The charge on the capacitor after 2 s is 0.43 C.
Explanation:
The formula for finding the voltage while charging a capacitor is
[tex]V = V_{0}(1-e^{\frac{-t}{RC} })[/tex]
Here , V₀ is the initial potential before charging and t is the time at which we have to determine the voltage, R is the resistance and C is the capacitance for the given circuit.
The given problem have given us the values for V₀ = 50 V and maximum current I is given as 500 mA.
Then, resistance can be determined using Ohm's law: [tex]R = \frac{V}{I} =\frac{50}{500*10^{-3} } = 100 Ohm[/tex]
The capacitance is defined as the ratio of charge to the unit voltage.
[tex]C = \frac{Q}{V} = \frac{IT}{IR} = \frac{T}{R}[/tex]
Here T is the time constant which is given as 1 s and R is found to be 100 ohm, then capacitance will be [tex]\frac{1}{100} = 10 mF[/tex]
So, the values for parameters like V₀ = 50 V, R = 100 Ω, C = 10 mF and t = 2 s.
Then, [tex]V =50*(1-e^{\frac{-2}{100*10*10^{-3} } }) =50*(1-e^{-2}) = 50*(1-0.1353)[/tex]
V= 43 V.
Then, [tex]Q = CV = 10*10^{-3} * 43 V = 0.43 C[/tex]
Thus, the charge on the capacitor after 2 s is 0.43 C.
The unit of charge is Coulomb.
The charge on the capacitor after 2 second is 0.43 Coulomb.
To finding the voltage while charging a capacitor is given as,
[tex]V=V_{0}(1-e^{-\frac{t}{RC} } )[/tex]
Here , V₀ is the initial potential, R is the resistance and C is the capacitance
It is given that V₀ = 50 V and maximum current I is given as 500 mA.
So, [tex]R=\frac{V}{I}=\frac{50}{500*10^{-3} } =100ohm[/tex]
Time constant = RC = 1
So, [tex]C=\frac{1}{R} =\frac{1}{100}F[/tex]
Now, we have V₀ = 50 V, R = 100 Ω, C = 1/100 F and t = 2 s.
[tex]V=50(1-e^{-\frac{2}{100*\frac{1}{100} } } )=50(1-e^{-2} )=50(1-0.1353)=43V[/tex]
We know that, [tex]Q=CV=\frac{1}{100}*43=0.43C[/tex]
Thus, the charge on the capacitor after 2 s is 0.43 C.
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Which front formed widespread clouds rain or snow
Answer: A cold front occurs when cold, denser air replaces the rising, less dense air mass. The reason this front brings in the rain is that as the rising warm air cools (as it rises to the cooler upper atmosphere) the moisture in it condenses into clouds that precipitate down as rain or snow.
Answer:
cold front
Explanation:
Let's return to our friend the sphere, with surface charge density σ(θ, f) = σ0sinθcos2f . Find the net polarization of this sphere, assuming the charge density is bound charge. You can assume the sphere is not rotating (f is azimuthal angle). What is the electric potential a large distance from the sphere?
Answer:
Explanation:
solution solved below
You stare at a bright red screen for so long that your red cones become saturated and no longer function. The red screen is then rapidly replaced by a white screen. Immediately afterwards, what color do you see on the screen?
Answer: you'll see cyan color on the screen
Explanation:
Saturating the red cone causes them to stop functioning, hence you can't perceive the red part of white light. White light is made up of three main colors which are blue, red and green. When one can no longer perceive the red part of light, one is left with the grean and blue part. The green and blue part of light will superimpose to give a cyan color.
Answer:
Cyan color
Explanation:
When an individual stares at a red object and immediately look at a white area afterward, there will be an image formation almost immediately that is the same size and shape as the previous one. The only difference is there will be a color change to blue-green, or cyan.
This is due to the eyes generally using the red, blue and green cone cells to perceive white light. Since the red cone cells are fatigued due to long view of it and cone cells being saturated , the blue and green color which are the remaining color cone cells use to perceive white light is viewed. This gives rise to the blue green or cyan color.
Two friction disks A and B are brought into contact when the angular velocity of disk A is 240 rpm counterclockwise and disk B is at rest. A period of slipping follows and disk B makes 2 revolutions before reaching its final angular velocity. Assuming that the angular acceleration of each disk is constant and inversely proportional to the cube of its radius, determine (a) the angular acceleration of each disk, (b) the time during which the disks slip.
Answer:
a) αA = 4.35 rad/s²
αB = 1.84 rad/s²
b) t = 3.7 rad/s²
Explanation:
Given:
wA₀ = 240 rpm = 8π rad/s
wA₁ = 8π -αA*t₁
The angle in B is:
[tex]\theta _{B} =4\pi =\frac{1}{2} \alpha _{B} t_{1}^{2} =\frac{1}{2} (\frac{r_{A} }{r_{B} } )^{3} \alpha _{A} t_{1}^{2}=\frac{1}{2} (\frac{0.15}{0.2} )^{3} \alpha _{A} t_{1}^{2}[/tex]
[tex]\alpha _{A} =8\pi (\frac{0.2}{0.15} )^{3} =59.57rad[/tex]
[tex]w_{B,1} =\alpha _{B} t_{1}=(\frac{0.15}{0.2} )^{3} \alpha _{A} t_{1}=0.422\alpha _{A} t_{1}[/tex]
The velocity at the contact point is equal to:
[tex]v=r_{A} w_{A} =0.15*(8\pi -\alpha _{A} t_{1})=1.2\pi -0.15\alpha _{A} t_{1}[/tex]
[tex]v=r_{B} w_{B} =0.2*(0.422\alpha _{A} t_{1})=0.0844\alpha _{A} t_{1}[/tex]
Matching both expressions:
[tex]1.2\pi -0.15\alpha _{A} t_{1}=0.0844\alpha _{A} t_{1}\\\alpha _{A} t_{1}=16.09rad/s[/tex]
b) The time during which the disks slip is:
[tex]t_{1} =\frac{\alpha _{A} t_{1}^{2}}{\alpha _{A} t_{1}} =\frac{59.574}{16.09} =3.7s[/tex]
a) The angular acceleration of each disk is
[tex]\alpha _{A}=\frac{\alpha _{A} t_{1}}{t_{1} } =\frac{16.09}{3.7} =4.35rad/s^{2} (clockwise)[/tex]
[tex]\alpha _{B}=(\frac{0.15}{0.2} )^{3} *4.35=1.84rad/s^{2} (clockwise)[/tex]
The angular acceleration of each disk and the time during which the disks slip cannot be determined with the given information.
