Answer:
z = 1.16m
Explanation:
The electric field in a point of the axis of a charged ring, and perpendicular to the plane of the ring is given by:
[tex]E_z=\frac{Qz}{(z^2+r^2)^{\frac{3}{2}}}[/tex]
z: distance to the plane of the ring
r: radius of the ring
Q: charge of the ring
you have that:
E_{z->0} = 0
E_{z->∞} = 0
To find the value of z that maximizes E you use the derivative respect to z, and equals it to zero:
[tex]\frac{dE_z}{dz}=Q[\frac{1}{(z^2+r^2)^{3/2}}+z(-\frac{3}{2})\frac{1}{(z^2+r^2)^{5/2}}(2z)]=0\\\\(z^2+r^2)^{5/2}=3z^2(z^2+r^2)^{3/2}\\\\(z^2+r^2)^2=3z^4\\\\z^4+2z^2r^2+r^4=3z^4\\\\2z^4-2z^2r^2-r^4=0\\\\z^2_{1,2}=\frac{-(-2)+-\sqrt{4-4(2)(-1)}}{2(2)}=\frac{2\pm 3.464}{4}\\\\[/tex]
you take the positive value:
[tex]z^2=\frac{2+3.464}{4}=1.366\\\\z=1.16m[/tex]
hence, the distance in which the magnitude if the electric field is maximum is 1.16m
A few years ago, an X-ray telescope detected a source called Cygnus X-3, whose intensity changed with a period of 4.8 hours. This type of astronomical object emitting periodic signals could be a binary X-ray source, which is a star that is in orbit around a much more massive black hole. The period of the X-ray signal is then the period of the star’s orbit. If the distance between the centers of the star and the black hole is one-fiftieth of the distance between the centers of the Earth and our Sun, then determine how many times more massive the black hole is than our Sun. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Answer:
Explanation:
The star is revolving the black hole like earth revolves around the sun .so time period of rotation T is given by the following relation
T² = [tex]\frac{4\pi^2\times R^3}{GM }[/tex] , R is distance between black hole and star , M is mass of black hole
Given T = 4.8 hours
4.8² = [tex]\frac{4\pi^2\times R^3}{GM }[/tex]
Using the same equation for earth sun system
24² = [tex]\frac{4\pi^2\times (50R)^3}{GM_s }[/tex] , Ms is mass of the sun and 50R is distance between the sun and the earth .
Dividing the equation
[tex](\frac{4.8}{24})^2[/tex] = [tex]\frac{M_s}{M}\times\frac{1^3}{50^3}[/tex]
[tex]\frac{M}{M_s}[/tex] = 2x 10⁻⁴
The mass of the black hole is [tex]2\times 10^{-4}[/tex] times the mass of our sun.
Given information:
An X-ray telescope detected a source called Cygnus X-3, whose intensity changed with a period of 4.8 hours (T).
The star is orbiting around the black hole.
The distance between the centers of the star and the black hole is one-fiftieth of the distance between the centers of the Earth and our Sun.
Let R be the center to center distance between the black hole and the sun.
So, the time period T of the sun around the black hole will be,
[tex]T^2=\dfrac{2\pi^2R^3}{GM}\\4.8^2=\dfrac{2\pi^2R^3}{GM}[/tex]
where M is the mass of the black hole.
The distance between the earth and our sun will be 50R.
So, the time period of the earth will be,
[tex]T^2=\dfrac{2\pi^2R^3}{GM}\\24^2=\dfrac{2\pi^2(50R)^3}{GM_s}[/tex]
where [tex]M_s[/tex] is the mass of our sun.
Now, compare the above two relations to get the mass of black hole in terms of mass of our sun as,
[tex]4.8^2=\dfrac{2\pi^2R^3}{GM}\\24^2=\dfrac{2\pi^2(50R)^3}{GM_s}\\\dfrac{24^2}{4.8^2}=\dfrac{50^3}{1}\times \dfrac{M}{M_s}\\\dfrac{M}{M_s}=0.0002\\\dfrac{M}{M_s}=2\times 10^{-4}[/tex]
Therefore, the mass of black hole is [tex]2\times 10^{-4}[/tex] times the mass of our sun.
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How are the electric field lines around a positive charge affected when a second positive charge is rear it?
Answer:they repel
Explanation:
Like charges repels, unlike charges attracts
A meat baster consists of a squeeze bulb attached to a plastic tube. When the bulb is squeezed and released, with the open end of the tube under the surface of the basting sauce, the sauce rises in the tube to a distance h, as the drawing shows. Using 1.013 × 105 Pa for the atmospheric pressure and 1490 kg/m3 for the density of the sauce, find the absolute pressure PB in the bulb when the distance h is (a) 0.15 m and (b) 0.10 m.
Answer:[tex]P_{bulb}=99.11\ kPa[/tex]
Explanation:
Given
When the bulb is squeezed then liquid level rises to height h i.e. pressure decreases inside the bulb which causes rises in the tube
for common elevation i.e. at liquid level pressure must be equal therefore
[tex]P_{bulb}+\rho gh=P_{atm}[/tex]
[tex]P_{bulb}=1.013\times 10^5-1490\times 9.8\times h[/tex]
for [tex]h=0.15\ m[/tex]
[tex]P_{bulb}=1.013\times 10^5-1490\times 9.8\times 0.15[/tex]
[tex]P_{bulb}=101.3\times 10^3-2.1903\times 10^3[/tex]
[tex]P_{bulb}=99.11\ kPa[/tex]
for [tex]h=0.1\ m[/tex]
[tex]P_{bulb}=101.3\times 10^3-1.4602\times 10^3[/tex]
[tex]P_{bulb}=99.83\ kPa[/tex]
If 10 cal of energy are added to 2 g of ice at -30°C calculate the final temperature of the ice
Answer:
The final temperature of the ice is [tex]-40^{\circ}\ C[/tex].
Explanation:
It is given that,
Energy, Q = 10 cal
Mass of ice, m = 2 g
Initial temperature, [tex]T_1=-30^{\circ}\ C[/tex]
We need to find the final temperature of the ice. We know that the specific heat of ice is [tex]0.5\ cal\ g^{-1} ^{\circ} C^{-1}[/tex]
The heat added in terms of specific heat is given by :
[tex]Q=mc(T_1-T_2)[/tex]
[tex]T_2[/tex] final temperature of the ice
c is specific heat of ice
[tex]T_2=T_1-\dfrac{Q}{mc}\\\\T_2=(-30)-\dfrac{10}{2\times 0.5}\\\\T_2=-40^{\circ}[/tex]
So, the final temperature of the ice is [tex]-40^{\circ}\ C[/tex].
