Explanation:
In Single Slit Experiment:
The width of the central diffraction maximum is inversely proportional to the width of the slit.
Therefore, if we make the slit width smaller, the angle T(representing the angle between the wave ray to a point on the screen and the normal line between the slit and the screen) increases, giving a wider central band.
Water has a specific heat capacity of 1.00 cal/g °C, and copper has a specific heat capacity of 0.092 cal/g °C. If both are heated to 100 °C, which takes longer to cool?
Answer:
The water takes longer, because it is the better insulator here.
Explanation:
Conductors and insulators work similarly in "reverse".
If something is a good heat conductor, then it's good at both absorbing heat energy and giving it away. Insulators are good at resisting temperature changes, but also take longer to cool down once they are heated up.
So because copper is the better conductor here, it will cool faster than the water at the same temperature.
The place you get your hair cut has two nearly parallel mirrors 6.5 m apart. As you sit in the chair, your head is
Complete question is;
The place you get your hair cut has two nearly parallel mirrors 6.50 m apart. As you sit in the chair, your head is 3.00 m from the nearer mirror. Looking toward this mirror, you first see your face and then, farther away, the back of your head. (The mirrors need to be slightly nonparallel for you to be able to see the back of your head, but you can treat them as parallel in this problem.) How far away does the back of your head appear to be?
Answer:
13 m
Explanation:
We are given;
Distance between two nearly parallel mirrors; d = 6.5 m
Distance between the face and the nearer mirror; x = 3 m
Thus, the distance between the back-head and the mirror = 6.5 - 3 = 3.5m
Now, From the given values above and using the law of reflection, we can find the distance of the first reflection of the back of the head of the person in the rear mirror.
Thus;
Distance of the first reflection of the back of the head in the rear mirror from the object head is;
y' = 2y
y' = 2 × 3.5
y' = 7
The total distance of this image from the front mirror would be calculated as;
z = y' + x
z = 7 + 3
z = 10
Finally, the second reflection of this image will be 10 meters inside in the front mirror.
Thus, the total distance of the image of the back of the head in the front mirror from the person will be:
T.D = x + z
T.D = 3 + 10
T.D = 13m
3. A very light bamboo fishing rod 3.0 m long is secured to a boat at the bottom end. It is
held in equilibrium by an 18 N horizontal force while a fish pulls on a fishing line
attached to the rod shown below. How much force F does the fishing line exert on the
rod? (3)
18 N
pivot
30°
1.8 m
3.0 in
The image in the attachment describes the situation of the fishing rod.
Answer: F = 10.8 N
Explanation: The image shows a fishing rod attached to an axis. To stay in equilibrium, Torque must be equal for the force of magnitude 18N and for the unknow force.
Torque (τ) is a measure of a force's tendency to cause rotation and, in physics, defined as:
τ = F.r.sin(θ)
F is the force acting on the object;
r is distance between where the torque is measured to where the force is applied;
θ is the angle between F and r;
For the fishing rod:
[tex]\tau_{1} = \tau_{2}[/tex]
[tex]F_{1}.r_{1}.sin(\theta) = F_{2}.r_{2}.sin(\theta)[/tex]
Assuming part (1) is related to unknown force:
[tex]F = \frac{F_{2}.r_{2}.sin(\theta}{r_{1}.sin(\theta) }[/tex]
Replacing the corresponding values:
[tex]F = \frac{18*1.8*sin(30)}{3*sin(30)}[/tex]
[tex]F = \frac{18*1.8}{3}[/tex]
F = 10.8
The fishing line exert on the the rod a force of 10.8N.
A goldfish bowl is spherical, 8.0 cm in radius. A goldfish is swimming 3.0 cm from the wall of the bowl. Where does the fish appear to be to an observer outside? The index of refraction of water is 1.33. Neglect the effect of the glass wall of the bowl.
Answer:
41.5 cm to the left of the observer
Explanation:
See attached file
Each proton-proton cycle generates 26.7 MeV of energy. If 9.9 Watts are generated via by the proton-proton cycle, how many billion neutrinos are produced
Answer:
4.635 *10^12 Neutrinos
Explanation:
Here in this question, we are to determine the number of neutrinos in billions produced, given the power generated by the proton-proton cycle.
We proceed as follows;
In proton-proton cycle generates 26.7 MeV of energy and in this cycle two neutrinos are produced.
