Answer:
Explanation:
Using the formula for finding the Z score
Z = x-µ/σ
x is the sample size
µ is the sample mean
σ is the standard deviation
For percentage of the resistors will have resistance between 148 Ω and 152 Ω, or is calculated as shown
P(148≤x152) = Z(152-150/5) - Z(148-150/5)
P(148≤x152) = Z(0.4)-Z( - 0.4)
P(148≤x152) = 0.6554-0.3446
The Z values are from the normal distribution table.
P(148≤x152) = 0.3108
The percentage of resistor that will have between 148 and 152 ohms is 0.3108×100% = 31.08%
Similarly for resistances between 140 Ω and 160 Ω
P(140≤x160) = Z(160-150/5) - Z(140-150/5)
P(140≤x160) = Z(2.0)-Z( - 2.0)
P(140≤x160) = 0.9775-0.02275
The Z values are from the normal distribution table.
P(140≤x160) = 0.9547
The percentage of resistor that will have between 140 and 160ohms is 0.9547×100% = 95.47%
The percentage of the resistors will have resistance between 148 Ω and 152 Ω is 31.08%
The percentage of the resistors will have resistance between 140 Ω and 160 Ω is 95.47%
Percentage of resistors:To solve for the probability we will use the standard score of the Z score, which is given by:
Z = (x - µ)/σ
where x is the sample size
µ is the sample mean = 150Ω
σ is the standard deviation = 5Ω
The probability of the resistors with resistance between 148 Ω and 152 Ω, will be:
P(148 ≤ 152) = Z((152-150)/5) - Z((148-150)/5)
P(148 ≤ 152) = Z(0.4)-Z( - 0.4)
P(148 ≤ 152) = 0.6554-0.3446
P(148 ≤ 152) = 0.3108
So, the percentage will be:
0.3108×100% = 31.08%
Similarly for resistances between 140 Ω and 160 Ω
P(140 ≤ 160) = Z((160-150)/5) - Z((140-150)/5)
P(140 ≤ 160) = Z(2.0)-Z( - 2.0)
P(140 ≤ 160) = 0.9775-0.02275
P(140 ≤ 160) = 0.9547
The percentage of resistor that will have between 140 and 160ohms is 0.9547×100% = 95.47%
Learn more about standard score:
https://brainly.com/question/25875323?referrer=searchResults
Which scientist proved experimentally that a shadow of the circular object illuminated 18. with coherent light would have a central bright spot?
A. Young
B. Fresnel
C. Poisson
D. Arago
Answer:
Your answer is( D) - Arago
5.
Find the equation of the circle tangential to the line 3x-4y+1=0 and with
centre at (4,7).
20
Answer: (x - 4)² + (y - 7)² = 9
Explanation:
The equation of a circle is: (x - h)² + (y - k)² = r² where
(h, k) is the centerr is the radiusGiven: (h, k) = (4, 7)
Find the intersection of the given equation and the perpendicular passing through (4, 7).
3x - 4y = -1
-4y = -3x - 1
[tex]y=\dfrac{3}{4}x-1[/tex]
[tex]m=\dfrac{3}{4}[/tex] --> [tex]m_{\perp}=-\dfrac{4}{3}[/tex]
[tex]y-y_1=m_{\perp}(x-x_1)\\\\y-7=-\dfrac{4}{3}(x-4)\\\\\\y=-\dfrac{4}{3}x+\dfrac{16}{3}+7\\\\\\y=-\dfrac{4}{3}x+\dfrac{37}{3}[/tex]
Use substitution to find the point of intersection:
[tex]x=\dfrac{29}{5}=5.8,\qquad y=\dfrac{23}{5}=4.6[/tex]
Use the distance formula to find the distance from (4, 7) to (5.8, 4.6) = radius
[tex]r=\sqrt{(5.8-4)^2+(4.6-7)^2}\\\\r=\sqrt{3.24+5.76}\\\\r=\sqrt9\\\\r=3[/tex]
Input h = 4, k = 7, and r = 3 into the circle equation:
(x - 4)² + (y - 7)² = 3²
(x - 4)² + (y - 7)² = 9
What is the inductance of a coil if the coil produces an emf of 2.40 V when the current in it changes from -27.0 mA to 33.0 mA in 11.0 ms
Answer:
Inductance of a coil(L) = 0.44 H (Approx)
Explanation:
Given:
coil produces emf = 2.40 V
Old current = -27 mA
New current = 33 mA
Time taken = 11 mS
Find:
Inductance of a coil(L)
Computation:
Inductance of a coil(L) = -emf / [Δi / Δt]
Inductance of a coil(L) = -2.4 / [(-33 - 27) / 11]
Inductance of a coil(L) = -2.4 / [-5.4545]
Inductance of a coil(L) = 0.44 H (Approx)
•• A metal sphere carrying an evenly distributed charge will have spherical equipotential surfaces surrounding it. Suppose the sphere’s radius is 50.0 cm and it carries a total charge of (a) Calculate the potential of the sphere’s surface. (b)You want to draw equipotential surfaces at intervals of 500 V outside the sphere’s surface. Calculate the distance between the first and the second equipotential surfaces, and between the 20th and 21st equipotential surfaces. (c) What does the changing spacing of the surfaces tell you about the electric field?