Explanation:(a) To determine the angular acceleration of each disk, we can use the formula: angular acceleration = (final angular velocity - initial angular velocity) / time. For disk A, the initial angular velocity is 240 rpm counterclockwise and the final angular velocity is 0 rpm. The time during which disk A stops is not given, so we cannot determine its angular acceleration at this time. For disk B, the initial angular velocity is 0 rpm and the final angular velocity is also 0 rpm. Since disk B is at rest initially and then starts moving, it undergoes an angular acceleration that brings it to rest. However, the specific time during which disk B slips is not given, so we cannot determine its angular acceleration during slipping.
(b) The time during which the disks slip is not provided in the question, so we cannot determine the exact time. However, we know that disk B makes 2 revolutions before reaching its final angular velocity, so during this period, disk B is slipping.
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12. A woman lifts a 35-kg child a distance of 1.5 m and carries her forward for 6.5 m.
How much work does the woman do in lifting the child?
Answer:
515.025J
A woman lifts a 35-kg child a distance of 1.5 m and carries her forward for 6.5 m.
How much work does the woman do in lifting the child?
Explanation:
Given;
Mass m = 35kg
Distance of lifting l = 1.5m
Acceleration due to gravity g = 9.81 m/s^2
Workdone = mgl = 35×1.5 × 9.81 = 515.025 J
The change in potential energy is equal to the work that must be done, to lift the child a certain distance. Therefore, the work done by the woman in lifting the child is:
[tex]W=\Delta U\\W=mgh_f-mgh_0\\W=mg(h_f-h_0)\\W=35kg*1.5m(1.5m-0m)\\W=515.03J[/tex]
A circular saw spins at 6000 rpm , and its electronic brake is supposed to stop it in less than 2 s. As a quality-control specialist, you're testing saws with a device that counts the number of blade revolutions. A particular saw turns 75 revolutions while stopping.Does it meet its specs?
a.yes
b.no
Answer:
a. yes
Explanation:
The initial speed of the circular saw is:
[tex]\dot n_{o} = \frac{6000}{60} \,\frac{rev}{s}[/tex]
[tex]\dot n_{o} = 100\,\frac{rev}{s}[/tex]
Deceleration rate needed to stop the circular saw is:
[tex]\ddot n = -\frac{100\,\frac{rev}{s} }{2\,s}[/tex]
[tex]\ddot n = - 50\,\frac{rev}{s^{2}}[/tex]
The number of turns associated with such deceleration rate is:
[tex]\Delta n = \frac{\dot n^{2}}{2\cdot \ddot n}[/tex]
[tex]\Delta n = \frac{\left(100\,\frac{rev}{s} \right)^{2}}{2\cdot \left(50\,\frac{rev}{s^{2}} \right)}[/tex]
[tex]\Delta n = 100\,rev[/tex]
Since the measured number of revolutions is lesser than calculated number of revolution, the circular saw meets specifications.
A 38000-Mg ocean liner has an initial velocity of 4 km/h. Neglecting the frictional resistance of the water, determine the time required to bring the liner to rest by using a single tugboat which exerts a constant force of 195 kN. The time required is ____ min ___ s.
Answer:
216.59 s.
Explanation:
Using,
F = ma................. Equation 1
Where F = force exerted by the tugboat, m = mass of the ocean liner, a = acceleration of the ocean liner
make a the subject of the equation
a = F/m.................. Equation 2
Given: F = 195 kN = 195000 N, m = 38000 Mg = 38000000 kg.
Substitute into equation 2
a = 195000/38000000
a = 5.13×10⁻³ m/s²
Also using,
Assuming the liner is decelerating
a = (u-v)/t............ Equation 3
Where v = final velocity, u = initial velocity, t = time
make t the subject of the equation
t = (u-v)/a............. Equation 4
Given: u = 4 km/h = 4(1000/3600) = 1.111 m/s, v = 0 m/s, a = 0.00513 m/s²
Substitute into equation 4
t = (1.111-0)/0.00513
t = 216.59 s.
In Case 1, a force F is pushing perpendicular on an object a distance L/2 from the rotation axis. In Case 2 the same force is pushing at an angle of 30 degrees a distance L from the axis.1)In which case is the torque due to the force about the rotation axis biggest?
a. Case 1b. Case 2c. Same
Explanation:
Case 1,
A force F is pushing perpendicular on an object a distance L/2 from the rotation axis. Torque is given by :
[tex]\tau_1=Fd\sin \theta\\\\\tau_1=\dfrac{FL}{2}\sin (90)\\\\\tau_1=\dfrac{FL}{2}\ ..........(1)[/tex]
Case 2,
The same force is pushing at an angle of 30 degrees a distance L from the axis. New torque is given by :
[tex]\tau_2=Fd\sin \theta\\\\\tau_2=FL\sin (30)\\\\\tau_2=\dfrac{FL}{2}\ ..........(2)[/tex]
From equation (1) and (2), it is clear that the force in both cases is same.
how large would a coin be with diameter of 3/4 in appear through a 5x lens
Answer:
h ’= 3.75 inch
Explanation:
For this exercise we use the definition of magnification which is the height of the image over the height of the object
m = h ’/ h
h’= m h
Let's calculate
h ’= 5 3/4
h ’= 3.75 inch
A constant force of 45 N directed at angle θ to the horizontal pulls a crate of weight 100 N from one end of a room to another a distance of 4 m. Given that the vertical component of the pulling force is 12 N, calculate the work done by the force in moving the crate.