From the question, we are given that
Power P = 9.9 watts = 9.9 J/s
Watts is same as J/s
The number of proton-proton cycles required to generate E energy is N = E / E '
Where E ' = Energy generated in proton-proton cycle which is given as 26.7 Mev in the question
Converting Mev to J, we have
= 26.7 x1.6 x10 -13 J
To get the number N which is the number of proton-proton cycle required, we have;
N = 9.9 /(26.7 x1.6 x10^-13) = 2.32 * 10^12
Since we have two proton cycles( proton-proton), it automatically means 2 neutrinos will be produced.
Therefore number of neutrions produced = 2 x Number of proton-proton cycles = 2 * 2.32 * 10^12 = 4.635 * 10^12 neutrinos
Which object forms when a supergiant explodes? a red giant a black hole a white dwarf a neutron star
Answer:
a neutron star
Explanation:
Answer:
d
Explanation:
Let surface S be the boundary of the solid object enclosed by x^2+z^2=4, x+y=6, x=0, y=0, and z=0. and, let f(x,y,z)=(3x)i+(x+y+2z)j + (3z)k be a vector field (for example, the velocityfaild of a fluid flow). the solid object has five sides, S1:bottom(xy-plane), S2:left side(xz-plane), S3 rear side(yz-plane), S4:right side, and S5:cylindrical roof.
a. Sketch the solid object.
b. Evaluate the flux of F through each side of the object (S1,S2,S3,S4,S5).
c. Find the total flux through surface S.
a. I've attached a plot of the surface. Each face is parameterized by
• [tex]\mathbf s_1(x,y)=x\,\mathbf i+y\,\mathbf j[/tex] with [tex]0\le x\le2[/tex] and [tex]0\le y\le6-x[/tex];
• [tex]\mathbf s_2(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf k[/tex] with [tex]0\le u\le2[/tex] and [tex]0\le v\le\frac\pi2[/tex];
• [tex]\mathbf s_3(y,z)=y\,\mathbf j+z\,\mathbf k[/tex] with [tex]0\le y\le 6[/tex] and [tex]0\le z\le2[/tex];
• [tex]\mathbf s_4(u,v)=u\cos v\,\mathbf i+(6-u\cos v)\,\mathbf j+u\sin v\,\mathbf k[/tex] with [tex]0\le u\le2[/tex] and [tex]0\le v\le\frac\pi2[/tex]; and
• [tex]\mathbf s_5(u,y)=2\cos u\,\mathbf i+y\,\mathbf j+2\sin u\,\mathbf k[/tex] with [tex]0\le u\le\frac\pi2[/tex] and [tex]0\le y\le6-2\cos u[/tex].
b. Assuming you want outward flux, first compute the outward-facing normal vectors for each face.
[tex]\mathbf n_1=\dfrac{\partial\mathbf s_1}{\partial y}\times\dfrac{\partial\mathbf s_1}{\partial x}=-\mathbf k[/tex]
[tex]\mathbf n_2=\dfrac{\partial\mathbf s_2}{\partial u}\times\dfrac{\partial\mathbf s_2}{\partial v}=-u\,\mathbf j[/tex]
[tex]\mathbf n_3=\dfrac{\partial\mathbf s_3}{\partial z}\times\dfrac{\partial\mathbf s_3}{\partial y}=-\mathbf i[/tex]
[tex]\mathbf n_4=\dfrac{\partial\mathbf s_4}{\partial v}\times\dfrac{\partial\mathbf s_4}{\partial u}=u\,\mathbf i+u\,\mathbf j[/tex]
[tex]\mathbf n_5=\dfrac{\partial\mathbf s_5}{\partial y}\times\dfrac{\partial\mathbf s_5}{\partial u}=2\cos u\,\mathbf i+2\sin u\,\mathbf k[/tex]
Then integrate the dot product of f with each normal vector over the corresponding face.