Answer:
Explanation:
For this exercise we will use that the potential is created by the charge inside the equinoctial surface and just like in Gauss's law we can consider all the charge concentrated in the center.
Therefore the potential on the ferric surface is
V = k Q / r
where k is the Coulomb constant, Q the charge of the sphere and r the distance from the center to the point of interest
a) On the surface the potential
V = 9 10⁹ Q / 0.5
V = 18 10⁹ Q
Unfortunately you did not write the value of the load, suppose a value to complete the calculations Q = 1 10⁻⁷ C, with this value the potential on the surfaces V = 1800 V
b) The equipotential surfaces are concentric spheres, let's look for the radii for some potentials
for V = 1300V let's find the radius
r = k Q / V
r = 9 109 1 10-7 / 1300
r = 0.69 m
other values are shown in the following table
V (V) r (m)
1800 0.5
1300 0.69
800 1,125
300 3.0
In other words, we draw concentric spheres with these radii and each one has a potential difference of 500V
C) The spacing of the spheres corresponds to lines of radii of the electric field that have the shape
E = k Q / r²
A proton that is initially at rest is accelerated through an electric potential difference of magnitude 500 V. What speed does the proton gain? (e = 1.60 × 10-19 C , mproton = 1.67 × 10-27 kg)
Answer:
[tex]3.1\times 10^{5}m/s[/tex]
Explanation:
The computation of the speed does the proton gain is shown below:
The potential difference is the difference that reflects the work done as per the unit charged
So, the work done should be
= Potential difference × Charge
Given that
Charge on a proton is
= 1.6 × 10^-19 C
Potential difference = 500 V
[tex]v= \sqrt{\frac{2.q.\Delta V}{m_{p}}} \\\\\\= \sqrt{\frac{2\times 1.6\times 10^{-19}\times 5\times 10^{2}}{1.67\times 10^{-27}}}[/tex]
[tex]v= \sqrt{9.58\times 10^{10}}m/s \\\\= 3.095\times 10^{5}m/s\\\\\approx 3.1\times 10^{5}m/s[/tex]
Simply we applied the above formulas
Consider the uniform electric field E = (8.0ĵ + 2.0 ) ✕ 103 N/C. What is its electric flux (in N · m2/C) through a circular area of radius 9.0 m that lies in the xy-plane? (Enter the magnitude.)
Answer:
5.09 x 10⁵ Nm²/C
Explanation:
The electric flux φ through a planar area is defined as the electric field Ε times the component of the area Α perpendicular to the field. i.e
φ = E A
From the question;
E = (8.0j + 2.0k) ✕ 10³ N/C
r = radius of the circular area = 9.0m
A = area of a circle = π r² [Take π = 3.142]
A = 3.142 x 9² = 254.502m²
Now, since the area lies in the x-y plane, only the z-component of the electric field is responsible for the electric flux through the circular area.
Therefore;
φ = (2.0) x 10³ x 254.502
φ = 5.09 x 10⁵ Nm²/C
The electric flux is 5.09 x 10⁵ Nm²/C
what effect does condensation on a glass of ice water have on the rate at which the ice melts? Will the condensation speed up the melting process or slow it down?