Answer: the correct answer is -20
Explanation:
A metabolic waste of algae that can be recycled for use in cellular respiration is
Group of answer choices
oxygen
organic acid
sodium
carbon dioxide
Answer:oxygen
Explanation:algae are plants that lives in aquatic habitat,with a few of them occurring in land. Algae have chlorophyll and as a result are autotrophic in nutrition. Algae uses carbon dioxide as a raw material for photosynthesis which is the process where they produce food. They also give off oxygen which results from the splitting of water by light.
This oxygen given off is used by organisms for cellular respiration.the mitochondria is the organelle responsible it's utilization in respiration and carbon dioxide is given off.oxygen serves as an electron acceptor in the energy producing process in the mitochondria. It is an important gas for aerobic respiration.
Final answer:
Carbon dioxide is a metabolic waste of photosynthesis in algae that can be recycled back into the ecosystem through cellular respiration.
Explanation:
Among the metabolic wastes produced by algae during photosynthesis, carbon dioxide (CO₂) is a byproduct that can be recycled and used in cellular respiration. Photosynthesis is the process in which algae absorb light energy to create carbohydrates within chloroplasts, releasing oxygen as a byproduct. Alternatively, cellular respiration involves using oxygen to break down carbohydrates, primarily in the cytoplasm and mitochondria, releasing ATP and carbon dioxide. These two processes are synchronously linked in a biological cycle that allows organisms to harness energy from the sun, guide by electron transport chains that drive cellular reactions.
A 1.5-kg mass attached to an ideal massless spring with a spring constant of 20.0 N/m oscillates on a horizontal, frictionless track. At time t = 0.00 s, the mass is released from rest at x = 10.0 cm. (That is, the spring is stretched by 10.0 cm.) (a) Find the frequency of the oscillations and express the displacement x as a function of time t. (b) Determine the maximum speed and the maximum acceleration of the mass. (c) At what location are the kinetic energy and the potential energy of the system the same?
Answer:
Explanation:
a ) angular frequency ω = [tex]\sqrt{\frac{k}{m} }[/tex]
k is spring constant and m is mass attached
ω = [tex]\sqrt{\frac{20}{1.5} }[/tex]
= 3.6515 rad / s
frequency of oscillation n = 3.6515 / (2 x 3.14)
= .5814 s⁻¹
x = .1 mcos(ωt)
= .1 mcos(3.6515t)
b ) maximum speed = ωA , A is amplitude
= 3.6515 x .1
= .36515 m /s
36.515 cm /s
maximum acceleration = ω²A
= 3.6515² x .1
= 1.333 m / s²
c ) Kinetic energy at displacement x
= 1/2 m ω²( A²-x²)
potential energy =1/2 m ω²x²
so 1/2 m ω²( A²-x²) = 1/2 m ω²x²
A²-x² = x²
2x² = A²
x = A / √2
The frequency of the oscillation is 0.58 Hz and the displacement x varies as a cosine function. The maximum speed is 0.366 m/s and maximum acceleration is 0.427 m/s². The kinetic and potential energies are equal when the mass is 0.0707m away from the equilibrium point.
Explanation:(a) The frequency of the oscillations is given by the formula f = 1/2π √(k/m), where k is the spring constant and m is the mass. Substituting the given values we get f = 1/2π √(20.0 N/m / 1.5 kg) = 0.58 Hz. The displacement as a function of time is given by the equation x(t) = A cos(ωt + φ), where A is the amplitude, ω is the angular frequency, and φ is the phase angle. Given the mass was released from rest, the phase angle is 0 and the equation becomes x(t) = (0.10 m) cos(2πft).
(b) The maximum speed is given by the formula vmax = Aω, substituting the values, vmax = (0.10 m)(2πf) = 0.366 m/s. The maximum acceleration is given by the formula amax = Aω² which is amax = (0.10 m)(2πf)² = 0.427 m/s².
(c) The kinetic energy and potential energy of the system are the same when the displacement is equal to the amplitude divided by √2: x = A/√2, that is x = 0.10m /√2 = 0.0707 m from the equilibrium point.
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A typical human lens has an index of refraction of 1.430 . The lens has a double convex shape, but its curvature can be varied by the ciliary muscles acting around its rim. At minimum power, the radius of the front of the lens is 10.0 mm, whereas that of the back is 6.00 mm. At maximum power, the radii are 6.50 mm and 5.50 mm, respectively. If the lens were in air, what would be the maximum power and associated focal length of the lens
Answer:
Maximum Power = 144.3 D
The associated focal length of the lens = [tex]6.92*10^{-3} m[/tex]
Explanation:
According to the Lens maker's Formula:
[tex]\frac{1}{f} = (n-1) (\frac{1}{R_1}-\frac{1}{R_2} )[/tex]
where;
[tex]n_1[/tex] = the refractive index of the medium
[tex]R_1[/tex] and [tex]R_2[/tex] = radius of curvature on each surface
For a convex lens, The radius of curvature in the front surface will be positive and that of the second surface will be negative . Therefore;
[tex]\frac{1}{f} = (n-1) (\frac{1}{R_1}-\frac{1}{-R_2} ) \\ \\ \frac{1}{f} = (n-1) (\frac{1}{R_1}+\frac{1}{R_2} )[/tex]
At maximum power
[tex]\frac{1}{f} = (1.430-1) (\frac{1}{6.50 \ mm}-\frac{1}{5.50 \ mm} )[/tex]
= [tex]0.144 \ mm^{-1}[/tex]
This Implies
[tex]f = 6.92 mm\\f = 6.92*10^{-3} \ m[/tex]
Therefore; the power is given by the formula:
[tex]P_{max} = \frac{1}{f}[/tex]
[tex]P_{max}= \frac{1}{6.92*10^{-3}}[/tex]
= 144.3 D
The maximum power is 144.3 D and the associated focal length of the lens is 6.92 [tex]\rm \times 10^{-3}[/tex] m and this can be determined by using the lens maker's formula.
Given :
A typical human lens has an index of refraction of 1.430.The lens has a double convex shape, but its curvature can be varied by the ciliary muscles acting around its rim.At minimum power, the radius of the front of the lens is 10.0 mm, whereas that of the back is 6.00 mm.At maximum power, the radii are 6.50 mm and 5.50 mm.The lens maker's formula can be used in order to determine the maximum power and associated focal length of the lens.
The lens maker's formula is given below:
[tex]\rm \dfrac{1}{f}=(n - 1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)[/tex]
where f is the focal length of the lens, n is the refractive index, and [tex]\rm R_1[/tex] and [tex]\rm R_2[/tex] are the radius of curvature on each surface.
The radius of curvature of the first surface is positive and the radius of curvature of the second surface is negative. So, the lens maker's formula becomes:
[tex]\rm \dfrac{1}{f}=(n - 1)\left(\dfrac{1}{R_1}+\dfrac{1}{R_2}\right)[/tex]
Now, substitute the values of the known terms in the above formula.