[tex]\displaystyle\iint_{S_1}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{6-x}f(x,y,0)\cdot\mathbf n_1\,\mathrm dy\,\mathrm dx[/tex]
[tex]=\displaystyle\int_0^2\int_0^{6-x}0\,\mathrm dy\,\mathrm dx=0[/tex]
[tex]\displaystyle\iint_{S_2}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{\frac\pi2}\mathbf f(u\cos v,0,u\sin v)\cdot\mathbf n_2\,\mathrm dv\,\mathrm du[/tex]
[tex]\displaystyle=\int_0^2\int_0^{\frac\pi2}-u^2(2\sin v+\cos v)\,\mathrm dv\,\mathrm du=-8[/tex]
[tex]\displaystyle\iint_{S_3}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^6\mathbf f(0,y,z)\cdot\mathbf n_3\,\mathrm dy\,\mathrm dz[/tex]
[tex]=\displaystyle\int_0^2\int_0^60\,\mathrm dy\,\mathrm dz=0[/tex]
[tex]\displaystyle\iint_{S_4}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{\frac\pi2}\mathbf f(u\cos v,6-u\cos v,u\sin v)\cdot\mathbf n_4\,\mathrm dv\,\mathrm du[/tex]
[tex]=\displaystyle\int_0^2\int_0^{\frac\pi2}-u^2(2\sin v+\cos v)\,\mathrm dv\,\mathrm du=\frac{40}3+6\pi[/tex]
[tex]\displaystyle\iint_{S_5}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^{\frac\pi2}\int_0^{6-2\cos u}\mathbf f(2\cos u,y,2\sin u)\cdot\mathbf n_5\,\mathrm dy\,\mathrm du[/tex]
[tex]=\displaystyle\int_0^{\frac\pi2}\int_0^{6-2\cos u}12\,\mathrm dy\,\mathrm du=36\pi-24[/tex]
c. You can get the total flux by summing all the fluxes found in part b; you end up with 42π - 56/3.
Alternatively, since S is closed, we can find the total flux by applying the divergence theorem.
[tex]\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\iiint_R\mathrm{div}\mathbf f(x,y,z)\,\mathrm dV[/tex]
where R is the interior of S. We have
[tex]\mathrm{div}\mathbf f(x,y,z)=\dfrac{\partial(3x)}{\partial x}+\dfrac{\partial(x+y+2z)}{\partial y}+\dfrac{\partial(3z)}{\partial z}=7[/tex]
The integral is easily computed in cylindrical coordinates:
[tex]\begin{cases}x(r,t)=r\cos t\\y(r,t)=6-r\cos t\\z(r,t)=r\sin t\end{cases},0\le r\le 2,0\le t\le\dfrac\pi2[/tex]
[tex]\displaystyle\int_0^2\int_0^{\frac\pi2}\int_0^{6-r\cos t}7r\,\mathrm dy\,\mathrm dt\,\mathrm dr=42\pi-\frac{56}3[/tex]
as expected.
A plano-convex glass lens of radius of curvature 1.4 m rests on an optically flat glass plate. The arrangement is illuminated from above with monochromatic light of 520-nm wavelength. The indexes of refraction of the lens and plate are 1.6. Determine the radii of the first and second bright fringes in the reflected light.
Given that,
Radius of curvature = 1.4 m
Wavelength = 520 nm
Refraction indexes = 1.6
We know tha,
The condition for constructive interference as,
[tex]t=(m+\dfrac{1}{2})\dfrac{\lambda}{2}[/tex]
Where, [tex]\lambda=wavelength[/tex]
We need to calculate the radius of first bright fringes
Using formula of radius
[tex]r_{1}=\sqrt{2tR}[/tex]
Put the value of t
[tex]r_{1}=\sqrt{2\times(m+\dfrac{1}{2})\dfrac{\lambda}{2}\times R}[/tex]
Put the value into the formula
[tex]r_{1}=\sqrt{2\times(0+\dfrac{1}{2})\dfrac{520\times10^{-9}}{2}\times1.4}[/tex]
[tex]r_{1}=0.603\ mm[/tex]
We need to calculate the radius of second bright fringes
Using formula of radius
[tex]r_{2}=\sqrt{2\times(m+\dfrac{1}{2})\dfrac{\lambda}{2}\times R}[/tex]
Put the value into the formula
[tex]r_{1}=\sqrt{2\times(1+\dfrac{1}{2})\dfrac{520\times10^{-9}}{2}\times1.4}[/tex]
[tex]r_{1}=1.04\ mm[/tex]
Hence, The radius of first bright fringe is 0.603 mm
The radius of second bright fringe is 1.04 mm.
