 35.04
Disk
I = mr^2/2
= 93.1(1.93)^2/2
= 173.39
L = Iω
= 173.39*0.945
= 163.85
Total
L = 109.4+ 7.93+ 36.04+ 163.85
= 317.22 kg m^2/s
Discuss the phase change condition due to reflection of light from a surface. Summarize equations of interference for thin film.
Answer:
if this surface has a higher index than in the medium where the light travels, the reflected wave has a phase change of 180º
Explanation:
When a ray of light falls on a surface if this surface has a higher index than in the medium where the light travels, the reflected wave has a phase change of 180º this can be explained by Newton's third law, the light when arriving pushes the atoms of the medium that is more dense, and these atoms respond with a force of equal magnitude, but in the opposite direction.
When the fractional index is lower than that of the medium where the reflacted beam travels, notice a change in phase.
Also, when light penetrates the medium, it modifies its wavelength
λ = λ₀ / n
We take these two aspects into account, the condition for contributory interference is
d sin θ = (m + 1/2) λ
for destructive interference we have
d sin θ = m λ
in general this phenomenon is observed at 90º
2 d = (m +1/2) λ° / n
2nd = (m + ½) λ₀
Calculate the wavelength of light that has its third minimum at an angle of 30.0º when falling on double slits separated by 3.00 µm.
Answer:
λ = 428.6 nm
Explanation:
Hello,
In this case, we must remember that the Young's double slit experiment is described by the expression :
d sin θ = m λ
For constructive interference , and:
d sin θ = (m + ½) λ
For destructive interference , whereas d accounts for the distance between the slits, λ for the wavelength and m for an integer that describes the order of interference . Thus, for the given angle 30º, the distance between the slits is 3.00 μm or 3.00 10⁻⁶ m and the order of interference is 3; we therefore use the destructive interference equation to compute the wavelength as shown below:
λ = 3x10⁻⁶ sin (30) / (3 +1/2)
λ = 4.286 10⁻⁷ m
Or in manometers:
λ = 428.6 nm
Best regards.
A step-down transformer is used for recharging the batteries of portable devices. The turns ratio N2/N1 for a particular transformer used in a CD player is 2:29. When used with 120-V (rms) household service, the transformer draws an rms current of 180 mA.
Find the rms output voltage of the transformer
Answer:
8.28 V
Explanation:
Using,
N2/N1 = V2/V1.................. Equation 1
Where N2/N1 = Turn ratio of the transformer, V1 = primary/input voltage, V2 = output/secondary voltage
make V2 the subject of the equation
V2 = (N2/N1)V1............ Equation 2
Given: N2/N1 = 2:29 = 2/29, V1 = 120 V
Substitute these values into equation 2
V2 = (2/29)120
V2 = 8.28 V
Hence the rms output voltage of the transformer = 8.28 V
In a physics lab, Asha is given a 11.5 kg uniform rectangular plate with edge lengths 62.9 cm by 46.9 cm . Her lab instructor requires her to rotate the plate about an axis perpendicular to its plane and passing through one of its corners, and then prepare a report on the project. For her report, Asha needs the plate's moment of inertia ???? with respect to given rotation axis. Calculate ???? .
