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Ket [755]
1 year ago
8

The distance between two slits is 1.50 *10-5 m. A beam of coherent light of wavelength 600 nm illuminates these slits, and the d

istance between the slit and the screen is 2.00 m. What is the distance on the screen between the central bright fringe and the fourth-order bright fringe?
Physics
1 answer:
Fed [463]1 year ago
8 0

Answer: y = 2.4×10^-6m or y= 2.4μm

Explanation: The formulae for the distance between the central bright fringe to any other fringe in pattern is given as

y = R×mλ/d

Where y = distance between nth fringe and Central bright spot fringe.

m = position of fringe = 4

λ = wavelength of light= 600nm = 600×10^-9 m

d = distance between slits = 1.50×10^-5m

R = distance between slit and screen = 2m

y = 2 × 4 × 600×10^-9/2

y = 4800×10^-9/2

y = 2400 × 10^-9

y = 2.4×10^-6m or y= 2.4μm

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2 years ago
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A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the neg
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Answer:

=2,012,319.36 \ m/s

Explanation:

-The only relevant force is the electrostatic force

-The formula for the electrostatic force is:

F = Eq

E is the electric field and q is the magnitude of the charge.

#Since the electric field is the same in both cases, and the charge of the protons and electrons have the same magnitude, you can state that the magnitude of the electric forces acting in both proton and electron are the same.

F_e = F_p\\\\F_e= Force \ on \ electron\\F_p = Force \ on \ proton

-Applying Newton's 2nd Law:

F=ma

F_e=M_ea_e

F_p=M_pa_p

#equate the two forces:

F_e = F_p\\\\M_ea_e=M_pa_p\\\\a_e=\frac{M_pa_p}{M_e}

#The equations for velocity in uniform acceleration:

V_f^2=V_o^2+2ad\\\\V_o^2=0\\\\\therefore V_f^2=2ad

#For the proton:

V_f^2=2a_pd\\\\a_p=\frac{V_f^2}{2d}\\\\a_p=\frac{47000m/s)^2}{2d}

#For the electron:

V_f^2=2{a_e}^2\times 2d\\\\A_e=M_p\times A_p/M_e\\\\V_f^2=M_p\times (47000m/s)^2/2d\times2d/M_e\\\\V_f^2=M_p\times (47000m/s)^2/M_e\\\\V_f=47000m/s\times\sqrt{\frac{M_p}{M_e}}

The mass values of the proton and electron are:

M_p=1.67\times 10^{-27} kg\\\\M_e=9.11\times10^{-31}kg

The speed of the ion is therefore calculated as:

V_f=47000m/s\times\sqrt{\frac{M_p}{M_e}}\\\\=47000m/s\times\sqrt{\frac{1.67\times10^{-27}}{9.11\times10^{-31}}\\\\=2,012,319.36 \ m/s

Hence, the ion's speed at the negative plate is =2,012,319.36 \ m/s

7 0
2 years ago
A basketball rolls off a deck that is 3.2 m from the pavement. The basketball lands 0.75 m from the edge of the deck. How fast w
astra-53 [7]

Answer:

d). The value of y should be -32m

Vx=0.92 m/s

Explanation:

Using equation of motion uniform to y motion

x_{f}=x_{o}+v_{o}*t+\frac{1}{2}*a*t^{2}\\x_{o}=0m\\v_{o}=0\\x_{f}=\frac{1}{2}*a*t^{2}\\x_{f}=-3.2m \\a=g=-9.8\frac{m}{s^{2}} \\

So to find t that is the same time for all the motion

t^{2}=\frac{2*x_{f}}{a} \\t=\sqrt{\frac{2*x_{f}}{a}}=\sqrt{\frac{2*-3.2m}{-9.8\frac{m}{s^{2}}}}\\t=0.808s

The value of Xf=-3.2m because the g is negative from the axis

Now in the axis 'x' to find Vx

Vx=\frac{0.75m}{0.808S}\\ Vx=0.92 \frac{m}{s}

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How much energy does a 50 kg rock have if it is sitting on the edge of a 15 m cliff?
noname [10]

Answer:

7350 J

Explanation:

The gravitational potential energy of the rock sitting on the edge of the cliff is given by:

U=mgh

where

m is the mass of the rock

g is the gravitational acceleration

h is the height of the cliff

In this problem, we have

m = 50 kg

g = 9.8 m/s^2

h = 15 m

Substituting numbers into the formula, we find:

U=(50 kg)(9.8 m/s^2)(15 m)=7350 J

3 0
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