A double-convex thin lens is made of glass with an index of refraction of 1.52. The radii of curvature of the faces of the lens are 60 cm and 72 cm. What is the focal length of the lens
Answer:
63 cm
Explanation:
Mathematically;
The focal length of a double convex lens is given as;
1/f = (n-1)[1/R1 + 1/R2]
where n is the refractive index of the medium given as 1.52
R1 and R2 represents radius of curvature which are given as 60cm and 72cm respectively.
Plugging these values into the equation, we have:
1/f = (1.52-1)[1/60 + 1/72)
1/f = 0.0158
f = 1/0.0158
f = 63.29cm which is approximately 63cm
In state-of-the-art vacuum systems, pressures as low as 1.00 10-9 Pa are being attained. Calculate the number of molecules in a 1.90-m3 vessel at this pressure and a temperature of 28.0°C. molecules
Answer:
The number of molecules is 4.574 x 10¹¹ Molecules
Explanation:
Given;
pressure in the vacuum system, P = 1 x 10⁻⁹ Pa
volume of the vessel, V = 1.9 m³
temperature of the system, T = 28°C = 301 K
Apply ideal gas law;
[tex]PV= nRT = NK_BT[/tex]
Where;
n is the number of gas moles
R is ideal gas constant = 8.314 J / mol.K
[tex]K_B[/tex] is Boltzmann's constant, = 1.38 x 10⁻²³ J/K
N is number of gas molecules
N = (PV) / ([tex]K_B[/tex]T)
N = (1 x 10⁻⁹ X 1.9) / ( 1.38 x 10⁻²³ X 301)
N = 4.574 x 10¹¹ Molecules
Therefore, the number of molecules is 4.574 x 10¹¹ Molecules
Two parallel metal plates, each of area A, are separatedby a distance 3d. Both are connected to ground and each plate carries no charge. A third plate carrying charge Qis inserted between the two plates, located a distance dfrom the upper plate. As a result, negative charge is induced on each of the two original plates. a) In terms of Q, find the amount of charge on the upper plate, Q1, and the lower plate, Q2. (Hint: it must be true that Q
Answer:
Upper plate Q/3
Lower plate 2Q/3
Explanation:
See attached file
Can you come up with a mathematical relationship, based on your data that shows the relationship between distance from the charges and electric field strength?
Answer:
Explanation:
This question appears incomplete because of the absence of the data been talked about in the question. However, there is a general ruling here and it can be applied to the data at hand.
If an increase in the distance of charges (let's denote with "d") causes the electric field strength (let's denote with"E") to increase, then the mathematical representation can be illustrated as d ∝ E (meaning distance of charge is directly proportional to electric field strength).
But if an increase in the distance of the charges causes the electric field strength to decrease, then the mathematical representation can be illustrated as d ∝ 1/E (meaning distance of charge is inversely proportional to electric field strength).
A scatterplot can also be used to determine this. If there is a positive correlation (correlation value is greater than zero but less than or equal to 1) on the graph, then it is illustrated as "d ∝ E" BUT if there is a negative correlation (correlation value is less than zero but greater than or equal to -1), then it can be illustrated as "d ∝ 1/E".
Charge of uniform density (0.30 nC/m2) is distributed over the xy plane, and charge of uniform density (−0.40 nC/m2) is distributed over the yz plane. What is the magnitude of the resulting electric field at any point not in either of the two charged planes?
Answer: E = 39.54 N/C
Explanation: Electric field can be determined using surface charge density:
[tex]E = \frac{\sigma}{2\epsilon_{0}}[/tex]
where:
σ is surface charge density
[tex]\epsilon_{0}[/tex] is permitivitty of free space ([tex]\epsilon_{0} = 8.85.10^{-12}[/tex][tex]C^{2}/N.m^{2}[/tex])
Calculating resulting electric field:
[tex]E=E_{1} - E_{2}[/tex]
[tex]E = \frac{\sigma_{1}-\sigma_{2}}{2\epsilon_{0}}[/tex]
[tex]E = \frac{[0.3-(-0.4)].10^{-9}}{2.8.85.10^{-12}}[/tex]
[tex]E=0.03954.10^{3}[/tex]
E = 39.54
The resulting Electric Field at any point is 39.54N/C.
The magnitude of the resulting electric field at any point should be 28.2 N/C.
Calculation of the magnitude:Since the Charge of uniform density (0.30 nC/m2) should be allocated over the xy plane, and charge of uniform density (−0.40 nC/m2)should be allocated over the yz plane.