Answer:
Explanation:
When water droplet condenses on the outer wall of glass of ice , it releases heat equal to mass x latent heat of condensation of water . This heat reaches the ice melting inside glass . Due to this heat , the melting process is accelerated .
Hence the process of melting gets accelerated when water droplet condenses on the outer wall of glass containing mixture of ice and water .
Charge of uniform density (40 pC/m2) is distributed on a spherical surface (radius = 1.0 cm), and a second concentric spherical surface (radius = 3.0 cm) carries a uniform charge density of 60 pC/m2. What is the magnitude of the electric field at a point 4.0 cm from the center of the two surfaces?
Answer:
4.1 N/C
Explanation:
First of all, we know from maths that the surface area of a sphere = 4πr²
Charge on inner sphere ..
Q(i) = 40.0*10^-12C/m² x 4π(0.01m)²
Q(i) = 5.03*10^-14 C
Charge on outer sphere
Q(o) = 60*10^-12 x 4π(0.03m)²
Q(o) = 6.79*10^-13 C
Inner sphere has a - 5.03*10^-14C charge (-Qi) on inside of the outer shell. As a result, there is a net zero charge within the outer shell.
For the outer shell to show a NET charge +6.79*10^-13C, it's must have a +ve charge
= +6.79*10^-13C + (+5.03*10^-14C)
= +7.29*10^-13 C
Now again, we have
E = kQ /r²
E = (9.0*10^9)(+7.29*10^-13 C) / (0.04)²
E = 6.561*10^-3 / 1.6*10^-3
E = 4.10 N/C
Thus, the magnitude of the electric field is 4.1 N/C
A tiger leaps horizontally out of a tree that is 3.30 m high. He lands 5.30 m from the base of the tree. (Neglect any effects due to air resistance.)
Calculate the initial speed. (Express your answer to three significant figures.)
m/s Submit
Answer:
The initial velocity is [tex]v_h = 8.66 \ m/s[/tex]
Explanation:
From the question we are told that
The height of the tree is [tex]h = 3.30\ m[/tex]
The distance of the position of landing from base is [tex]d = 5.30 \ m[/tex]
According to the second equation of motion
[tex]h = u_o * t + \frac{1}{2} at^2[/tex]
[tex]Where\ u_o[/tex] is the initial velocity in the vertical axis
a is equivalent to acceleration due to gravity which is positive because the tiger is downward
So
[tex]3 = 0 + 0.5 * 9.8 *t^2[/tex]
=> [tex]t = \frac{3 }{9.8 * 0.5}[/tex]
[tex]t = 0.6122\ s[/tex]
Now the initial velocity in the horizontal direction is mathematically evaluated as
[tex]v_h = \frac{5.30}{0.6122}[/tex]
[tex]v_h = 8.66 \ m/s[/tex]
Estimate the distance (in cm) between the central bright region and the third dark fringe on a screen 6.3 m from two double slits 0.49 mm apart illuminated by 739-nm light. (give answer in millimeters)
Answer:
Explanation:
distance of third dark fringe
= 2.5 x λ D / d
where λ is wavelength of light , D is screen distance and d is slit separation
putting the given values
required distance = 2.5 x 739 x 10⁻⁹ x 6.3 / .49 x 10⁻³
= 23753.57 x 10⁻⁶
= 23.754 x 10⁻³ m
= 23.754 mm .
Suppose a space vehicle with a rest mass of 150 000 kg travels past the International Space Station at a constant speed of 2.6 x 108 m/s with respect to the I.S.S. When an observer on the I.S.S. measures the moving vehicle, her measurement of the space vehicle length is 25.0 m. Determine the relativistic mass of the space vehicle. Determine the length of the space vehicle as measured by an astronaut on the space vehicle.
Answer:
m = 300668.9 kg
L₀ = 12.47 m
Explanation:
The relativistic mass of the space vehicle is given by the following formula:
[tex]m = \frac{m_{0}}{\sqrt{1-\frac{v^{2} }{c^{2}} } }[/tex]
where,
m = relativistic mass = ?
m₀ = rest mass = 150000 kg
v = relative speed = 2.6 x 10⁸ m/s
c = speed of light = 3 x 10⁸ m/s
Therefore
[tex]m = \frac{150000kg}{\sqrt{1-\frac{(2.6 x 10^{8}m/s)^{2} }{(3 x 10^{8}m/s)^{2}} } }[/tex]
m = 300668.9 kg
Now, for rest length of vehicle:
L = L₀√(1 - v²/c²)
where,
L = Relative Length of Vehicle = 25 m
L₀ = Rest Length of Vehicle = ?