Answer:
[tex]W=173.48J[/tex]
Explanation:
information we know:
Total force: [tex]F=45N[/tex]
Weight: [tex]w=100N[/tex]
distance: [tex]4m[/tex]
vertical component of the force: [tex]F_{y}=12N[/tex]
-------------
In this case we need the formulas to calculate the components of the force (because to calculate the work we need the horizontal component of the force).
horizontal component: [tex]F_{x}=Fcos\theta[/tex]
vertical component: [tex]F_{y}=Fsen\theta[/tex]
but from the given information we know that [tex]F_{y}=12N[/tex]
so, equation these two [tex]F_{y}=Fsen\theta[/tex] and [tex]F_{y}=12N[/tex]
[tex]Fsen\theta =12N[/tex]
and we know the force [tex]F=45N[/tex], thus:
[tex]45sen\theta=12[/tex]
now we clear for [tex]\theta[/tex]
[tex]sen\theta =12/45\\\theta=sin^{-1}(12/45)\\\theta =15.466[/tex]
the angle to the horizontal is 15.466°, with this information we can calculate the horizontal component of the force:
[tex]F_{x}=Fcos\theta[/tex]
[tex]F_{x}=45cos(15.466)\\F_{x}=43.37N[/tex]
whith this horizontal component we calculate the work to move the crate a distance of 4 m:
[tex]W=F_{x}*D\\W=(43.37N)(4m)\\W=173.48J[/tex]
the work done is W=173.48J
To calculate the work done by the force in moving the crate, we use the formula Work = Force x Distance x cos(theta). We are given the force, distance, and the vertical component of the force, so we can find the horizontal component using trigonometry. Substituting the values into the formula gives us the work done as 174 J.
Explanation:To calculate the work done by a force, we use the formula:
Work = Force x Distance x cos(theta)
In this case, the force is 45 N, the distance is 4 m, and the vertical component of the force is 12 N. We need to find the horizontal component of the force, which can be calculated using the given information.
Given that the vertical component of the force is 12 N, we can use the trigonometric relationship between the horizontal and vertical components of a force to find the horizontal component as follows:
sin(theta) = vertical component / total force
sin(theta) = 12 N / 45 N
sin(theta) = 0.267
theta = sin^(-1)(0.267)
theta = 15.1 degrees
Once we have the horizontal component of the force, we can calculate the work done as follows:
Work = Force x Distance x cos(theta)
Work = 45 N x 4 m x cos(15.1 degrees)
Work = 180 J x cos(15.1 degrees)
Work = 180 J x 0.966
Work = 174 J
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An ideal spring is mounted horizontally, with its left end fixed. The force constant of the spring is 170 N/m. A glider of mass 1.2 kg is attached to the free end of the spring. The glider is pulled toward the right along a frictionless air track, and then released. Now the glider is moving in simple harmonic motion with amplitude 0.045 m. The motion is horizontal (one-dimensional).
Suddenly, Slimer holding an apple flies in and approaches the glider. Slimer drops the apple vertically onto the glider from a very small height. The apple sticks to the glider. The mass of the apple is 0.18 kg.
Recall that the total mechanical energy is E = 1/2 mv^2 + 1/2 kx^2 = 1/2 kA^2 = constant
(a) Calculate the new amplitude of the motion of the glider with apple if the apple is dropped at the moment when the glider passes through its equilibrium position, x = 0 m.
Hints: The total energy of the glider just before the collision is E = 1/2mglider v^2 = 1/2KA^2
The apple sticks to the glider in a completely inelastic collision. The glider is now moving with the apple but at a lower speed. The linear momentum is conserved. Write a corresponding equation.
Also, assume that the collision is very short, so just before the collision the glider is at x = 0 m, and just after the collision the glider and apple are still at x = 0 m. Therefore, the total energy of the glider just after the collision is Enew = 1/2mglider + applev^2new = 1/2kA^2new
(b) Calculate the period of the motion of the glider and the period of the motion of the glider with apple.
Hint: it's a very simple question.
Answer:
A) The new amplitude = 0.048 m
B) Period T = 0.6 seconds
Explanation: Please find the attached files for the solution
dam is used to block the passage of a river and to generate electricity. Approximately 58.4 x 103 kg of water falls each second through a height of 20.1 m. If one half of the gravitational potential energy of the water were converted to electrical energy, how much power (in MW) would be generated
Answer:
8.049 MW
Explanation:
The expression for gravitational potential energy is given as
Ep = mgh............. Equation 1
Ep = gravitational potential energy, m = mass of water, h = height, g = acceleration due to gravity.
Given: m = 58.4×10³ kg, h = 20.1 m, g = 9.81 m/s²
Substitute into equation 1
Ep = 58.4×10³(20.1)(9.81)
Ep = 1.6098×10⁷ J.
If one half the gravitational potential energy of the water were converted to electrical energy
Electrical energy = Ep/2
Electrical energy = (1.6098×10⁷)/2
Electrical energy = 8.049×10⁶ J
In one seconds,
The power generated = 8.049×10⁶ W
Power generated = 8.049 MW
An iron wire and a copper wire of the same length have the same potential difference applied to them. What must be the ratio of their radii if the currents in the two wires are to be the same? (Ratio of the radius of iron wire to that of copper wire). For the resistivities of iron and copper use: rhoiron = 1.0 times 10-7 capital omega m; rhocopper = 1.7 times 10-8 capital omega m;
Answer: The ratio of their radii (Ratio of the radius of iron wire to that of copper wire) is 2.43
Explanation: Please see the attachments below
Final answer:
The ratio of the radius of an iron wire to that of a copper wire that have the same potential difference and current is approximately 1.9.
Explanation:
To find the ratio of the radii of an iron wire and a copper wire with the same potential difference and current, we can use Ohm's law and the formulas for resistance and resistivity.