So,
E1
= σ1/2ε0
= 0.30e-9/(2*8.85e-12)
= 16.949 N/C
So, direction of E1 is +z
Now
E2 = σ2/2ε0
= 0.40e-9/(2*8.85e-12)
= 22.6 N/C
So, direction of E2 is -x
Now
E = √(E1*E1+E2*E2)
= √(16.949*16.949+22.6*22.6)
= 28.2 N/C
Learn more about magnitude here: https://brainly.com/question/14576767
In a velocity selector having electric field E and magnetic field B, the velocity selected for positively charged particles is v= E/B. The formula is the same for a negatively charged particles.
a. True
b. False
Answer:
True or False
Explanation:
Because.....
easy 50% chance you are right
The molecules in Tyler are composed of carbon and other atoms that share one or more electrons between two atoms, forming what is known as a(n) _____ bond.
Answer:
covalent
Explanation:
covalent bonds share electrons
A brick is resting on a smooth wooden board that is at a 30° angle. What is one way to overcome the static friction that is holding the brick in place?
Answer:
We apply force to move the brick.
Explanation:
Let me first of define a force .
A force is something applied to an object or thing to change it's internal or external state.
Now if a brick is resting on smooth wood inclined at 30° to the horizontal for us to overcome the friction which is also a force we have to apply a force greater than the gravity force acting on the body and then depending on the direction of the applied force the angle to apply it also.
You shine unpolarized light with intensity 52.0 W/m2 on an ideal polarizer, and then the light that emerges from this polarizer falls on a second ideal polarizer. The light that emerges from the second polarizer has intensity 15.0 W/m2. Find the intensity of the light that emerges from the first polarizer.
Answer:
The intensity of light from the first polarizer is [tex]I_1 = 26 W/m^2[/tex]
Explanation:
The intensity of the unpolarized light is [tex]I_o = 52.0 \ W/m^2[/tex]
Generally the intensity of light that emerges from the first polarized light is
[tex]I_1 = \frac{I_o}{2 }[/tex]
substituting values
[tex]I_1 = \frac{52. 0}{2 }[/tex]
[tex]I_1 = 26 W/m^2[/tex]
Which statement about kinetic and static friction is accurate?
Static friction is greater than kinetic friction, and they both act in conjunction with the applied force.
Kinetic friction is greater than static friction, and they both act in conjunction with the applied force.
Kinetic friction is greater than static friction, but they both act opposite the applied force.
Static friction is greater than kinetic friction, but they both act opposite the applied forcr
Answer:
Static friction is greater than kinetic friction, but they both act opposite the applied force.
Explanation:
Newton's 3rd law states that every action has and equal but opposite reaction.
If an object has static friction, that means it stays in one spot, and it takes a great amount of force to get it moving.
Once the object is moving it has kinetic friction, but it's easier to keep it moving unless you are trying to stop it.
The equal but opposite reaction to something moving it is stopping it, and the equal but opposite reaction to stopping something is moving it.
The same amount of force used to move/stop something is used to stop/move it.
Static friction is greater than kinetic friction, but they both act opposite the applied force.
Friction is the force that opposes motion. Frictional force always acts in opposition to the direction of motion.
There are two kinds of friction;
Static frictionDynamic frictionSince more forces tend to act on a body at rest and prevent it from getting into motion than the forces that tend to stop an already moving body, it follows that static friction is greater than kinetic friction. Both act in opposite direction to the applied force.
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Suppose there are only two charged particles in a particular region. Particle 1 carries a charge of +q and is located on the positive x-axis a distance d from the origin. Particle 2 carries a charge of +2q and is located on the negative x-axis a distance d from the origin.
Required:
Where is it possible to have the net field caused by these two charges equal to zero?