[tex]\rm \dfrac{1}{f}=(1.430 - 1)\left(\dfrac{1}{6.50}+\dfrac{1}{5.50}\right)[/tex]
Now, simplify the above equation in order to determine the value of 'f'.
[tex]\rm \dfrac{1}{f}=0.144\;mm^{-1}[/tex]
f = 6.92 mm = 6.92 [tex]\rm \times 10^{-3}[/tex] m
Now, the maximum power is given by the formula:
[tex]\rm P_{max} =\dfrac{1}{f}[/tex]
[tex]\rm P_{max} = 144.3 \;D[/tex]
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Monochromatic light of variable wavelength is incident normally on a thin sheet of plastic film in air. The reflected light is a maximum only for 477.1 nm and 668.0 nm in the visible spectrum.What is the minimum thickness of the film (n=1.58)?
Answer:
thickness t = 528.433 nm
Explanation:
given data
wavelength λ1 = 477.1 nm
wavelength λ2 = 668.0 nm
n = 1.58
solution
we know for constructive interference condition will be
2 × t × μ = (m1+0.5) × λ1 ....................1
2 × t × μ = (m2+0.5) × λ2 ....................2
so we can say from equation 1 and 2
(m1+0.5) × λ1 = (m2+0.5) × λ2
so
[tex]\frac{\lambda 2}{\lambda 1} = \frac{m1+0.5}{m2+0.5}[/tex] ..............3
put here value and we get
[tex]\frac{668.0}{477.1} = \frac{m1+0.5}{m2+0.5}[/tex]
[tex]\frac{m1+0.5}{m2+0.5}[/tex] = 1.4
[tex]\frac{m1+0.5}{m2+0.5} = \frac{7}{5}[/tex] ...................4
so we here from equation 4
m1+0.5 = 7
m1 = 3 .................5
m2+0.5 = 4
m2 = 2 .................6
so now put value in equation 1
2 × t × μ = (m1+0.5) × λ1
2 × t × 1.58 = (3+0.5) × 477.1
solve it we get
thickness t = 528.433 nm
Using the formula for thin film interference, the minimum thickness of the plastic film that would create a condition of maximum reflection in the visible spectrum is approximately 151 nm.
Explanation:The question is asking for the minimum thickness of the film for which maximum light is reflected for the provided wavelengths. This is a classic example of thin film interference.
For constructive interference (maximum reflection), the thickness of the film (t) is given by the formula: t = mλ/2n. Here 'm' is the order of the bright fringe, 'λ' is wavelength, and 'n' is the index of refraction.
Considering the first-order maximum, we find the thickness for each wavelength:
t1 = (1)(477.1 x 10^-9 m)/2(1.58) = 1.51 x 10^-7 m (or 151 nm for 477.1 nm light)
t2 = (1)(668.0 x 10^-9 m)/2(1.58) = 2.12 x 10^-7 m (or 212 nm for 668.0 nm light)
Therefore, the minimum thickness of the film that would create a condition of maximum reflection in the visible spectrum is approximately 151 nm.
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Why is it easier to slide a heavy box over a floor that it is to start it sliding in the first place?
Answer:
This is because of the inertia of the object.
Explanation:
This is because of the inertia of the object.
When you push a static object you must overcome the static friction force of the object. Once you have overcome the static friction the inertia law demands that the object tend to conserve its motion. That is the reason why you need less force when the object is already in motion. In other words, the inertia gained by the object with the initial force, "helps" you with the work of moving the object. This is also the reason why the kinetic friction of an object in motion over a surface is lower than the static friction.
Which statement best explains the pattern that causes people on Earth to see only one side of the moon?
The moon rotates on its axis at the same rate at which it orbits Earth so that the side of the moon that faces Earth remains the same as it orbits.
The moon orbits on its axis at the same rate at which Earth orbits the sun so that the side of the moon that faces Earth remains the same as it orbits.
The moon rotates on its axis at the same rate at which Earth orbits the sun so that the side of the moon that faces Earth remains the same as it orbits.
The correct answer is A. The moon rotates on its axis at the same rate at which it orbits Earth so that the side of the moon that faces Earth remains the same as it orbits.
Explanation
The moon is the most popular and well-known natural satellite on planet earth, it is a satellite widely studied by humans, they have even walked on the moon. However, from Earth, at night when you see the moon you always see the same face of the moon, which has caused people to wonder why this phenomenon. The answer to this phenomenon is that we always see the same face of the moon because it takes the same time to rotate once on itself as it does to go around the Earth (about 27 days). The result is that the same part of the moon always points towards the earth. According to the above, the correct answer is A. The moon rotates on its axis at the same rate at which it orbits Earth so that the side of the moon that faces Earth remains the same as it orbits.
"If a ball of mass M is dropped from a height h onto a spring with spring constant k (whose equilibrium positions is at height 0), compresses the spring an additional distance (L/2), and then rebounds, what height will the ball reach? Express your answer symbolically."
So I was thinking something along the lines of h= M*k*(L/2) but I'm not sure :/ Any guidiance?
Answer:
h ’= k (h +L/2)²/ (2 M g )
Explanation:
For this exercise, one of the best methods to solve it is with energy conservation.
Starting point. Lower, spring with maximum compression
Em₀ = Ke = ½ k x²
Final point. Higher after bounce
= U = M g h ’
Em₀ = Em_{f}
½ k x² = M g h’
the compressed distance is
x = h+ L / 2
where h is the distance that compresses the spring by the height where it comes from and L/2 the additional compression
h ’= ½ k x² / M g
we calculate
h ’= k (h +L/2)²/ (2 M g )
h' = (k/2Mg) h2 (1 + L/2h)2
Suppose a police officer is 1/2 mile south of an intersection, driving north towards the intersection at 40 mph. At the same time, another car is 1/2 mile east of the intersection, driving east (away from the intersection) at an unknown speed. The officer's radar gun indicates 25 mph when pointed at the other car (that is, the straight-line distance between the officer and the other car is increasing at a rate of 25 mph). What is the speed of the other car?