The resistance is given by R = rho * (L/A), where rho is the resistivity, L is the length, and A is the cross-sectional area of the wire. Since the potential difference and current are the same for both wires, we can equate the resistances and cross-sectional areas, resulting in rho_iron * (L/A_iron) = rho_copper * (L/A_copper). Rearranging the equation, we get (A_iron/A_copper) = (rho_iron/rho_copper).
Since the radii are related to the areas of the wires by the equation A = pi * r^2, we can substitute A_iron = pi * (r_iron)^2 and A_copper = pi * (r_copper)^2 into the equation. This gives us (pi * (r_iron)^2)/(pi * (r_copper)^2) = (rho_iron/rho_copper). By canceling out the pi terms, we find (r_iron/r_copper)^2 = (rho_iron/rho_copper). Simplifying further, we get r_iron/r_copper = sqrt(rho_iron/rho_copper).
Substituting the given resistivities into the equation, we have r_iron/r_copper = sqrt((1.0 * 10^-7) / (1.7 * 10^-8)). Evaluating the expression, we find r_iron/r_copper ≈ 1.9. Therefore, the ratio of the radius of the iron wire to that of the copper wire should be approximately 1.9.
A time-dependent but otherwise uniform magnetic field of magnitude B0(t) is confined in a cylindrical region of radius 6.5 cm. Initially the magnetic field in the region is pointed out of the page and has a magnitude of 5.5 T, but it is decreasing at a rate of 24.5 G/s. Due to the changing magnetic field, an electric field will be induced in this space which causes the acceleration of charges in the region. What is the direction of acceleration of a proton placed in at 1.5 cm from the center?
Answer:
The acceleration is [tex]a = 3.45*10^{3} m/s^2[/tex]
Explanation:
From the question we are told that
The radius is [tex]d = 6.5 cm = \frac{6.5}{100} = 0.065 m[/tex]
The magnitude of the magnetic field is [tex]B = 5.5 T[/tex]
The rate at which it decreases is [tex]\frac{dB}{dt} = 24.5G/s = 24.5*10^{-4} T/s[/tex]
The distance from the center of field is [tex]r = 1.5 cm = \frac{1.5}{100} = 0.015m[/tex]
According to Faraday's law
[tex]\epsilon = - \frac{d \o}{dt}[/tex]
and [tex]\epsilon = \int\limits {E} \, dl[/tex]
Where the magnetic flux [tex]\o = B* A[/tex]
E is the electric field
dl is a unit length
So
[tex]\int\limits {E} \, dl = - \frac{d}{dt} (B*A)[/tex]
[tex]{E} l = - \frac{d}{dt} (B*A)[/tex]
Now [tex]l[/tex] is the circumference of the circular loop formed by the magnetic field and it mathematically represented as [tex]l = 2\pi r[/tex]
A is the area of the circular loop formed by the magnetic field and it mathematically represented as [tex]A= \pi r^2[/tex]
So
[tex]{E} (2 \pi r)= - \pi r^2 \frac{dB}{dt}[/tex]
[tex]E = \frac{r}{2} [ - \frac{db}{dt} ][/tex]
Substituting values
[tex]E = \frac{0.015}{2} (24*10^{-4})[/tex]
[tex]E = 3.6*10^{-5} V/m[/tex]
The negative signify the negative which is counterclockwise
The force acting on the proton is mathematically represented as
[tex]F_p = ma[/tex]
Also [tex]F_p = q E[/tex]
So
[tex]ma = qE[/tex]
Where m is the mass of the the proton which has a value of [tex]m = 1.67 *10^{-27} kg[/tex]
[tex]q = 1.602 *10^{-19} C[/tex]
So
[tex]a =\frac{1.60 *10^{-19} *(3.6 *10^{-5}) }{1.67 *10^{-27}}[/tex]
[tex]a = 3.45*10^{3} m/s^2[/tex]
A cube of linear elastic material is again subjected to a vertical compressive stress s1 in the 1-direction, but is now constrained (ε ¼0) in both the 2 and the 3 directions. (a) Findexpressionsfortheinducedtransversestresses, s2 and s3 intermsof s1. Hence, derive an expression for the ‘effective modulus’ (s1/ε1) in this case. (b) Sketch the variation of effective modulus with n, and comment on the limiting values when n¼0 (foam) and n z 0.5 (rubber). (c) Explain why the rubber soles of running shoes are designed with some combination of air or gel pockets, partially foamed rubber, and a tread
Answer:
Explanation:
Solution is attached below
Jim is driving a 2268-kg pickup truck at 30.0m/sand releases his foot from the accelerator pedal. The car eventually stops due to an effective friction force that the road, air, and other things exert on the car. The friction force has an average magnitude of 700N .Part A Determine the initial kinetic energy of the truck.Part B Determine the stopping distance of the truck.
Explanation:
Given that,
Mass of the truck, m = 2268 kg
Speed of the truck, v = 30 m/s
The friction force has an average magnitude of 700 N, f = -700 N
(a) The initial kinetic energy of the truck is given by :
[tex]K=\dfrac{1}{2}mv^2\\\\K=\dfrac{1}{2}\times 2268\times (30)^2\\\\K=1.02\times 10^6\ J[/tex]
(b) Finally, the stops due to an effective friction force that the road, final speed is 0. Let d is the stopping distance of the truck. using third equation of motion to find it as :
[tex]v^2-u^2=2ad\\\\d=\dfrac{-u^2}{2a}[/tex]
Since, f = ma
[tex]d=\dfrac{-u^2m}{2f}\\\\d=\dfrac{-(30)^2\times 2268}{-2\times 700}\\\\d=1458\ m[/tex]
So, the stopping distance of the truck is 1458 meters.
Martha is measuring the barometric pressure in her back yard every thirty minutes. She records her data in the table below.