Answer:
6.9kgm²
Explanation:
For an axis through the center of the rectangle, I = m[(w²+L²)/12
Using the parallel axis theorem, the added value of I = mR² = m[(w²/4 + L²/4]
Adding the 2 expressions,
I = (m/3)*(w²+L²)
I =6.95 kg∙m²
The only force acting on a 3.4 kg canister that is moving in an xy plane has a magnitude of 3.0 N. The canister initially has a velocity of 2.5 m/s in the positive x direction, and some time later has a velocity of 4.8 m/s in the positive y direction. How much work is done on the canister by the 3.0 N force during this time
Answer:
16.79JExplanation:
Given data
mass of canister= 3.4 kg
force acting on canister= 3 N
initial velocity u= 2.5 m/s
final velocity v= 4.8 m/s
The work done on the canister is the change in kinetic energy on the canister
change in [tex]KE= Kfinal- Kinitial[/tex]
K.E initial
[tex]Kintial= \frac{1}{2} mv^2\\\\Kintial= \frac{1}{2}*2*2.5^2\\\\KInitial= \frac{1}{2} *2*6.25\\\\Kinitial= 6.25J[/tex]
K.E final
[tex]Kfinal= \frac{1}{2} mv^2\\\\ Kfinal= \frac{1}{2}*2*4.8^2\\\\ Kfinal= \frac{1}{2} *2*23.04\\\\ Kfinal= 23.04J[/tex]
The net work done is [tex]KE= Kfinal- Kinitial[/tex]
[tex]W net= 23.04-6.25= 16.79J[/tex]
g Calculate the maximum wavelength of light that will cause the photoelectric effect for potassium. Potassium has work function 2.29 eV = 3.67 x 10–19 J.
Answer:
λ = 5.4196 10⁻⁷m, λ = 541.96 nm this is green ligh
Explanation:
The photoelectric effect was explained by Eintein assuming that the light was made up of particles called photons and these collided with the electrons taking them out of the material.
K = h f -Ф
where K is the kinetic energy of the ejected electrons, hf is the energy of the light quanta and fi is the work function of the material.
The speed of light is related to wavelength and frequency
c = λ / f
f = c /λ
we substitute
K = h c / λ - Φ
for the case that they ask us the kinetic energy of the electons is zero (K = 0)
h c / λ = Ф
λ = h c / Ф
we calculate
λ = 6.63 10⁻³⁴ 3 10⁸ / 3.67 10⁻¹⁸
λ = 5.4196 10⁻⁷m
let's take nm
lam = 541.96 nm
this is green light
zeugen and yardang differences
Answer:
Yardangs are formed on vertical strata while zeugen on horizontal strata. ... Yardangs are formed on vertical hard/soft layers of rock, while zeugen (this is its plural form) are formed on horizontal bands of hard/soft rocks giving it a more mushroom-like shape. The Great Sphinx of Giza has been sculpted in a yardang
A polarized laser beam of intensity 285 W/m2 shines on an ideal polarizer. The angle between the polarization direction of the laser beam and the polarizing axis of the polarizer is 16.0 ∘. What is the intensity of the light that emerges from the polarizer?
Answer:
The intensity is [tex]I_1 = 263.35 \ W/m^2[/tex]
Explanation:
From the question we are told that
The intensity of the beam is [tex]I = 285\ W/m^2[/tex]
The angle is [tex]\theta = 16^o[/tex]
The intensity of the light that emerges from the polarizer is mathematically represented by Malus' law as
[tex]I_1 = I * cos^2 (\theta )[/tex]
substituting values
[tex]I_1 = 285 * [cos(16)]^2[/tex]
substituting values
[tex]I_1 = 285 * [cos(16)]^2[/tex]
[tex]I_1 = 263.35 \ W/m^2[/tex]
Some stove tops are smooth ceramic for easy cleaning. If the ceramic is 0.630 cm thick and heat conduction occurs through an area of 1.45 ✕ 10−2 m2 at a rate of 500 J/s, what is the temperature difference across it (in °C)? Ceramic has the same thermal conductivity as glass and concrete brick.
Answer:
The temperature difference [tex]\Delta T = 258.6 \ ^ o\ C[/tex]
Explanation:
From the question we are told that
The thickness is [tex]\Delta x = 0.630 cm = 0.0063 m[/tex]
The area is [tex]A = 1.45 *10^{-2 } \ m^2[/tex]
The rate is [tex]P = 500 J/s[/tex]
The thermal conductivity is [tex]\sigma = 0.84J[\cdot s \cdot m \cdot ^oC ][/tex]
Generally the rate heat conduction mathematically represented as
[tex]P = \sigma * A * \frac{\Delta T}{\Delta x }[/tex]
=> [tex]\Delta T = \frac{P * \Delta x }{\sigma * A }[/tex]
=> [tex]\Delta T = \frac{ 500 * 0.00630 }{ 0.84 * 1.45 *10^{-2} }[/tex]
=> [tex]\Delta T = 258.6 \ ^ o\ C[/tex]
The distance from the Sun to Earth is approximately 149,600,000 km. The distance from the Sun to Venus is approximately 108,200,000 km. The elongation angle αα is the angle formed between the line of sight from Earth to the Sun and the line of sight from Earth to Venus. Suppose that the elongation angle for Venus is 10∘.10 ∘. Use this information to find the possible distances between Earth and Venus.