Answer:
75.36 mph
Explanation:
The distance between the other car and the intersection is,
[tex]x=x_{0}+V t \\ x=\frac{1}{2}+V t[/tex]
The distance between the police car and the intersection is,
[tex]y=y_{0}+V t[/tex]
[tex]y=\frac{1}{2}-40 t[/tex]
(Negative sign indicates that he is moving towards the intersection)
Therefore the distance between them is given by,
[tex]z^{2}=x^{2}+y^{2}(\text { Using Phythogorous theorem })[/tex]
[tex]z^{2}=\left(\frac{1}{2}+V t\right)^{2}+\left(\frac{1}{2}-40 t\right)^{2} \ldots \ldots \ldots(1)[/tex]
The rate of change is,
[tex]2 z \frac{d z}{d t}=2\left(\frac{1}{2}+V t\right) V+2\left(\frac{1}{2}-40 t\right)(-40)[/tex]
[tex]2 z \frac{d z}{d t}=V+2 V^{2} t-40+3200 t \ldots \ldots \ldots[/tex]
Now finding [tex]z[/tex] when [tex]t=0,[/tex] from (1) we have
[tex]z^{2}=\left(\frac{1}{2}+V(0)\right)^{2}+\left(\frac{1}{2}-40(0)\right)^{2}[/tex]
[tex]z^{2}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2} \\ z=\sqrt{\frac{1}{2}} \approx 0.7071[/tex]
The officer's radar gun indicates 25 mph pointed at the other car then, [tex]\frac{d z}{d t}=25[/tex] when [tex]t=0,[/tex] from
From (2) we get
[tex]2(0.7071)(25)=V+2 V^{2}(0)-40+3200(0)[/tex]
[tex]2(0.7071)(25)=V+2 V^{2}(0)-40[/tex]
[tex]35.36=V-40[/tex]
[tex]V=35.36+40=75.36[/tex]
Hence the speed of the car is [tex]75.36 mph[/tex]
To find the speed of the other car, we can use the concept of relative velocity. The police officer's radar gun indicates a speed of 25 mph, which is the relative velocity between the two cars. Since the police car is moving north at 40 mph, the other car must be moving east at a speed that results in a relative velocity of 25 mph.
Explanation:To find the speed of the other car, we can use the concept of relative velocity. The police officer's radar gun indicates a speed of 25 mph, which is the relative velocity between the two cars. Since the police car is moving north at 40 mph, the other car must be moving east at a speed that results in a relative velocity of 25 mph.
To calculate the speed of the other car, we can use the Pythagorean theorem. The police car is moving in a straight line, so the relative velocity is the hypotenuse of a right triangle. The northward velocity of the police car is one side of the triangle, and the eastward velocity of the other car is the other side. Using the equation a² + b² = c², where a is the northward velocity, b is the eastward velocity, and c is the relative velocity, we can solve for b. Rearranging the equation, we get b = √(c² - a²). Plugging in the known values, we have b = √(25² - 40²) = √(625 - 1600) = √(-975)
The square root of a negative number is not a real number, so it is not possible to determine the speed of the other car using only the given information.
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Suppose that a parallel-plate capacitor has circular plates with radius R = 26 mm and a plate separation of 4.0 mm. Suppose also that a sinusoidal potential difference with a maximum value of 220 V and a frequency of 76 Hz is applied across the plates; that is, V = (220 V) sin[2π(76 Hz)t]. Find Bmax(R), the maximum value of the induced magnetic field that occurs at r = R.
Answer:
B(max) = 3.7971 × [tex]10^{-12}[/tex] T
Explanation:
given data
radius R = 26 mm
plate separation d = 4.0 mm
potential difference Vm = 220 V
frequency f = 76 Hz
V = (220 V) sin[2π(76 Hz)t]
solution
we know that E will be
E = V ÷ d ............1
put here value
E = [tex]\frac{220 \times sin(2\pi 76\times t)}{d}[/tex]
and here we take as given r = R
so A = π R² .................2
and
ФE = E × A
ФE = [tex]\frac{\pi R^2 \times 220 \times sin(2\pi 76 \times t)}{d}[/tex] .....................3
so use use here now Ampere's Law that is
∫ B ds = [tex]\mu_o \times \epsilon_o \times \frac{d\Phi E}{dt} + \mu_o \times I_{encl}[/tex] .....................4
and
here [tex]I_{encl}[/tex] is = 0 and r = R
so
[tex]2B \times \pi \times R = \mu_o \times \epsilon_o \times \frac{d\Phi E}{dt}[/tex] .....................5
and put here value we get
B = [tex]\frac{\mu_o \times \epsilon_o \times \pi \times f \times R \times V_m cos(2\pi f t)}{d}[/tex] .....................6
put here value for B maximum cos(2πft) = 1
and we get B (max)
B(max) = [tex]\frac{\mu_o\times \epsilon_o\times \pi \times f\times R\times V_m}{d}[/tex] ....................7
put here all value
B(max) = [tex]\frac{4\pi \times10^{-7} \times 8.85\times 10^{-12}\times \pi \times 76 \times 0.026\times 220 }{4\times 10^{-3}}[/tex]
solve it we get
B(max) = 3.7971 × [tex]10^{-12}[/tex] T
When ultraviolet light with a wavelength of 400 nm falls on a certain metal surface, the maximum kinetic energy of the emitted photoelectrons is 1.10 eV.
What is the maximum kinetic energy K_0 of the photoelectrons when light of wavelength 330 nm falls on the same surface?
Use h = 6.63×10-34 J*s for Planck's constant and c = 3.00×108 m/s for the speed of light and express your answer in electron volts.
Answer:
1.76 eV
Explanation:
Maximum kinetic energy of the emitted photo electrons = energy of the photon- work function of the metal
K.E' = (hc/λ)-∅.................. Equation 1
Where K.E' = maximum kinetic energy of the emitted photo electrons, h = Planck's constant, c = speed, λ = wave length, ∅ = work function of the metal.
make ∅ the subject of the equation
∅ = (hc/λ)-K.E'.................. Equation 2
Given: h = 6.63×10⁻³⁴ J.s, c = 3×10⁸ m/s, λ = 400 nm = 400×10⁻⁹ m, K.E' = 1.1 eV = 1.1(1.602×10⁻¹⁹) J = 1.7622×10⁻¹⁹ J
Substitute into equation 2
∅ = [6.63×10⁻³⁴(3×10⁸)/400×10⁻⁹ ]-1.7622×10⁻¹⁹
∅ = (4.973×10⁻¹⁹)-(1.7622×10⁻¹⁹)
∅ = 3.21×10⁻¹⁹ J.
The maximum kinetic energy of the photo electrons when the wave length is 330 nm is
K.E' = [6.63×10⁻³⁴( 3×10⁸ )/330×10⁻⁹]-(3.21×10⁻¹⁹)
K.E' = (6.03×10⁻¹⁹)-(3.21×10⁻¹⁹)
K.E' = 2.82×10⁻¹⁹ J
K.E' = 2.82×10⁻¹⁹/1.602×10⁻¹⁹
K.E' = 1.76 eV
A large electromagnet draws 300A at 270V. The coils of the electromagnet are cooled by a flow of mystery liquid passing over them. The liquid enters the electromagnet at a temperature of 15 deg C, absorbs the Joule heat, and leaves at a higher temperature. If the liquid is to leave at a temperature no higher than 81 deg C, and the maximum flow rate of liquid through the electromagnet is 0.307kg/s. What is the specific heat of the mystery liquid? Pick the closest value. The solution should include the correct units.