1. At the origin.
2. Somewhere on the x-axis between the two charges, but not at the origin.
3. Somewhere on the x-axis to the right of q2.
4. Somewhere on the y-axis.
5. Somewhere on the x-axis to the left of q1.
Answer:
x₂ = 0.1715 d
1) false
2) True
3) True
4) false
5) True
Explanation:
The field electrifies a vector quantity, so we can add the creative field by these two charges
E₂-E₁ = 0
k q₂ / r₂² - k q₁ / r1₁²= 0
q₂ / r₂² = q₁ / r₁²
suppose the sum of the fields is zero at a place x to the right of zero
r₂ = d + x
r₁ = d -x
we substitute
q₂ / (d + x)² = q₁ / (d-x)²
we solve the equation
q₂ / q₁ (d-x)² = (d + x) ²
let's replace the value of the charges
q₂ / q₁ = + 2q / + q = 2
2 (d²- 2xd + x²) = d² + 2xd + x²
x² -6xd + d² = 0
we solve the quadratic equation
x = [6d ± √ (36d² - 4 d²)] / 2
x = [6d ± 5,657 d] / 2
x₁ = 5.8285 d
x₂ = 0.1715 d
with the total field value zero it is between the two loads the correct solution is x₂ = 0.1715 d
this value remains on the positive part of the x axis, that is, near charge 1
now let's examine the different proposed outcomes
1) false
2) True
3) True
4) false
5) True
If you stood on a planet having a mass four times higher than Earth's mass, and a radius two times 70) lon longer than Earth's radius, you would weigh:________
A) four times more than you do on Earth.
B) two times less than you do on Earth.
C) the same as you do on Earth
D) two times more than you do on Earth.
CHECK COMPLETE QUESTION BELOW
you stood on a planet having a mass four times that of earth mass and a radius two times of earth radius , you would weigh?
A) four times more than you do on Earth.
B) two times less than you do on Earth.
C) the same as you do on Earth
D) two times more than you do on Earth
Answer:
OPTION C is correct
The same as you do on Earth
Explanation :
According to law of gravitation :
F=GMm/R^2......(a)
F= mg.....(b)
M= mass of earth
m = mass of the person
R = radius of the earth
From law of motion
Put equation b into equation a
mg=GMm/R^2
g=GMm/R^2
g=GM/R^2
We know from question a planet having a mass four times that of earth mass and a radius two times of earth radius if we substitute we have
m= 4M
r=(2R)^2=4R^2
g= G4M/4R^2
Then, 4in the denominator will cancel out the numerator we have
g= GM/R^2
Therefore, g remain the same
which category would a person who has an IQ of 84 belong ?
Three ideal polarizing filters are stacked, with the polarizing axis of the second and third filters at 29.0and 58.0, respectively, to that of the first. If unpolarized light is incident on the stack, the light has intensity 110 after it passes through the stack.
If the incident intensity is kept constant, what is the intensity of the light after it has passed through the stack if the second polarizer is removed?
Answer:
I₂ = 143.79
Explanation:
To solve this problem, work them in two parts. A first one where we look for the intensity of the incident light in the set and a second one where we silence the light transmuted by the other set,
Let's start with the set of three curling irons
Beautiful light falls on the first polarized is not polarized, therefore only half the radiation passes
I₁ = I₀ / 2
this light reaches the second polarized and must comply with the Mule law
I₂ = I₁ cos² tea
The angle between the first polarized and the second is Tea = 29.0º
I₂ = I / 2 cos² 29
The light that comes out of the third polarized is
I₃ = I₂ cos² tea
the angle between the third - second polarizer is
tea = 58-29
tea = 29th
I3 = (I₀ / 2 cos² 29) cos² 29
indicate the output intensity
I3 = 110
we clear
I₀ = 2I3 / cos4 29
I₀ = 2 110 / cos4 29
I₀ = 375.96 W / cm²
Now we have the incident intensity in the new set of three polarizers
back to the for the first polarizer
I₁ = I₀ / 2
when passing the second polarizer
I₂ = I1 cos² 29
I2 = IO /2 cos²29
let's calculate
I₂ = 375.96 / 2 cos² 29
I₂ = 143.79
A charge of uniform density (0.74 nC/m) is distributed along the x axis from the origin to the point x = 10 cm. What is the electric potential (relative to zero at infinity) at a point, x = 23 cm, on the x axis? Hint: Use Calculus to solve this problem.
Answer:
V = - 3.85 V
Explanation:
The electric potential of a continuous charge distribution is
V = k ∫ dq / r
to find charge differential let's use the concept of linear density
λ = dq / dx
dq = λ dx
the distance from a load element to the point of interest
x₀ = 23 cm = 0.23 m
r = √ (x-x₀)² = x - x₀
we substitute
v = k ∫ λ dx / (x-x₀)
we integrate and evaluate between x = 0 and x = l = 0.10 cm
V = k λ [ln (x-x₀) - ln (-x₀)]
V = k λ ln ((x-x₀) / x₀)
let's calculate
V = 9 10⁹ 0.74 10⁻⁹ ln ((0.23 - 0.10) / 0.23)
V = - 3.85 V
An electron experiences a force of magnitude F when it is 5 cm from a very long, charged wire with linear charge density, lambda. If the charge density is doubled, at what distance from the wire will a proton experience a force of the same magnitude F?