Answer:a low pressure system is moving into the area
Explanation:
This is the only reasonable answer
Answer: a low pressure system is moving into the area
Explanation: It says it on study island
Light of wavelength 588.0 nm illuminates a slit of width 0.74 mm. (a) At what distance from the slit should a screen be placed if the first minimum in the diffraction pattern is to be 0.93 mm from the central maximum(b) Calculate the width of the central maximum.
Answer:
a) L = 1.17 m
b) width of central maxima = 1.28 mm
Explanation:
Given:-
- The wavelength of light, λ = 588.0 nm
- The slit of width, a = 0.74 mm
Find:-
(a) At what distance from the slit should a screen be placed if the first minimum in the diffraction pattern is to be 0.93 mm from the central maximum
(b) Calculate the width of the central maximum.
Solution:-
- The results of Young's single slit experiment are given in form of a relation as the angle of separation between fringes ( θ ) as function of fringe order ( m ) and wavelength ( λ ).
- Destructive interference produces the dark fringes ( minimum ). Dark fringes in the diffraction pattern of a single slit are found at angles θ for which:
a*sin ( θ ) = m*λ
Where,
m : The order number for the minimum ( dark fringe ).
- We are to investigate for the first (m = 1 )) dark fringe (minima) which is y = 0.93 mm from central order ( m = 0 ) for which the screen must be placed at a distance L:
a*y/L = m*λ
L = a*y / m*λ
L = (0.74*0.93) / (1*588*10^-9)
L = 1.17 m
- The distance L of screen should be 1.17 m away from slit.
- The central maximum - central bright fringe. The maxima lie between the minima and the width of the central maximum is simply the distance between the 1st order minima from the centre of the screen on both sides of the centre.
tanθ ≈ θ ≈ y/L = w*λ
y = w*λ*L
The width of the central maximum is simply twice this value:
- Width of central maximum = 2λLw = 2*588*10^-6*1170*0.93 = 1.28 mm
Answer:
a) 1.17 m
b) 0.929mm
Explanation:
(a) to find the distance to the screen you use
[tex]y=\frac{m\lambda D}{d}[/tex]
m: order of the fringe
lambda: wavelength of the light = 588*10^{-9} m
D: distance to the screen
d: distance between slits = 0.74*10^-3 m
by doing D the subject of the formula and replacing the values of the other parameters you obtain:
[tex]D=\frac{dy}{m\lambda}=\frac{(0.74*10^{-3}m)(0.93*10^{-3})}{(1)(588*10^{-9}m)}=1.17m[/tex]
the distance to the screen is 1.17m
(b) to find the width of the central maximum you calculate the position of the first dark fringe:
[tex]y=\frac{(1)(588*10^{-9}m/2)(1.17m)}{0.74*10^{-3}m}=4.64*10^{-4}m=0.464mm[/tex]
[tex]y=\frac{m(\lambda/2)D}{d}[/tex]
2y = 2(0.464mm)=0.929mm is the width of the central maximum
For a certain optical medium the speed of light varies from a low value of 1.80 × 10 8 m/s for violet light to a high value of 1.92 × 10 8 m/s for red light. Calculate the range of the index of refraction n of the material for visible light.
Answer:
1.56 - 1.67
Explanation:
Refractive index of any material is given as the ratio of the speed of light in a vacuum to the speed of light in that medium.
Mathematically, it is given as:
n = c/v
Where c is the speed of light in a vacuum and v is the speed of light in the medium.
Given that the speed of light in the optical medium varies from 1.8 * 10^8 m/s to 1.92 * 10^8 m/s, we can find the range of the refractive index.
When the speed is 1.8 * 10^8 m/s, the refractive index is:
n = (3 * 10^8) / (1.8 * 10^8)
n = 1.67
When the speed is 1.92 * 10^8 m/s, the refractive index is:
n = (3 * 10^8) / (1.92 * 10^8)
n = 1.56
Therefore, the range of values of the refractive index of the optical medium is 1.56 - 1.67.
Final answer:
The range of the index of refraction for visible light in this optical medium is approximately 1.56 to 1.67.
Explanation:
The index of refraction, n, of a material can be calculated using the equation n = c/v, where c is the speed of light in a vacuum and v is the speed of light in the material.
For this specific optical medium, the speed of light varies from 1.80 × 10^8 m/s for violet light to 1.92 × 10^8 m/s for red light. To calculate the range of the index of refraction, we need to determine the ratio of the speed of light in a vacuum to the speed of light in the material for both violet and red light.
The range of the index of refraction, n, can be calculated as:
For violet light: n = c/v = (3.00 × 10^8 m/s) / (1.80 × 10^8 m/s) ≈ 1.67
For red light: n = c/v = (3.00 × 10^8 m/s) / (1.92 × 10^8 m/s) ≈ 1.56
Therefore, the range of the index of refraction for visible light in this optical medium is approximately 1.56 to 1.67.
Consider a thin film surrounded with two other transparent materials. Index of refraction of the top material is 1.2; index of refraction of a film 1.3 and of the bottom material 1.4 For normal incidence this film will look bright for light which wavelength satisfies the following condition a. Thickness of a film is equal to of wavelength of light in bottom material b. Thickness of a film is equal ½ of wavelength in the film c. Thickness of a film is equal ¼ of wavelength in the film. d. Thickness of a film is equal to ½ of wavelength of light in upper material e. Thickness of a film is equal ¼ of wavelength of light in upper material f. Thickness of a film is equal 4 of wavelength in bottom material
Answer:
Answer: The thickness of film is equal to 1/2 of wavelength in the film
Explanation:
Consider a thin film surrounded with two other transparent materials. Index of refraction of the top material is 1.2; index of refraction of a film 1.3 and of the bottom material 1.4 For normal incidence this film will look bright for light which wavelength satisfies the following condition a. Thickness of a film is equal to of wavelength of light in bottom material b. Thickness of a film is equal ½ of wavelength in the film c. Thickness of a film is equal ¼ of wavelength in the film. d. Thickness of a film is equal to ½ of wavelength of light in upper material e. Thickness of a film is equal ¼ of wavelength of light in upper material f. Thickness of a film is equal 4 of wavelength in bottom material
For constructive interference
Zt = ∧/n
t=∧/n
2
thickness = 1/2 x wavelength in the film
The thickness of film is equal to 1/2 of wavelength in the film
The condition for a thin film to look bright is when its thickness is ½ of the wavelength in the film itself, due to constructive interference resulting from the phase shift upon reflection.