Answer:[tex]c=3.99\approx 4\ kJ/kg-K[/tex]
Explanation:
Given
Current [tex]I=300\ A[/tex]
Voltage [tex]V=270\ V[/tex]
Mass flow rate [tex]m=0.307\ kg/s[/tex]
Inlet temperature is [tex]T_i=15^{\circ}C\approx 288\ K[/tex]
[tex]T_{out}\leq 81^{\circ}C[/tex]
Here heat of Electromagnet is absorbed by liquid
Heat rate of Electromagnet [tex]\dot{Q}=VI[/tex]
[tex]\dot{Q}=270\times 300=81\ kJ/s[/tex]
Heat absorbed by liquid [tex]\dot{Q}=mc(\Delta T)[/tex]
[tex]\dot{Q}=0.307\times c\times (81-15)[/tex]
[tex]81\times 10^3=0.307\times c\times (81-15)[/tex]
[tex]c=3.99\ kJ/kg-K[/tex]
In a physics laboratory experiment, a coil with 200 turns enclosing an area of 11.8 cm^2 is rotated during the time interval 4.90×10^-2 s from a position in which its plane is perpendicular to Earth's magnetic field to one in which its plane is parallel to the field. The magnitude of Earth's magnetic field at the lab location is 5.60×10-5 T.
a) What is the total magnitude of the magnetic flux through the coil before it is rotated?
b) What is the magnitude of the total magnetic flux through the coil after it is rotated?
c) What is the magnitude of the average emf induced in the coil?
Answer:
Explanation:
Flux through the coil = nBA , n is no of turns , B is magnetic flux and A a is area of the coli
= 200 x 5.6 x 10⁻⁵ x 11.8 x 10⁻⁴
= 13216 x 10⁻⁹ weber .
b ) When the coil becomes parallel to magnetic field , flux through it will become zero.
c ) e m f induced = change in flux / time
= 13216 x 10⁻⁹ / 4.9 x 10⁻²
= 2697.14 x 10⁻⁷ V
= 269.7 x10⁻⁶
269.7 μV.
The LIGO experiment, which historically detected gravitational waves for the first time in September 2015, uses a pair of highly sensitive Michelson interferometers. These have arms that are 4.00 km long and use powerful Nd:Yag lasers with 1064 nm wavelength. The beams traverse the arms both ways 280 times before recombining, which effectively lengthens the arm length to 1120 km. The devices are tuned so that the beams destructively interfere when they recombine if no gravitational wave is present.
The beam has power of 100 kW, concentrated into area of square centimeter. Calculate the amplitude of the electric field in the beam.
Answer:
867755.73 V/m
Explanation:
Detailed explanation and calculation is shown in the image below
A submarine is 3.00 x 10^2 m horizontally from shore and 120.0 m beneath the surface of the water. A laser beam is sent from the submarine so that the beam strikes 2.10 x 10^2 m from the buildingstanding on the shoreand the laser beam hits the target on the top of the building. Use n=1.333 for the refractive index of water.A)Calculate the angle of incidence ?1 on the water/air interface.
B)Calculate the angle of refraction ?2.
C)Find angle ? with the horizontal.
D)Find the height h of the building
Answer:
Explanation:
a ) Let angle of incidence and angle of refraction be i and r respectively .
submarine is 300 m away from the shore . The point where laser strikes the surface of sea is 90 m horizontally away .
Tan r = 90 / 120
= 3 / 4
.75
r = 37 degree
c ) sini / sin37 = 1.333
sini = .8
i = 53 degree
Tan 53 = 210 / h , h is height of the building .
h = 210 / tan 53
= 158 m
Water is pumped from a stream at the rate of 90kg every 30s and sprayed
unto a farm at a velocity of 15m/s. Calculate the power of the pump
Answer:
The power of the pump is 337.5 W.
Explanation:
Power is the rate of transfer of energy or the rate at which work is done. It is measured in watts (W).
Power = [tex]\frac{Energy}{Time}[/tex]
But the energy here is the kinetic energy (K.E), which is the energy possessed by a moving body or object.
K.E = [tex]\frac{1}{2}[/tex] m[tex]v^{2}[/tex]
where m is the mass = 90 kg and v is the velocity = 15 m/s.
K.E = [tex]\frac{1}{2}[/tex] × 90 × [tex]15^{2}[/tex]
= 10125 Joules
K.E = 10125 J = 10.125 KJ
So that the power of the pump when t = 30 s is;
Power = [tex]\frac{10125}{30}[/tex]
= 337.5 W
The power of the pump is 337.5 W.
Alberta is going to have dinner at her grandmother's house, but she is running a bit behind schedule. As she gets onto the highway, she knows that she must exit the highway within 55
min
min
if she is not going to arrive late. Her exit is 43
mi
miHow much time would it take at the posted 60 mph speed?
away.
Complete Question
Alberta is going to have dinner at her grandmother's house, but she is running a bit behind schedule. As she gets onto the highway, she knows that she must exit the highway within 55 min if she is not going to arrive late. Her exit is 43 mi away. How much time would it take at the posted 60 mph speed?
Answer:
The time it would take at the given speed is [tex]x = 43.00 \ minutes[/tex]
Explanation:
From the question we are told that
The time taken to exist the highway is [tex]t = 55 min[/tex]
The distance to the exist is [tex]d = 43\ mi[/tex]
Alberta speed is [tex]v = 60 mph[/tex]
The time it would take travelling at the given speed is mathematically represented as
[tex]t_z = \frac{d}{v}[/tex]
substituting values
[tex]t_z = \frac{43}{60}[/tex]
[tex]t_z = 0.71667\ hrs[/tex]
Converting to minutes
1 hour = 60 minutes
So 0.71667 hours = x minutes
Therefore
[tex]x = 0.71667 * 60[/tex]
[tex]x = 43.00 \ minutes[/tex]
At a speed of 60 mph, it would take Alberta approximately 43 minutes to travel the 43 miles to her grandmother's house, which is within the 55 minutes time frame she has to avoid being late.
Explanation:To determine how long it will take Alberta to reach her grandmother's house if she travels at a constant speed of 60 mph, we need to use the formula for time which is time = distance ÷ speed. Alberta's exit is 43 miles away and the speed limit is 60 mph.
First, we calculate the time it would take her to travel 43 miles at 60 mph:
Time = Distance ÷ Speed
= 43 miles ÷ 60 mph
= 0.7167 hours
Since time in hours is not always intuitive, let's convert it to minutes by multiplying by 60 (since there are 60 minutes in one hour):
Time in minutes = 0.7167 hours × 60 minutes/hour
= 43 minutes
Thus, it will take Alberta approximately 43 minutes to reach her exit at the posted speed of 60 mph.