Answer:
335206922km
Explanation:
Pls see attached file
What do Earth scientists do?
Answer:
Study Earth as a whole
Explanation:
ex. oxygen around Earth, layers, formations, temperature, mountains and how they form etc.
Answer:
Geologists study rocks and help to locate useful minerals. Earth scientists often work in the field—perhaps climbing mountains, exploring the seabed, crawling through caves, or wading in swamps. They measure and collect samples (such as rocks or river water), then they record their findings on charts and maps.
An 1,820 W toaster, a 1,420 W electric frying pan, and a 55 W lamp are plugged into the same outlet in a 15 A, 120 V circuit. (The three devices are in parallel when plugged into the same socket.)
Required:
a. What current is drawn by each device?
b. Will this combination blow the 15-A fuse?
Answer:
toaster- 15.1A
electric frying pan- 11.8 A
lamp- 0.5 A
b) The combination will blow the fuse.
Explanation:
When devices are connected in parallel, the potential difference across each of the devices is the same but the current through each is different. Hence;
V= 120 V
Power= IV
For the toaster;
I= 1820/120 = 15.1 A
For the electric frying pan;
I= 1420/120 = 11.8 A
For the lamp;
55/120 = 0.5 A
Total current = 15.1 +11.8 + 0.5 = 27.4 A
The combination will blow the fuse.
Explanation:
step one:
Given data
power of toaster= 1,820 W
power of electric frying pan= 1,420 W
power of lamp= 55 W
current of the outlet= 15 A
voltage of outlet = 120 V
step two
since all three appliances are connected in parallel to the socket outlet, they will use the same voltage of 120 V and the currents will be different across each appliance,
Hence the current across the Toaster will be I₁
using P=I₁V we have
I₁= P/V
I₁= 1820/120 = 15.16 A
A. The current drawn by each device
the current across the electric frying pan will be I₂
using P=I₂V we have
I₂= P/V
I₂= 1420/120 = 11.83 A
the current across the lamp will be I₃
using P=I₃V we have
I₃= P/V
I₃= 55/120 = 0.45 A
therefore the total current drawn by all appliances will be
Total current = I₁+I₂+I₃= 15.16 +11.83+ 0.45= 27.44
B. Will this combination blow the 15-A fuse?
27.44 A > 15 A by 45% ...and this will make fuse to blow
If this is the only water being used in your house, how fast is the water moving through your house's water supply line, which has a diameter of 0.021 m (about 3/4 of an inch)?
Answer:
0.273m/s
Explanation:
first find out the meaning of 0.90×10−4m3/s
literally, that is 0.9x6 = 5.4m3/s = 3•5.4m/s or 16.2 m/s
1.5 gal/min = 0.00009464 m³/s, perhaps that is what you mean?
cross-sectional area of pipe is πr² = 0.0105²π = 0.0003464 m²
so you have a a flow of 0.00009464 m³/s flowing through an area of 0.0003464 m²
they divide to 0.00009464 m³/s / 0.0003464 m² = 0.273 m/s
Specular reflection occurs where the light ray in the glass strikes the reflector. If no light is to enter the water, we require that there be reflection only. Which phenomenon prevents the light from entering the water?
Answer:
The critical angle phenomenon.
Explanation:
Critical angle in optics is the smallest angle of incidence of a wave, that will give total reflection of the wave. This phenomenon occurs at the boundary of two medium, where light will normally move from one medium to another.