The question deals with thin film interference, a phenomenon where light waves reflected off the top and bottom surfaces of a thin film interfere with each other. For a thin film to appear bright for light of a certain wavelength, the condition for constructive interference must be satisfied. This often involves the film's thickness being a multiple or a fraction of the wavelength of light in the film itself or in the surrounding media.
In the context of thin films surrounded by materials with different indices of refraction, the correct condition for the film to look bright is when its thickness is equal to ½ of the wavelength in the film itself. This accounts for the phase shift that occurs upon reflection, leading to constructive interference when the optical path difference is a multiple of the wavelength.
It's important to note that the index of refraction of the material influences the effective wavelength of light within it. Therefore, the wavelength in the material can be determined by dividing the wavelength in vacuum by the material's index of refraction.
A dielectric cube of side a, centered at the origin, carries a "frozen-in" polarization P = kr, where k is a constant. Find all the bound charges and check that they add up to zero.
The total volume of bound charge is zero.
Explanation:
We have to the volume and surface bounded charge densities.
ρb = - Δ . p = - Δ .k ([tex]x^{X}[/tex] +[tex]y^{Y}[/tex] +[tex]x^{Y}[/tex])
= - 3k
On the top of the cube the surface charge density is
σb = p . z
= [tex]\frac{ka}{2}[/tex]
By symmetry this holds for all the other sides. The total bounded charge should be zero
Qtot = (-3k)a³ + 6 . [tex]\frac{ka}{2}[/tex] . a² = 0
σb = -3K σb = [tex]\frac{ka}{2}[/tex]
Qtot = 0
Hence, the total volume of bound charge is zero.
Consider a spring with a force constant of 965 N/m. Suppose the mass of the block is 1.70 kg, but its initial speed can be varied. What initial speed is required to give a maximum spring compression of 3.70 cm.
Answer: 0.88 m/s
Explanation:
given
Force constant of the spring, k = 965 N/m
Compression of the spring, x = 3.7 cm = 0.037m
Mass of the block, m = 1.7 kg
To solve this, we would be using law of conservation of energy, where initial energy is all kinetic and elastic energy is elastic but potential.
E(f) = E(i)
1/2kx² = 1/2mv²
kx² = mv²
965 * 0.037² = 1.7 * v²
1.321 = 1.7 v²
v² = 1.321 / 1.7
v² = 0.777
v = 0.88 m/s
Thus, the initial speed of the block is v = 0.88 m/s
A projectile proton with a speed of 1100 m/s collides elastically with a target proton initially at rest. The two protons then move along perpendicular paths, with the projectile path at 45° from the original direction. After the collision, what are the speeds of (a) the target proton and (b) the projectile proton?
Answer:
So, the target proton's speed is 777.82 m/s
And, the projectile proton's speed is 777.82 m/s
Explanation:
as per the system, it conserves the linear momentum,
so along x axis :
Mp V1 (i) = Mp V1 (f) cos θ1 + Mp V2 (f) cos θ2
along y axis :
0 = -Mp V1 (f) sin θ1 + Mp V2 (f) sin θ2
let us assume before collision it was moving on positive x axis, hence target angle will be θ2 = 45° from x axis
V2(f) = V1 (i) sin θ1 / ( cosθ2 sin θ1 + cos θ1 sin θ2)
= 1100 * sin 45 / ( cos 45 sin 45 + cos 45 sin 45 )
= 1100 * 0.7071 /( 0.7071 * 0.7071 +0 .7071 * 0.7071 )
= 777.82 /( 0.5 + 0.5)
= 777.82 m/s
(b)
the speed of projectile , V1 (f) = sinθ2 * V2(f) / sinθ1
= sin 45 * 777.82 / sin 45
= 777.82 m/s
So, the target proton's speed is 777.82 m/s
And, the projectile proton's speed is 777.82 m/s
An object with a mass of 3 kg has a force of 9 newtons applied to it. What is the resulting acceleration of the object?
Answer: 3 m/s²
Explanation:
m = 3 kg
F = 9 N
a = ?
F = m × a
9 = 3 × a
[tex]\frac{9}{3}[/tex] = a
a = 3 m/s²
An air-plane has an effective wing surface area of 17.0 m² that is generating the lift force. In level flight the air speed over the top of the wings is 66.0 m/s, while the air speed beneath the wings is 40.0 m/s. What is the weight of the plane?(The density of air is 1.29 kg/m³)
Answer:
Explanation:
Given that,
Surface area A= 17m²
The speed at the top v" = 66m/s
Speed beneath is v' =40 m/s
The density of air p =1.29kg/m³
Weight of plane?
Assuming that,
the height difference between the top and bottom of the wind is negligible and we can ignore any change in gravitational potential energy of the fluid.
Using Bernoulli equation
P'+ ½pv'²+ pgh' = P'' + ½pv''² + pgh''
Where
P' is pressure at the bottom in N/m²
P" is pressure at the top in N/m²
v' is velocity at the bottom in m/s
v" is velocity at the top in m/s
Then, Bernoulli equation becomes
P'+ ½pv'² = P'' + ½pv''²
Rearranging
P' — P'' = ½pv"² —½pv'²
P'—P" = ½p ( v"² —v'²)
P'—P" = ½ × 1.29 × (66²-40²)
P'—P" = 1777.62 N/m²
Lift force can be found from
Pressure = force/Area
Force = ∆P ×A
Force = (P' —P")×A
Since we already have (P'—P")
Then, F=W = (P' —P")×A
W = 1777.62 × 17
W = 30,219.54 N
The weight of the plane is 30.22 KN
Answer:
Weight of plane ; W = 30219.54 N
Explanation:
For us to determine the lift force of the system, let's multiply the pressure difference with the effective wing surface area given that the area is obtained by Bernoulli equation. Thus,
P_b + (1/2)ρ(v_b)² + ρg(y_b) = P_t + (1/2)ρ(v_t)² + ρg(y_t)
Now, since the flight is level, the height is constant.