At what depth of lake water is the pressure equal to 201kpa?
A. 10m
B. 20m
C. 5.1m
D. none
Answer:
20m
Explanation:
Pressure = pgh
p = density of water 1000
kg/m^3
g = acceleration due to gravity 9.81 m/s^2
h is the depth of water
Pressure = 201 kPa = 201 x 10^3 Pa
201 x 10^3 = 1000 x 9.81 x h
201 x 10^3 = 9810h
h = 20.49 m
Approximately 20 m
How much current will pass through a 12.5 ohm resistor when it is connected to ta 115 volt source of electrical potential?
Answer:
9.2 A
Explanation:
V = IR
115 V = I (12.5 Ω)
I = 9.2 A
An experiment is set up to test the angular resolution of an optical device when red light (wavelength λrλr) shines on an aperture of diameter DD.
Which aperture diameter will give the best resolution?
a. D=0.5λ
b. D=λr
c. D=2λr
Answer:
Option C ⇒ D=2λr is the correct answer, since it has the largest aperture diameter.
Explanation:
Regarding Rayleigh's criterion, the angular resolution is given as below:
θ = 1.22λ/D
From this expression, it is observed that the larger the aperture size, the smaller will be the value of the angular resolution, and the better the device will be.
This signifies that at very high angular differences, the precision for distinguishing two points is high.
Therefore, option C ⇒ D=2λr is the correct answer, since it has the largest aperture diameter.
A student sits on a rotating stool holding two 5 kg objects. When his arms are extended horizontally, the objects are 0.9 m from the axis of rotation, and he rotates with angular speed of 0.66 rad/sec. The moment of inertia of the student plus the stool is 8 kg m2 and is assumed to be constant. The student then pulls the objects horizontally to a radius 0.31 m from the rotation axis. Calculate the final angular speed of the student.
Answer:[tex]\omega _f=1.185\ rad/s[/tex]
Explanation:
Given
mass of objects [tex]m=5\ kg[/tex]
Initially mass is at [tex]r=0.9\ m[/tex]
Initial angular speed [tex]\omega_i=0.66\ rad/s[/tex]
Moment of inertia of student and stool is [tex]I_s=8\ kg-m^2[/tex]
Finally masses are at a distance of [tex]r_f=0.31\ m[/tex] from axis
[tex]I_i=I_p+I_m[/tex]
[tex]I_i=8+2\times 5\times (0.9)^2[/tex]
[tex]I_i=16.1\ kg-m^2[/tex]
Final moment of inertia of the system
[tex]I_f=I_s+I_m[/tex]
[tex]I_f=8+2\times 5\times (0.31)^2[/tex]
[tex]I_f=8+0.961=8.961\ kg-m^2[/tex]
As there is no external torque therefore moment of inertia is conserved
[tex]I_i\omega _i=I_f\omega _f[/tex]
[tex]\omega _f=\frac{16.1}{8.96}\times 0.66[/tex]
[tex]\omega _f=1.796\times 0.66[/tex]
[tex]\omega _f=1.185\ rad/s[/tex]
Final answer:
When the student pulls the objects closer, his angular speed increases to conserve angular momentum.
Explanation:
Angular momentum is conserved in this scenario. Initially, the student with the objects has a certain angular speed and moment of inertia. When he pulls the objects closer, his moment of inertia decreases, resulting in an increase in angular speed to conserve angular momentum.
g 1. A mass undergoing simple harmonic motion along the x-axis has a period of T = 0.5 s and an amplitude of 25 mm. Its position is x = 14 mm when t = 0. a. Write down x(t) = A cos(ωt + φ0) for this oscillator, filling in A, ω, and φ0. b. What is the magnitude of the maximum velocity vmax? At what value(s) of x does it occur? c. What is the magnitude of the maximum acceleration amax? At what value(s) of x does it occur?
Answer:
Explanation:
a )
Amplitude A = 14 mm , angular frequency ω = 2π / T
= 2π / .5
ω = 4π rad /s
φ₀ = initial phase
Putting the given values in the equation
14 = 25 cos(ωt + φ₀ )
14/25 = cosφ₀
φ₀ = 56 degree
x(t) = 25cos(4πt + 56° )
b )
maximum velocity = ω A
= 4π x 25
100 x 3.14 mm /s
= 314 mm /s
At x = 0 ( equilibrium position or middle point , this velocity is achieved. )
maximun acceleration = ω² A
= 16π² x A
= 16 x 3.14² x 25
= 3943.84 mm / s²
3.9 m / s²
It occurs at x = A or at extreme position.
The flywheel of a steam engine runs with a constant angular velocity of 150 rev/min. When steam is shut off, the friction of the bearings and of the air stops the wheel in 1.6 h. (a) What is the constant angular acceleration, in revolutions per minute-squared, of the wheel during the slowdown? (b) How many revolutions does the wheel make before stopping? (c) At the instant the flywheel is turning at 75.0 rev/min, what is the tangential component of the linear acceleration of a flywheel particle that is 37 cm from the axis of rotation? (d) What is the magnitude of the net linear acceleration of the particle in (c)?
Answer:
a) [tex]\ddot n = -1.563\,\frac{rev}{min^{2}}[/tex], b) [tex]\Delta n = 7197.697\,rev[/tex], c) [tex]a_{t} = 1.009\times 10^{-3}\,\frac{m}{s^{2}}[/tex], d) [tex]a = 22.823\,\frac{m}{s^{2}}[/tex]
Explanation:
a) Constant angular acceleration is:
[tex]\ddot n = \frac{\dot n - \dot n_{o}}{\Delta t}[/tex]
[tex]\ddot n = \frac{0\,\frac{rev}{min} - 150\,\frac{rev}{min}}{(1.6\,h)\cdot \left(60\,\frac{min}{h} \right)}[/tex]
[tex]\ddot n = -1.563\,\frac{rev}{min^{2}}[/tex]
b) The amount of revolutions required to stop the flywheel is:
[tex]\Delta n = \frac{\dot n^{2}-\dot n_{o}^{2}}{2\cdot \ddot n}[/tex]
[tex]\Delta n = \frac{\left(0\,\frac{rev}{min} \right)^{2}-\left(150\,\frac{rev}{min} \right)^{2}}{2\cdot \left(-1.563\,\frac{rev}{min^{2}} \right)}[/tex]
[tex]\Delta n = 7197.697\,rev[/tex]
c) The tangential acceleration of the particle is:
[tex]a_{t} = \left(1.563\,\frac{rev}{min^{2}} \right)\cdot \left(\frac{1}{3600}\,\frac{min^{2}}{s^{2}}\right)\cdot \left(2\pi\,\frac{rad}{rev}\right)\cdot (0.37\,m)[/tex]
[tex]a_{t} = 1.009\times 10^{-3}\,\frac{m}{s^{2}}[/tex]
d) The radial acceleration of the particle is:
[tex]a_{r} = \left[\left(75\,\frac{rev}{min} \right)\cdot \left(\frac{1}{60}\,\frac{min}{s} \right)\cdot \left(2\pi\,\frac{rad}{rev} \right)\right]^{2}\cdot (0.37\,m)[/tex]
[tex]a_{r} = 22.823\,\frac{m}{s}[/tex]
The net linear acceleration is:
[tex]a = \sqrt{a_{r}^{2}+a_{t}^{2}}[/tex]
[tex]a = \sqrt{\left(22.823\,\frac{m}{s^{2}} \right)^{2}+\left(1.009\times 10^{-3}\,\frac{m}{s^{2}} \right)^{2}}[/tex]
[tex]a = 22.823\,\frac{m}{s^{2}}[/tex]
78kg*9.8m/s2 what is the force
Answer:764.4N
Explanation:
Mass=78kg
Acceleration=9.8m/s^2
force=mass x acceleration
Force=78 x 9.8
Force=764.4
Force=764.4N
A loud speaker has an input of 79 dB. If the volume of the sound is turned up so that the output intensity is 100 times greater, what is the new sound intensity level?