Answer:
The distance of the proton is [tex]r_p =10 \ cm[/tex]
Explanation:
Generally the force experience by the electron is mathematically represented as
[tex]F_e = \frac{q * \lambda_e }{ 2 \pi * \epsilon_o * r_e}[/tex]
Where [tex]\lambda _e[/tex] is the charge density of the charge wire before it is doubled
Also the force experience by the proton is mathematically represented as
[tex]F_p = \frac{q * \lambda_p }{ 2 \pi * \epsilon_o * r_p}[/tex]
Given that the charge density is doubled i.e [tex]\lambda_p = 2 \lambda_e[/tex] and that the the force are equal then
[tex]\frac{q * \lambda_e }{ 2 \pi * \epsilon_o * r_e} = \frac{q * 2 \lambda_e }{ 2 \pi * \epsilon_o * r_p}[/tex]
[tex]\frac{ \lambda_e }{ r_e} = \frac{ 2 \lambda_e }{ r_p}[/tex]
[tex]r_p * \lambda_e =2 \lambda_e * r_e[/tex]
[tex]r_p =2 r_e[/tex]
Now given from the question that [tex]r_e[/tex] the distance of the electron from the charged wire is 5 cm
Then
[tex]r_p =2 (5)[/tex]
[tex]r_p =10 \ cm[/tex]
what is defect of vision
Answer:
The vision becomes blurred due to the refractive defects of the eye. There are mainly three common refractive defects of vision. These are (i) myopia or near-sightedness, (ii) Hypermetropia or far – sightedness, and (iii) Presbyopia. These defects can be corrected by the use of suitable spherical lenses.
Which one of the following frequencies of a wave in the air can be heard as an audible sound by human ear
Answer:
1,000 Hz hope this helps.
Explanation:
The sound said to be audible if it comes in the range of audible sound range. The audible sound is the specific frequency range of sound, which can be heard by human ears. The audible sound frequency range is 20Hz−20,000H
The roller coaster car reaches point A of the loop with speed of 20 m/s, which is increasing at the rate of 5 m/s2. Determine the magnitude of the acceleration at A if pA
Answer and Explanation:
Data provided as per the question is as follows
Speed at point A = 20 m/s
Acceleration at point C = [tex]5 m/s^2[/tex]
[tex]r_A = 25 m[/tex]
The calculation of the magnitude of the acceleration at A is shown below:-
Centripetal acceleration is
[tex]a_c = \frac{v^2}{r}[/tex]
now we will put the values into the above formula
= [tex]\frac{20^2}{25}[/tex]
After solving the above equation we will get
[tex]= 16 m/s^2[/tex]
Tangential acceleration is
[tex]= \sqrt{ac^2 + at^2} \\\\ = \sqrt{16^2 + 5^2}\\\\ = 16.703 m/s^2[/tex]
The hot glowing surfaces of stars emit energy in the form of electromagnetic radiation. It is a good approximation to assume that the emissivity eee is equal to 1 for these surfaces.
Required:
a. Find the radius RRigel of the star Rigel, the bright blue star in the constellation Orion that radiates energy at a rate of 2.7 x 10^31 W and has a surface temperature of 11,000 K.
b. Find the radius RProcyonB of the star Procyon B, which radiates energy at a rate of 2.1 x 10^23 W and has a surface temperature of 10,000 K. Assume both stars are spherical. Use σ=5.67 x 10−8^ W/m^2*K^4 for the Stefan-Boltzmann constant.