Therefore,
25 m = L₀√[1 - (2.6 x 10⁸ m/s)²/(3 x 10⁸ m/s)²]
L₀ = (25 m)(0.499)
L₀ = 12.47 m
Find the absolute value of the change of the gravitational potential energy (GPE) of the puck-Earth system from the moment the puck begins to move to the moment it hits the spring. Use 0.253 m for the displacement of the puck along the ramp and 9.80 m/s2 for the acceleration due to gravity. Assume that the mass of the puck is 0.180 kg. Express your answer using SI units to three significant figures.
Answer:
0.16joules
Explanation:
Using the relation for The gravitational potential energy
E= Mgh
Where,
E= Potential energy
h = Vertical Height
M = mass
g = Gravitational Field Strength
To find the vertical component of angle of launch Where the angle is 22°
h= sin theta
So E = mghsintheta
= 0.18 x 0.98 x 0.253 sin22
=0.16joules
Explanation:
A 5.00-kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical oscillations having a period of 1.60 s. Find the force constant of the spring.
Answer:A7.50kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical oscillations having a period of2.30s. Find the force constant of the spring.
N/m
Explanation:
A −3.0 nC charge is on the x-axis at x=−9 cm and a +4.0 nC charge is on the x-axis at x=16 cm. At what point or points on the y-axis is the electric potential zero?
Answer:
y = 10.2 m
Explanation:
It is given that,
Charge, [tex]q_1=-3\ nC[/tex]
It is placed at a distance of 9 cm at x axis
Charge, [tex]q_2=+4\ nC[/tex]
It is placed at a distance of 16 cm at x axis
We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,
[tex]\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0[/tex]
Here,
[tex]r_1=\sqrt{y^2+9^2} \\\\r_2=\sqrt{y^2+15^2}[/tex]
So,
[tex]\dfrac{kq_1}{r_1}=-\dfrac{kq_2}{r_2}\\\\\dfrac{q_1}{r_1}=-\dfrac{q_2}{r_2}\\\\\dfrac{-3\ nC}{\sqrt{y^2+81} }=-\dfrac{4\ nC}{\sqrt{y^2+225} }\\\\3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}[/tex]
Squaring both sides,
[tex]3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}\\\\9(y^2+225)=16\times (y^2+81)\\\\9y^2+2025=16y^2-+1296\\\\2025-1296=7y^2\\\\7y^2=729\\\\y=10.2\ m[/tex]
So, at a distance of 10.2 m on the y axis the electric potential equals 0.
According to the question,
Charge,
[tex]q_1 = -3 \ nC[/tex] (9 cm at x-axis)[tex]q_2 = +4 \ nC[/tex] (16 cm at x-axis)Now,
→ [tex]\frac{kq_1}{r_1} +\frac{kq_2}{r_2} =0[/tex]
or,
→ [tex]\frac{kq_1}{r_1} =-\frac{kq_2}{r_2}[/tex]
→ [tex]\frac{q_1}{r_1} = \frac{q_2}{r_2}[/tex]
here,
[tex]r_1 = \sqrt{y^2+81}[/tex]
[tex]r^2 = \sqrt{y^2+225}[/tex]
By substituting the values,
→ [tex]\frac{-3 }{\sqrt{y^2+225} } = -\frac{4}{\sqrt{y^2+225} }[/tex]
By applying cross-multiplication,
[tex]3\times \sqrt{y^2+225} = 4\times \sqrt{y^2+81}[/tex]
By squaring both sides, we get
→ [tex]9(y^2+225) = 16(y^2+81)[/tex]
[tex]9y^2+2025 = 16 y^2+1296[/tex]
[tex]2025-1296=7y^2[/tex]
[tex]7y^2=729[/tex]
[tex]y = 10.2 \ m[/tex]
Thus the solution above is correct.