Thus, (y_b) = (y_t)
So, we now have;
P_b + (1/2)ρ(v_b)² = P_t + (1/2)ρ(v_t)²
Rearranging, we have ;
P_b - P_t = (1/2)ρ(v_t)² - (1/2)ρ(v_b)²
P_b - P_t = (1/2)ρ[(v_t)² - (v_b)²]
Now, weight is given by the formula;
W = (P_b - P_t) •A
Thus,
W = (1/2)ρ[(v_t)² - (v_b)²] •A
From the question,
Density; ρ = 1.29 kg/m³
Velocity over top of wings; v_t = 66 m/s
Velocity beneath the wings; v_b = 40 m/s
Surface Area; A = 17 m²
Thus;
W = (1/2)1.29[(66)² - (40)²] •17
W = (1/2)•1.29•17[2756]
W = 30219.54 N
In fission processes, which of the following statements is true? The total number of protons and the total number of neutrons both remain the same. The total number of protons and the total number of mass nuclei both remain the same. Only the total number of protons remains the same. Only the total number of mass nuclei remains the same. Only the total number of neutrons remains the same.
Answer: A. The total number of protons and the total number of neutrons both remain the same.
Explanation:
Nuclear fission is the process in which a large nucleus splits into two smaller nuclei with the release of energy. In other words, fission is the process in which a nucleus is divided into two or more fragments, and neutrons and energy are released.
In fission processes, the total number of protons and neutrons both remain the same because nucleons are neither created nor destroyed, they just rearrange into new nuclei.
Explanation:In fission processes, it is true that the overall number of protons and the total number of neutrons remains the same after the event. This is because nucleons, which include both protons and neutrons, are neither created nor destroyed during fission; they simply rearrange themselves into new nuclei. The mass number, which is the sum of protons and neutrons, is preserved, and though there can be a conversion between protons and neutrons (e.g., via beta decay), the total count of nucleons stays constant. However, the total mass of the product nuclei is less than the mass of the reactants due to the release of nuclear energy as a result of the conversion of mass to energy according to Einstein's equation E=mc2.
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A 0.065 kg ingot of metal is heated to 210◦C and then is dropped into a beaker containing 0.377 kg of water initially at 26◦C. If the final equilibrium state of the mixed system is 28.4 ◦C, find the specific heat of the metal. The specific heat of water is 4186 J/kg · ◦ C. Answer in units of J/kg · ◦ C.
Answer:
Explanation:
Given that,
Metal of mass
M = 0.065kg
Initial temperature of metal
θm = 210°C
The metal is drop into a beaker which contain liquid of mass
m = 0.377 kg
Initial temperature of water
θw = 26°C
The final mixture temperature is
θf = 28.14°C
Specific heat capacity of water
Cw = 4186 J/kg°C
Since the metal is hotter than the water, then the metal will lose heat, while the water will gain heat, we assume that no heat is loss by the beaker.
So,
Heat Loss = Heat gain
Now, heat loss by metal
H(loss) = M•Cm•∆θ
Where M is mass of meta
Cm is specific capacity of metal, which we are looking fro
So,
H(loss) = 0.065 × Cm × (θi - θf)
H(loss) = 0.065 × Cm × (210-28.14)
H(loss) = 11.821 •Cm
Now, Heat gain by water
H(gain) = m•Cw•∆θ
H(gain) = m•Cw•(θf - θi)
Where
m is mass of water and Cw is specific heat capacity of water
H(gain) = 0.377 ×4286 × (28.14-26)
H(gain) = 3457.86
So, H(loss) = Heat(gain)
11.821 •Cm = 3457.86
Cm = 3457.86/11.821
Cm = 292.52 J/Kg°C
The specific heat capacity of the metal ball is 292.52 J/Kg°C
Answer:
320.86J/kgC
Explanation:
To find the specific heat of the substance you take into account that the heat lost by the metal is gained by the water, that is:
[tex]Q_1=-Q_2[/tex]
Furthermore the heat is given by:
[tex]Q_1=m_1c(T_1-T)\\\\Q_2=m_2c(T_2-T)[/tex]
m1: mass of the metal
m2: mass of the water
c: specific heat
T: equilibrium temperature
T1: temperature of the metal
T2: temperature of water
By replacing all these values you can calculate c of the metal:
[tex]m_1c_1(T_1-T)=-m_2c_2(T_2-T)\\\\c_1(0.065kg)(210-28.4)\°C=-(0.377kg)(4186J/kg\°C)(26-28.4)\°C\\\\c_1=320.86\frac{J}{kg\°C}[/tex]
Hence, the specific heat of the metal is 320.86J/kgC
A disk is initially at motionless, then a motor is turned on and the disk begins to spin. If second the angular acceleration of the disk is a constant 3.5 r a d / s 2 , how many complete revolutions does the disk make in 9 seconds?
The number of revolutions is 301
Explanation:
Angular acceleration, α = 3.5 rad/s
Time, t = 9s
Number of revolutions, n = ?
We know,
[tex]\alpha = \frac{w}{t}[/tex]
where,
ω = angular velocity
t = time
It can also be written as:
[tex]\alpha = \frac{\pi n}{t X 30}[/tex]
On substituting the value, we get:
[tex]3.5 = \frac{\pi X n}{9 X 30} \\\\n = 300.8[/tex]
Therefore, the number of revolutions is 301
The answer is 22 complete revolutions.