Answer:
99 dB
Explanation:
To find the new sound intensity level you calculate first the initial intensity by using the following formula:
[tex]\beta=10log(\frac{I}{I_o})\\\\10^{\frac{\beta}{10}}=10^{log(\frac{I}{I_o})}\\\\10^{\frac{\beta}{10}}=\frac{I}{I_o}\\\\I=I_o10^{\frac{\beta}{10}}[/tex]
where β is the sound level of 79dB and Io is the hearing threshold of 10^-12 W/m^2. By replacing you obtain:
[tex]I=(10^{-12}W/m^2)10^{\frac{79}{10}}=7.94*10^{-5}W/m^2[/tex]
The new sound intensity level is given by:
[tex]\beta'=10log(\frac{100I}{I_o})=10log(\frac{100(7.94*10^{-5}W/m^2)}{10^{-12}W/m^2})\\\\\beta'=99\ dB[/tex]
hence, the answer is 99 dB
The electric potential in a certain region is given by the equation V(x,y,z) = 3αx2y3 - 2γx2y4z2 where the potential is in volts when the positions are given in meters. The constants in this equation are α = 2.5 V/m5 and γ = 1.33 V/m8
(a) Write an equation for the x-component of the electric field.
(b) Write an equation for the y-component of the electric field.
(c) Write an equation for the z-component of the electric field.
(d) Calculate the magnitude of the electric field at the point x13-(-5.0, 2.0, 1.5) m in units of newtons per coulomb.
Answer:
a) An equation for the x-component of the electric field.
Eₓ = (-15xy³ + 5.32xy⁴z²) N/C
b) An equation for the y-component of the electric field.
Eᵧ = (-22.5x²y² + 10.64x²y³z²) N/C
c) An equation for the z-component of the electric field.
Ez = (5.32x²y⁴z) N/C
d) At (-5.0, 2.0, 1.5) m, the electric field is given as
E = (-357.6î + 2,538ĵ + 3,192ķ) N/C
Magnitude of the electric field = 4,093.7 N/C
Explanation:
The electric field is given by the negative of the gradient of the electric potential,
E = −grad V
E = - ∇V
The electric potential is given as
V(x,y,z) = 3αx²y³ - 2γx²y⁴z²
α = 2.5 V/m⁵ and γ = 1.33 V/m⁸
V(x,y,z) = 7.5x²y³ - 2.66x²y⁴z²
grad = ∇ = (∂/∂x)î + (∂/∂y)ĵ + (∂/∂z)ķ
E = -grad V = -∇V
= -[(∂V/∂x)î + (∂V/∂y)ĵ + (∂V/∂z)ķ
E = -(∂V/∂x)î - (∂V/∂y)ĵ - (∂V/∂z)ķ
E = Eₓî + Eᵧĵ + Ez ķ
a) An equation for the x-component of the electric field.
Eₓ = -(∂V/∂x) = -(∂/∂x)(V)
= -(∂/∂x)(7.5x²y³ - 2.66x²y⁴z²)
= -(15xy³ - 5.32xy⁴z²)
= (-15xy³ + 5.32xy⁴z²)
b) An equation for the y-component of the electric field.
Eᵧ = -(∂V/∂y) = -(∂/∂x)(V)
= -(∂/∂y)(7.5x²y³ - 2.66x²y⁴z²)
= -(22.5x²y² - 10.64x²y³z²)
= (-22.5x²y² + 10.64x²y³z²)
c) An equation for the z-component of the electric field.
Ez = -(∂V/∂z) = -(∂/∂x)(V)
= -(∂/∂z)(7.5x²y³ - 2.66x²y⁴z²)
= -(0 - 5.32x²y⁴z)
= (5.32x²y⁴z)
d) E = Eₓî + Eᵧĵ + Ez ķ
E = (-15xy³ + 5.32xy⁴z²)î + (-22.5x²y² + 10.64x²y³z²)ĵ + (5.32x²y⁴z) ķ
At (-5.0, 2.0, 1.5) m
x = -5 m
y = 2 m
z = 1.5 m
Eₓ = (-15xy³ + 5.32xy⁴z²)
= (-15×-5×2³) + (5.32×-5×2⁴×1.5²)
= 600 - 957.6 = -357.6
Eᵧ = (-22.5x²y² + 10.64x²y³z²)
= (-22.5×(-5)²×2²) + (10.64×(-5)²×2³×1.5²)
= -2250 + 4788 = 2538
Ez = (5.32x²y⁴z) = (5.32×(-5)²×2⁴×1.5)
= 3192
E = -357.6î + 2,538ĵ + 3,192ķ
Magnitude = /E/ = √[(-357.6)² + 2538² + 3192²]
= 4,093.6763135353 = 4,093.7 N/C
Hope this Helps!!!!
The correct answers are as follows:
(a) The equation for the x-component of the electric field is[tex]\[ E_x = -15xy^3 + 5.32xy^4z^2 \][/tex]
(b) The equation for the y-component of the electric field is [tex]\[ E_y = -22.5x^2y^2 + 10.64x^2y^3z^2 \][/tex]
(c) The equation for the z-component of the electric field is [tex]\[ E_z = 5.32x^2y^4z \][/tex]
(d) the magnitude of the electric field at the point x13-(-5.0, 2.0, 1.5) m in units of newtons per coulomb is 14907.5 N/C.