Learn more about charge here:
https://brainly.com/question/12437696
A trash compactor can compress its contents to 0.350 times their original volume and 4 times denser than their original density. Neglecting the mass of air expelled, what factor is the old density of the rubbish
Answer:
2.8
Explanation:
Using p = m/v; (old density)
p' = m/v (new density)
=m/0.350 V
p'/p = (m/0.350V)/(m/v) = 1/0.350 = 2.86
16. In single-slit diffraction, the central band gets thicker as the distance to the screen increases. True False
Answer:
the right answer is true
Answer:
True
Explanation:
In a shipping yard, a crane operator attaches a cable to a 1,390 kg shipping container and then uses the crane to lift the container vertically at a constant velocity for a distance of 33 m. Determine the amount of work done (in J) by each of the following.
a) the tension in the cable.
b) the force of gravity.
Answer:
a) A = 449526 J, b) 449526 J
Explanation:
In this exercise they do not ask for the work of different elements.
Note that as the box rises at constant speed, the sum of forces is chorus, therefore
T-W = 0
T = W
T = m g
T = 1,390 9.8
T = 13622 N
Now that we have the strength we can use the definition of work
W = F .d
W = f d cos tea
a) In this case the tension is vertical and the movement is vertical, so the tension and displacement are parallel
A = A x
A = 13622 33
A = 449526 J
b) The work of the force of gravity, as the force acts in the opposite direction, the angle tea = 180
W = T x cos 180
W = - 13622 33
W = - 449526 J
From mechanics, you may recall that when the acceleration of an object is proportional to its coordinate, d2xdt2=−kmx=−ω2x , such motion is called simple harmonic motion, and the coordinate depends on time as x(t)=Acos(ωt+ϕ), where ϕ, the argument of the harmonic function at t=0, is called the phase constant. Find a similar expression for the charge q(t) on the capacitor in this circuit. Do not forget to determine the correct value of ϕ based on the initial conditions described in the problem. Express your answer in terms of q0 , L, and C. Use the cosine function in your answer.
Answer:
q = q₀ sin (wt)
Explanation:
In your statement it is not clear the type of circuit you are referring to, there are two possibilities.
1) The circuit of this problem is a system formed by an Ac voltage source and a capacitor, in this case all the voltage of the source is equal to the voltage at the terminals of the capacitor
ΔV = Δ[tex]V_{C}[/tex]
we assume that the source has a voltage of the form
ΔV = ΔV₀o sin wt
The capacitance of a capacitor is
C = q / ΔV
q = C ΔV sin wt
the current in the circuit is
i = dq / dt
i = c ΔV₀ w cos wt
if we use
cos wt = sin (wt + π / 2)
we make this change by being a resonant oscillation
we substitute
i = w C ΔV₀ sin (wt + π/2)
With this answer we see that the current in capacitor has a phase factor of π/2 with respect to the current
2) Another possible circuit is an LC circuit.
In this case the voltage alternates between the inductor and the capacitor
V_{L} + V_{C} = 0
L di / dt + q / C = 0
the current is
i = dq / dt
they ask us for a solution so that
L d²q / dt² + 1 / C q = 0
d²q / dt² + 1 / LC q = 0
this is a quadratic differential equation with solution of the form
q = A sin (wt + Ф)
to find the constant we derive the proposed solution and enter it into the equation
di / dt = Aw cos (wt + Ф)
d²i / dt²= - A w² sin (wt + Ф)
- A w² + 1 /LC A = 0
w = √ (1 / LC)
To find the phase factor, for this we use the initial conditions for t = 0
in the case of condensate for t = or the charge is zero
0 = A sin Ф
Ф = 0
q = q₀ sin (wt)
A piece of thin uniform wire of mass m and length 3b is bent into an equilateral triangle so that each side has a length of b. Find the moment of inertia of the wire triangle about an axis perpendicular to the plane of the triangle and passing through one of its vertices.
Answer:
Mb²/2
Explanation:
Pls see attached file
If the speed of a "cheetah" is 150 m / s. How long does it take to cover 800 m?
Answer:
5.33333... seconds
Explanation:
800 divided by 150 is equal to 5.33333... because it is per second that the cheetah moves at 150miles, the answer is 5.3333.....
A certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of What is the diameter of the wire in the fuse?
Answer:
0.45 mm
Explanation:
The complete question is
a certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of 620 A/ cm^2. What is the diameter of the wire in the fuse?