Given that,
Energy [tex]H=2.7\times10^{31}\ W[/tex]
Surface temperature = 11000 K
Emissivity e =1
(a). We need to calculate the radius of the star
Using formula of energy
[tex]H=Ae\sigma T^4[/tex]
[tex]A=\dfrac{H}{e\sigma T^4}[/tex]
[tex]4\pi R^2=\dfrac{H}{e\sigma T^4}[/tex]
[tex]R^2=\dfrac{H}{e\sigma T^4\times4\pi}[/tex]
Put the value into the formula
[tex]R=\sqrt{\dfrac{2.7\times10^{31}}{1\times5.67\times10^{-8}\times(11000)^4\times 4\pi}}[/tex]
[tex]R=5.0\times10^{10}\ m[/tex]
(b). Given that,
Radiates energy [tex] H=2.1\times10^{23}\ W[/tex]
Temperature T = 10000 K
We need to calculate the radius of the star
Using formula of radius
[tex]R^2=\dfrac{H}{e\sigma T^4\times4\pi}[/tex]
Put the value into the formula
[tex]R=\sqrt{\dfrac{2.1\times10^{23}}{1\times5.67\times10^{-8}\times(10000)^4\times4\pi}}[/tex]
[tex]R=5.42\times10^{6}\ m[/tex]
Hence, (a). The radius of the star is [tex]5.0\times10^{10}\ m[/tex]
(b). The radius of the star is [tex]5.42\times10^{6}\ m[/tex]
A point source emits sound waves with a power output of 100 watts. What is the sound level (in dB) at a distance of 10 m
Answer:
[tex]L = 109.01 db[/tex]
Explanation:
Given
Power, P = 100 W
Distance, d = 10 m
Required
Determine the Sound Level
First, the sound intensity as to be calculated; This is done, as follows;
[tex]I = \frac{P}{4\pi d^2}[/tex]
Substitute for P, d and take π as 3.14
[tex]I = \frac{100}{4 * 3.14 * 10^2}[/tex]
[tex]I = \frac{100}{4 * 3.14 * 100}[/tex]
[tex]I = \frac{100}{1256}[/tex]
[tex]I = 0.0796Wm^{-2}[/tex] --- Approximated
Next is to calculate the Sound Level, as follows
[tex]L = 10 * Log(\frac{I}{I_o})[/tex]
Where [tex]I_o = 10^{-12} Wm^{-2}[/tex]
Substitute for I and Io
[tex]L = 10 * Log(\frac{0.0796}{10^{-12}})[/tex]
[tex]L = 10 * Log(0.0796*10^{12)[/tex]
[tex]L = 10 * Log(0.0796*10^{12)[/tex]
[tex]L = 10 * 10.901[/tex]
[tex]L = 109.01 db[/tex]
Hence, the sound level is 109.01 decibels
One long wire carries a current of 30 A along the entire x axis. A second long wire carries a current of 40 A perpendicular to the xy plane and passes through the point (0, 4, 0) m. What is the magnitude of the resultant magnetic field at the point y
Complete question is;
One long wire carries a current of 30 A along the entire x axis. A second long wire carries a current of 40 A perpendicular to the xy plane and passes through the point (0, 4, 0) m. What is the magnitude of the resulting magnetic field at the point y = 2.0 m on the y axis?
Answer:
B_net = 50 × 10^(-7) T
Explanation:
We are told that the 30 A wire lies on the x-plane while the 40 A wire is perpendicular to the xy plane and passes through the point (0,4,0).
This means that the second wire is 4 m in length on the positive y-axis.
Now, we are told to find the magnitude of the resulting magnetic field at the point y = 2.0 m on the y axis.
This means that the position we want to find is half the length of the second wire.
Thus, at this point the net magnetic field is given by;
B_net = √[(B1)² + (B2)²]
Where B1 is the magnetic field due to the first wire and B2 is the magnetic field due to the second wire.
Now, formula for magnetic field due to very long wire is;
B = (μ_o•I)/(2πR)
Thus;
B1 = (μ_o•I_1)/(2πR_1)
Also, B2 = (μ_o•I_2)/(2πR_2)
Now, putting the equation of B1 and B2 into the B_net equation, we have;
B_net = √[((μ_o•I_1)/(2πR_1))² + ((μ_o•I_2)/(2πR_2))²]
Now, factorizing out some common terms, we have;
B_net = (μ_o/2π)√[((I_1)/R_1))² + ((I_2)/R_2))²]
Now,
μ_o is a constant and has a value of 4π × 10^(−7) H/m
I_1 = 30 A
I_2 = 40 A
Now, as earlier stated, the point we are looking for is 2 metres each from wire 2 end and wire 1.
Thus;
R_1 = 2 m
R_2 = 2 m
So, let's calculate B_net.
B_net = ((4π × 10^(−7))/2π)√[(30/2)² + (40/2)²]
B_net = 50 × 10^(-7) T