To find the components of the electric field, we need to take the negative gradient of the electric potential function V(x,y,z). The gradient of a function is a vector field whose components are the partial derivatives of the function with respect to each variable. The electric field [tex]\( \vec{E} \)[/tex] is related to the electric potential [tex]\( V \)[/tex] by the equation:
[tex]\[ \vec{E} = -\nabla V \][/tex]
where [tex]\( \nabla \)[/tex] is the gradient operator.
(a) The x-component of the electric field [tex]\( E_x \)[/tex] is given by the negative partial derivative of [tex]\( V \)[/tex] with respect to [tex]\( x \)[/tex] :
[tex]\[ E_x = -\frac{\partial V}{\partial x} = -\frac{\partial}{\partial x}(3αx^2y^3 - 2γx^2y^4z^2) \][/tex]
[tex]\[ E_x = -3α(2xy^3) + 2γ(2xy^4z^2) \][/tex]
[tex]\[ E_x = -6αxy^3 + 4γxy^4z^2 \][/tex]
Substituting the given values of [tex]\( α \)[/tex] and [tex]\( γ \)[/tex] :
[tex]\[ E_x = -6(2.5)xy^3 + 4(1.33)xy^4z^2 \][/tex]
[tex]\[ E_x = -15xy^3 + 5.32xy^4z^2 \][/tex]
(b) The y-component of the electric field [tex]\( E_y \)[/tex] is given by the negative partial derivative of [tex]\( V \)[/tex] with respect to [tex]\( y \)[/tex] :
[tex]\[ E_y = -\frac{\partial V}{\partial y} = -\frac{\partial}{\partial y}(3αx^2y^3 - 2γx^2y^4z^2) \][/tex] [tex]\[ E_y = -3αx^2(3y^2) + 2γx^2(4y^3z^2) \][/tex]
[tex]\[ E_y = -9αx^2y^2 + 8γx^2y^3z^2 \][/tex]
Substituting the given values of [tex]\( α \)[/tex] and [tex]\( γ \)[/tex] :
[tex]\[ E_y = -9(2.5)x^2y^2 + 8(1.33)x^2y^3z^2 \][/tex]
[tex]\[ E_y = -22.5x^2y^2 + 10.64x^2y^3z^2 \][/tex]
(c) The z-component of the electric field
[tex]\[ E_z = -\frac{\partial V}{\partial z} = -\frac{\partial}{\partial z}(3αx^2y^3 - 2γx^2y^4z^2) \][/tex]
[tex]\[ E_z = -\frac{\partial V}{\partial z} = -\frac{\partial}{\partial z}(3αx^2y^3 - 2γx^2y^4z^2) \][/tex]
is given by the negative partial derivative of [tex]\( V \)[/tex] with respect to [tex]\( z \)[/tex] :
[tex]\[ E_z = -\frac{\partial V}{\partial z} = -\frac{\partial}{\partial z}(3αx^2y^3 - 2γx^2y^4z^2) \][/tex]
[tex]\[ E_z = -2γx^2y^4(-2z) \][/tex]
[tex]\[ E_z = 4γx^2y^4z \][/tex]
Substituting the given value of[tex]\( γ \)[/tex] :
[tex]\[ E_z = 4(1.33)x^2y^4z \]\\ E_z = 5.32x^2y^4z \][/tex]
(d) To calculate the magnitude of the electric field at the point[tex]\( P(-5.0, 2.0, 1.5) \)[/tex] m, we first substitute [tex]\( x = -5.0 \)[/tex] m, [tex]\( y = 2.0 \)[/tex] m, and [tex]\( z = 1.5 \)[/tex] m into the equations for[tex]\( E_x \)[/tex], [tex]\( E_y \)[/tex] , and [tex]\( E_z \)[/tex] :
[tex]E_x = -15(-5.0)(2.0)^3 + 5.32(-5.0)(2.0)^4(1.5)^2 \]\\ E_x = 1500 - 5.32(-5.0)(16)(2.25) \]\\ E_x = 1500 - (-851.88) \]\\ E_x = 1500 + 851.88 \]\\ E_x = 2351.88 \text{ N/C} \][/tex]
[tex]\[ E_y = -22.5(-5.0)^2(2.0)^2 + 10.64(-5.0)^2(2.0)^3(1.5)^2 \]\\ E_y = -22.5(25)(4) + 10.64(25)(8)(2.25) \]\\ E_y = -2200 + 10.64(25)(8)(2.25) \]\\ E_y = -2200 + 4752 \]\\ E_y = 2552 \text{ N/C} \][/tex]
[tex]\[ E_z = 5.32(-5.0)^2(2.0)^4(1.5) \]\\ E_z = 5.32(25)(16)(1.5) \]\\ E_z = 5.32(600) \]\\ E_z = 3192 \text{ N/C} \][/tex]
Now, the magnitude of the electric field \( \vec{E} \) is given by:
[tex]\[ |\vec{E}| = \sqrt{E_x^2 + E_y^2 + E_z^2} \][/tex]
[tex]\[ |\vec{E}| = \sqrt{(2351.88)^2 + (2552)^2 + (3192)^2} \][/tex]
[tex]\[ |\vec{E}| = \sqrt{5527641.64 + 6512644 + 10180416} \][/tex]
[tex]\[ |\vec{E}| = \sqrt{22220502.64} \][/tex]
[tex]\[ |\vec{E}| \approx 14907.5 \text{ N/C} \][/tex]
Therefore, the magnitude of the electric field at the point [tex]\( P(-5.0, 2.0, 1.5) \)[/tex]m is approximately [tex]\( 14907.5 \)[/tex] N/C.
g Which of the following is true about magnetic field lines? A. All magnetic field lines are always parallel to the Earth’s magnetic field. B. Magnetic field lines form closed loops and never intersect. C. Magnetic field lines flow out of the south pole of a bar magnet, around, and into the north pole. D. Positive charges always feel a force along magnetic field line
Final answer:
The accurate statement about magnetic field lines is that they form closed loops and never intersect, originating at the north pole and terminating at the south pole of a magnet.
Explanation:
The correct answer regarding magnetic field lines is B. Magnetic field lines form closed loops and never intersect. This observation can be explained through several key principles:
Magnetic field lines are continuous and form closed loops, going from the north pole of a magnet to the south pole. This behavior indicates that the magnetic field is consistent and unbroken throughout its influence.It is also important to note that magnetic field lines are unique at each point in space, meaning they can never cross each other.The strength of the magnetic field (B-field) is proportional to the density of these lines. Closer lines indicate a stronger field.Therefore, options A, C, and D are not correct. The field lines start at the north pole and end at the south pole, not the other way around. Positive charges are affected by magnetic fields but not always along the magnetic field lines because the magnetic force is perpendicular to both the velocity of the charge and the magnetic field.