A) 0.45 mm
B) 0.63 mm
C.) 0.68 mm
D) 0.91 mm
Current in the fuse is 1.0 A
Current density of the fuse when it melts is 620 A/cm^2
Area of the wire in the fuse = I/ρ
Where I is the current through the fuse
ρ is the current density of the fuse
Area = 1/620 = 1.613 x 10^-3 cm^2
We know that 10000 cm^2 = 1 m^2, therefore,
1.613 x 10^-3 cm^2 = 1.613 x 10^-7 m^2
Recall that this area of this wire is gotten as
A = [tex]\frac{\pi d^{2} }{4}[/tex]
where d is the diameter of the wire
1.613 x 10^-7 = [tex]\frac{3.142* d^{2} }{4}[/tex]
6.448 x 10^-7 = 3.142 x [tex]d^{2}[/tex]
[tex]d^{2}[/tex] =[tex]\sqrt{ 2.05*10^-7}[/tex]
d = 4.5 x 10^-4 m = 0.45 mm
The value of the diameter of the wire is 0.45 mm.
Given that:
Current in the fuse = 1 Ampere
Current density = 620 A/cm²
Area of the wire = [tex]\dfrac{\text I}{\rho}[/tex]
Area = [tex]\dfrac{1}{620}[/tex]
Area = 1.613 x 10⁻³ cm²
Also, we know that:
10000 cm² = 1 m²
1.613 x 10⁻³ cm² = 1.613 x 10⁻⁷ m²
The area of a wire can be calculated as:
Area = [tex]\dfrac{\pi \text d^2}{4}[/tex]
where, d = diameter
Substituting the values in the above equation, we get:
1.613 x 10⁻⁷ = [tex]\dfrac{3.14 \times \text d^2}{4}[/tex]
6.448 x 10⁻⁷ = 3.142 x d²
Hence, the value of d will be:
d² = 2.05 x 10⁻⁴
d = [tex]\sqrt{2.05 \times 10^{-7}}[/tex]
d = 4.5 x 10⁻⁷
Thus, the value of the diameter of the wire is 0.45 mm.
To know more about current, refer to the following link:
https://brainly.com/question/2285102
Light emitted by element X passes through a diffraction grating that has 1200 slits/mm. The interference pattern is observed on a screen 77.0 cm behind the grating. First-order maxima are observed at distances of 58.0 cm , 65.4 cm , and 94.5 cm from the central maximum. What are the wavelengths of light emitted by element X?
Answer:
500 nm, 530 nm, 650 nm
Explanation:
Let's say that there is diffraction grating observed with a slit spacing of s. Respectively we must determine the angle θ which will help us determine the 3 wavelengths ( λ ) of the light emitted by element X. This can be done applying the following formulas,
s( sin θ ) = m [tex]*[/tex] λ, such that y = L( tan θ ) - where y = positioning, or the distance of the first - order maxima, and L = constant, of 77 cm
Now the grating has a slit spacing of -
s = 1 / N = 1 / 1200 = 0.833 [tex]*[/tex] 10⁻³ mm
The diffraction angles of the " positionings " should thus be -
θ = tan⁻¹ [tex]*[/tex] ( 0.58 / 0.77 ) = 37°,
θ = tan⁻¹ [tex]*[/tex] ( 0.654 / 0.77 ) = 40°,
θ = tan⁻¹ [tex]*[/tex] ( 0.945 / 0.77 ) = 51°
The wavelengths of these three bright fringes should thus be calculated through the formula : λ = s( sin θ ) -
λ = 0.833 [tex]*[/tex] 10⁻³ [tex]*[/tex] sin( 37° ) = ( 500 [tex]*[/tex] 10⁻⁹ m )
λ = 0.833 [tex]*[/tex] 10⁻³ [tex]*[/tex] sin( 40° ) = ( 530 [tex]*[/tex] 10⁻⁹ m )
λ = 0.833 [tex]*[/tex] 10⁻³ [tex]*[/tex] sin( 51° ) = ( 650 [tex]*[/tex] 10⁻⁹ m )
Wavelengths : 500 nm, 530 nm, 650 nm
This question will be solved using the "grating equation".