To prevent light from entering the water, the angle of incidence of the light incident on the water must exceed the critical angle.
A metal cube with sides of length a=1cm is moving at velocity v0→=1m/sj^ across a uniform magnetic field B0→=5Tk^. The cube is oriented so that four of its edges are parallel to its direction of motion (i.e., the normal vectors of two faces are parallel to the direction of motion).Find E, the magnitude of the induced electric field inside the cube. Express your answer numerically, in newtons per coulomb.
Answer:
the magnitude of the electric field is 1.25 N/C
Explanation:
The induced emf in the cube ε = LB.v where B = magnitude of electric field = 5 T , L = length of side of cube = 1 cm = 0.01 m and v = velocity of cube = 1 m/s
ε = LB.v = 0.01 m × 5 T × 1 m/s = 0.05 V
Also, induced emf in the cube ε = ∫E.ds around the loop of the cube where E = electric field in metal cube
ε = ∫E.ds
ε = Eds since E is always parallel to the side of the cube
= E∫ds ∫ds = 4L since we have 4 sides
= E(4L)
= 4EL
So,4EL = 0.05 V
E = 0.05 V/4L
= 0.05 V/(4 × 0.01 m)
= 0.05 V/0.04 m
= 1.25 V/m
= 1.25 N/C
So, the magnitude of the electric field is 1.25 N/C
The magnitude of the electric field is 1.25 N/C
Calculation of the magnitude of the electric field:But before that the following calculations need to be done.
ε = LB.v = 0.01 m × 5 T × 1 m/s
= 0.05 V
Now
ε = ∫E.ds
here ε = Eds because E is always parallel to the side of the cube
So,
= E∫ds ∫ds
= 4L so we have 4 sides
Now
= E(4L)
= 4EL
So,4EL = 0.05 V
Now
E = 0.05 V/4L
= 0.05 V/(4 × 0.01 m)
= 0.05 V/0.04 m
= 1.25 V/m
= 1.25 N/C
hence, The magnitude of the electric field is 1.25 N/C
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Which one of the following actions would make the maxima in the interference pattern from a grating move closer together?1. Increasing the wavelength of the laser.2. Increasing the distance to the screen.3. Increasing the frequency of the laser.4. Increasing the number of lines per length.
Answer:
Increase in frequency of the laser
Explanation:
Because An increase in frequency will result in more lines per centimeter and a smaller distance between each consecutive line. And a decrease in distance between each gratin
A spark is generated in an automobile spark plug when there is an electric potential of 3000 V across the electrode gap. If 60 W of power is generated in a single spark that delivers a total charge of 3 nC, how long does it take for the spark to travel across the gap?
A. 50 ns
B. 75 ns
C. 125 ns
D. 150 ns
E. 225 ns 5
Answer:
The correct option is d
Explanation:
From the question we are told that
The electric potential is [tex]V = 3000 \ V[/tex]
The power is [tex]P = 60 \ W[/tex]
The charge delivered is [tex]q = 3nC = 3.0 *10^{-9} \ C[/tex]
Generally the power generated is mathematically represented as
[tex]P = I V[/tex]
=> [tex]I = \frac{P}{V }[/tex]
=> [tex]I = \frac{60 }{3000 }[/tex]
=> [tex]I = 0.02 \ A[/tex]
This current flow is mathematically represented as
[tex]I = \frac{q -q_o}{\Delta t }[/tex]
Where [tex]q_o[/tex] is the charge delivered at t=0 s which is 0s
So
[tex]0.02 = \frac{ (3.0 *10^{-9}) -0 }{t - 0 }[/tex]
[tex]t = 1.50 *10^{-7 } \ s[/tex]
[tex]t = 150 *10^{-9 } \ s[/tex]
The target variable is the speed of light v in the glass, which you can determine from the index of refraction n of the glass. Which equations will you use to find n and v?
Answer:
n= speed of light in vacuum/ speed of light in the other medium.