The number of complete revolutions the disk makes in 9 seconds is given by the formula for angular displacement due to constant angular acceleration:
[tex]\[ \theta = \theta_0 + \omega_0 t + \frac{1}{2} \alpha t^2 \][/tex]
where:
-[tex]\( \theta \)[/tex] is the final angular displacement in radians,
-[tex]\( \theta_0 \)[/tex] is the initial angular displacement, which is zero since the disk starts from rest,
-[tex]\( \omega_0 \)[/tex] is the initial angular velocity, which is also zero since the disk starts from rest,
- [tex]\( \alpha \)[/tex] is the angular acceleration, which is 3.5 [tex]rad/s\( ^2 \)[/tex],
- [tex]\( t \)[/tex] is the time in seconds, which is 9 seconds.
Since the disk starts from rest, [tex]\( \theta_0 = 0 \)[/tex] and [tex]\( \omega_0 = 0 \)[/tex], the equation simplifies to:
[tex]\[ \theta = \frac{1}{2} \alpha t^2 \][/tex]
Substituting the given values:
[tex]\[ \theta = \frac{1}{2} \times 3.5 \, \text{rad/s}^2 \times (9 \, \text{s})^2 \][/tex]
[tex]\[ \theta = \frac{1}{2} \times 3.5 \times 81 \][/tex]
[tex]\[ \theta = 1.75 \times 81 \][/tex]
\[ \theta = 141.75 \, \text{radians} \]
To find the number of complete revolutions, we need to convert the angular displacement from radians to revolutions. Since there are [tex]\( 2\pi \)[/tex] radians in one revolution, we divide the angular displacement by [tex]\( 2\pi \)[/tex]:
[tex]\[ \text{Number of revolutions} = \frac{\theta}{2\pi} \][/tex]
[tex]\[ \text{Number of revolutions} = \frac{141.75}{2\pi} \][/tex]
[tex]\[ \text{Number of revolutions} = \frac{141.75}{6.2831853071796} \][/tex]
[tex]\[ \text{Number of revolutions} \approx 22.57 \][/tex]
Mopeds (small motorcycles with an engine capacity below 50 cm3) are very popular in Europe because of their mobility, ease of operation, and low cost. Suppose the maximum speed of a moped is normally distributed with mean value 46.6 km/h and standard deviation 1.75 km/h. Consider randomly selecting a single such moped.
Answer:
a) P ( X ≤ 49 ) = 0.9149
b) P ( X ≥ 48 ) = 0.2119
c) 0.9876
Explanation:
Solution:-
- Lets define a random variable X: the maximum speed of a moped.
- The random variable follows normal distribution with the following parameters:
X ~ Norm ( u , σ^2 )
Where,
u = Mean = 46.6 km/h
σ = Standard deviation 1.75 km/h
Hence,
X ~ Norm ( 46.6 , 1.75^2 ).
a) The probability that the maximum speed of mopeds is atmost 49 km/h?
- To evaluate the probability of P ( X ≤ 49 ). We will find the standard Z-score value using the following formula:
[tex]Z-score = \frac{x-u}{s} \\\\ Z-score = \frac{49-46.6}{1.75} \\\\Z-score = 1.37142[/tex]
- Now use the standard normal tables to determine the required probability of:
P ( Z < 1.37142 ) = 0.9149
Hence,
P ( X ≤ 49 ) = 0.9149
b) The probability that the maximum speed of mopeds is at-least 48 km/h?
- To evaluate the probability of P ( X ≥ 48 ). We will find the standard Z-score value using the following formula:
[tex]Z-score = \frac{x-u}{s} \\\\ Z-score = \frac{48-46.6}{1.75} \\\\Z-score = 0.8[/tex]
- Now use the standard normal tables to determine the required probability of:
P ( Z ≥ 0.8 ) = 0.2119
Hence,
P ( X ≥ 48 ) = 0.2119
c) The probability that the maximum speed of mopeds differs from mean by at-most 2.5 standard deviation.
- The required probability is the standard error from the mean value "u" of 2.5 standard deviation.
- We don't need to evaluate the test statistics as we are already given the standard error about mean.
- Using, the standard normal Z-score: The required probability is:
P ( -2.5 < Z < 2.5 ) = 2*P ( Z < 2.5 ) - 1
= 2*0.9938 - 1
= 0.9876
Why does the solar nebula theory imply planets are common? a. Planets formed from the spiral arms of matter encountered around the Sun. Most main-sequence stars pass spiral arms dozens of times in their lives. b. Planets formed from the Sun’s past binary partner star. Binary systems with different masses and evolution rates are common. c. Planets formed from the envelope of the Sun, ejected at the time of hydrogen fusion ignition. Stars commonly eject their envelopes. d. Planets formed from the disk of gas and dust that surrounded the Sun, and such disks are common around young stars.
Answer:
d. Planets formed from the disk of gas and dust that surrounded the Sun, and such disks are common around young stars.
Explanation:
A star passes through various stages before it becomes a full fledged star with its own planetary system. The same is with Sun as well. It was born out of a Nebula. Nebula is a cloud of dust and gases. The dust and gases start accumulating to form what we call as a protostar.
A lot of material at this stage is thrown out from the young star, this material forms a disk around the star. This disk is known as proto-planetary disk. The gases and dust of this disk then coalesce together to make planets and other objects of the star system.
We have observed such disks in the Orion nebula.
The solar nebula theory suggests that planets are formed from the disk of gas and dust that surrounds young stars. Such disks are commonly observed around stars, leading to the inference that planet formation is a universal occurrence.
Explanation:The solar nebula theory, or nebular hypothesis, explains the formation of our solar system from a nebular cloud of gas and dust. According to this theory, planets are formed from the disk of gas and dust that surrounds a young star, also known as protoplanetary disk. This leads to the implication that planets are common because such disks are commonly found around young stars.
When a star condenses from a nebula, it forms a hot, spinning disk of gas and dust around it. Over time, particles within this disk begin to collide and stick together, gradually forming planetesimals and eventually, planets. Since we observe many stars surrounded by these disks, we can infer that planet formation is a common process in the universe.
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