The wavelengths of the light emitted by element X are:
"1. 6.654 x 10⁻⁷ m = 665.4 nm
2. 6.349 x 10⁻⁷ m = 634.9 nm
3. 5.262 x 10⁻⁷ m = 526.2 nm"
The diffraction grating equation is given as follows:
[tex]m\lambda = d Sin\ \theta[/tex]
where,
m = order of maxima = 1
λ = wavelength of light = ?
d = grating element = [tex]\frac{1}{no.\ of\ slits\ per\ unit\ length} = \frac{1}{1200\ slits/mm}[/tex]
d = (8.33 x 10⁻⁴ mm/slit)(1 m/ 1000 mm) = 8.33 x 10⁻⁷ m/slit
θ = angle of diffraction = [tex]tan^{-1}(\frac{L}{y})[/tex]
where,
L = distance of grating from the screen = 77 cm
y = distance of maxima from central maxima
Hence, the general equation after substituting constant values becomes:
[tex]\lambda = (8.33\ x\ 10^{-7}\ m/slits)\ Sin(tan^{-1}(\frac{77\ cm}{y}))[/tex]
FOR y = 58 cm:
[tex]\lambda = (8.33\ x\ 10^{-7}\ m/slits)\ Sin(tan^{-1}(\frac{77\ cm}{58\ cm}))[/tex]
λ = 6.654 x 10⁻⁷ m = 665.4 nm
FOR y = 65.4 cm:
[tex]\lambda = (8.33\ x\ 10^{-7}\ m/slits)\ Sin(tan^{-1}(\frac{77\ cm}{65.4\ cm}))[/tex]
λ = 6.349 x 10⁻⁷ m = 634.9 nm
FOR y = 94.5 cm:
[tex]\lambda = (8.33\ x\ 10^{-7}\ m/slits)\ Sin(tan^{-1}(\frac{77\ cm}{94.5\ cm}))[/tex]
λ = 5.262 x 10⁻⁷ m = 526.2 nm
The attached picture shows the arrangement of the light rays in a diffraction grating.
Learn more about diffraction grating here:
https://brainly.com/question/17012571?referrer=searchResults
A tennis ball is thrown from ground level with velocity v0 directed 30 degrees above the horizontal. If it takes the ball 1.0s to reach the top of its trajectory, what is the magnitude of the initial velocity?
Answer:
vi = 19.6 m/s
Explanation:
Given:
final velocity vf = 0
gravity a = -9.8
time t = 1
Initial velocity vi = vf - at
vi = 0 + 9.8 (1.0)
vi = 9.8 m/s
the y component of velocity is the initial velocity.
therefore v sin 30 = 9.8
vi/2 = 9.8
vi = 19.6 m/s
A tennis ball is thrown from ground level with velocity v0 directed 30 degrees above the horizontal. If it takes the ball 1.0s to reach the top of its trajectory, the magnitude of the initial velocity vi = 19.62 m/s.
Given data to find the initial velocity,
final velocity vf = 0
gravity a = -9.8
time t = 1
What is deceleration?The motion of the tennis ball on the vertical axis is an uniformly accelerated motion, with deceleration of (gravitational acceleration).
The component of the velocity on the y-axis is given by the following law:
Initial velocity vi = vf - at
At the time t=0.5 s, the ball reaches its maximum height, and when this happens, the vertical velocity is zero (because it is a parabolic motion), Substituting into the previous equation, we find the initial value of the vertical component of the velocity:
vi = 0 + 9.8 (1.0)
vi = 9.8 m/s
the y component of velocity is the initial velocity.
However, this is not the final answer. In fact, the ball starts its trajectory with an angle of 30°. This means that the vertical component of the initial velocity is,
therefore, v sin 30° = 9.8
We found before the value of y component, so we can substitute to find the initial speed of the ball:
vi/2 = 9.8
vi = 19.6 m/s
Thus, the initial velocity can be found as 19.62 m/s.
Learn more about deceleration,
https://brainly.com/question/4403243
#SPJ2
Three resistors, 6.0-W, 9.0-W, 15-W, are connected in parallel in a circuit. What is the equivalent resistance of this combination of resistors?
Answer:
2.9Ω
Explanation:
Resistors are said to be in parallel when they are arranged side by side such that their corresponding ends are joined together at two common junctions. The combined resistance in such arrangement of resistors is given by;
1/Req= 1/R1 + 1/R2 + 1/R3 .........+ 1/Rn
Where;
Req refers to the equivalent resistance and R1, R2, R3 .......Rn refers to resistance of individual resistors connected in parallel.