Explanation:
If light is moving from medium 1 into medium 2 where medium 1 is vacuum (approximated to mean air) and we are required to find the velocity of light; then we can confidently write;
n= speed of light in vacuum/ speed of light in the other medium.
Hence;
n= c/v
Where;
n= refractive index of the material
c= speed of light in vacuum
v = speed of light in another medium.
Note that the refractive index is the amount by which a transparent medium decreases the speed of light.
A box is sliding down an incline tilted at a 12° angle above horizontal. The box is initially sliding down the incline at a speed of 1.5 m/s. The coefficient of kinetic friction between the box and the incline is 0.34. How far does the box slide down the incline before coming to rest?
Answer:
The box will cover a distance of 0.9199m before coming to rest
Explanation:
We are given;
Angle of tilt; θ = 12°
Speed of sliding down; u = 1.5 m/s
Coefficient of kinetic friction; μ = 0.34
We are told that the box is sliding down an incline tilted at a 12° angle above horizontal.
Thus,
The components of the weight of the block would be;
Fx = mg sinθ = mg sin 12
Fy = mg cosθ = mg cos 12
For, the normal force on the block, it will be counter balanced by the Y component of weight of block and so we have;
Normal force; Fn = mg cos 12
Now formula for the frictional force would be given by;
Ff = μmg cos 12
So, Ff = 0.34mg cos 12
So, the net force along the inclined plane is;
Fnet = Fx - Ff
Fnet = mg sin 12 - 0.34mg cos 12
Where Fnet = mass x acceleration.
Thus;
ma = mg sin 12 - 0.34mg cos 12
m will cancel out to give;
a = g sin 12 - 0.34g cos 12
a = 9.81(0.2079) - 0.34(9.81 × 0.9781)
a = -1.223 m/s²
According to Newton's equation of motion, we have;
(v² - u²) = 2as
s = (v² - u²)/2a
Final velocity is zero. Thus;
s = (0² - 1.5²)/(2 × -1.223)
s = -2.25/-2.446
s = 0.9199 m
Thus, the box will cover 0.9199m before coming to rest
A rod bent into the arc of a circle subtends an angle 2θ at the center P of the circle (see below). If the rod is charged uniformly with a total charge Q, what is the electric field at P? (Assume Q is positive. For the magnitude, use the following as necessary: ε0, Q, R, and θ.)
Answer:
Qsinθ/4πε₀R²θ
Explanation:
Let us have a small charge element dq which produces an electric field E. There is also a symmetric field at P due to a symmetric charge dq at P. Their vertical electric field components cancel out leaving the horizontal component dE' = dEcosθ = dqcosθ/4πε₀R² where r is the radius of the arc.
Now, let λ be the charge per unit length on the arc. then, the small charge element dq = λds where ds is the small arc length. Also ds = Rθ.
So dq = λRdθ.
Substituting dq into dE', we have
dE' = dqcosθ/4πε₀R²
= λRdθcosθ/4πε₀R²
= λdθcosθ/4πε₀R
E' = ∫dE' = ∫λRdθcosθ/4πε₀R² = (λ/4πε₀R)∫cosθdθ from -θ to θ
E' = (λ/4πε₀R)[sinθ] from -θ to θ
E' = (λ/4πε₀R)[sinθ]
= (λ/4πε₀R)[sinθ - sin(-θ)]
= (λ/4πε₀R)[sinθ + sinθ]
= 2(λ/4πε₀R)sinθ
= (λ/2πε₀R)sinθ
Now, the total charge Q = ∫dq = ∫λRdθ from -θ to +θ
Q = λR∫dθ = λR[θ - (-θ)] = λR[θ + θ] = 2λRθ
Q = 2λRθ
λ = Q/2Rθ
Substituting λ into E', we have
E' = (Q/2Rθ/2πε₀R)sinθ
E' = (Q/θ4πε₀R²)sinθ
E' = Qsinθ/4πε₀R²θ where θ is in radians