Note that;
R1= 6.0Ω
R2 = 9.0Ω
R3= 15.0 Ω
Therefore;
1/Req = 1/6 + 1/9 + 1/15
1/Req= 0.167 + 0.11 + 0.067
1/Req= 0.344
Req= (0.344)^-1
Req= 2.9Ω
The equivalent resistance of this combination of resistors is 2.9Ω.
Calculation of the equivalent resistance:The combined resistance in such arrangement of resistors is provided by;
1/Req= 1/R1 + 1/R2 + 1/R3 .........+ 1/Rn
here.
Req means the equivalent resistance and R1, R2, R3
.Rn means the resistance of individual resistors interlinked in parallel.
Also,
R1= 6.0Ω
R2 = 9.0Ω
R3= 15.0 Ω
So,
1/Req = 1/6 + 1/9 + 1/15
1/Req= 0.167 + 0.11 + 0.067
1/Req= 0.344
Req= (0.344)^-1
Req= 2.9Ω
learn more about resistance here: https://brainly.com/question/15047345
A square coil of wire with side 8.0 cm and 50 turns sits in a uniform magnetic field that is perpendicular to the plane of the coil. The coil is pulled quickly out of the magnetic field in 0.2 s. If the resistance of the coil is 15 ohm and a current of 12 mA is induced in the coil, calculate the value of the magnetic field.
Answer:
Explanation:
area of the coil A = .08 x .08 = 64 x 10⁻⁴ m ²
flux through the coil Φ = area of coil x no of turns x magnetic field
= 64 x 10⁻⁴ x 50 x B where B is magnetic field
emf induced = dΦ / dt = ( 64 x 10⁻⁴ x 50 x B - 0 ) / .2
= 1.6 B
current induced = emf induced / resistance
12 x 10⁻³ = 1.6 B / 15
B = 112.5 x 10⁻³ T .
Three cars (car 1, car 2, and car 3) are moving with the same velocity when the driver suddenly slams on the brakes, locking the wheels. The most massive car is car 1, the least massive is car 3, and all three cars have identical tires. For which car does friction do the largest amount of work in stopping the car
Answer:
Car 3
Explanation:
An electromagnetic ware has a maximum magnetic field strength of 10^-8 T at a specific place in vacuum. What is the intensity of the light at that place. μ0=4πx10^-7 WbA/m g
Answer:
[tex]I=1.19\times 10^{-2}\ W/m^2[/tex]
Explanation:
It is given that,
Maximum value of magnetic field strength, [tex]B=10^{-8}\ T[/tex]
We need to find the intensity of the light at that place.
The formula of the intensity of magnetic field is given by :
[tex]I=\dfrac{c}{2\mu _o}B^2[/tex]
c is speed of light
So,
[tex]I=\dfrac{3\times 10^8}{2\times 4\pi \times 10^{-7}}\times (10^{-8})^2\\\\I=1.19\times 10^{-2}\ W/m^2[/tex]
So, the intensity of the light is [tex]1.19\times 10^{-2}\ W/m^2[/tex].
Current folw in which dirction
If theta is 30 degrees and there is no friction, what would be the block's acceleration down the incline, in meters per second squared?
Answer:
a= 4.9m/s²
Explanation:
Using Fnet= mgsintheta = ma
But a= gsintheta
a= 9.8xsin 30
= 4.9m/s²
(a) Find the magnitude of an earthquake that has an intensity that is 37.25 (that is, the amplitude of the seismograph reading is 37.25 cm). (Round your answer to one decimal place.)
Answer:
The magnitude of an earthquake is 5.6.
Explanation:
The magnitude of an earthquake can be found as follows:
[tex] M = log(\frac{I}{S}) [/tex]
Where:
I: is the intensity of the earthquake = 37.25 cm
S: is the intensity of a standard earthquake = 10⁻⁴ cm
Hence, the magnitude is:
[tex]M = log(\frac{I}{S}) = log(\frac{37.25}{10^{-4}}) = 5.6[/tex]
Therefore, the magnitude of an earthquake is 5.6.
I hope it helps you!