1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Anton [14]
3 years ago
10

Using a 685 nm wavelength laser, you form the diffraction pattern of a 1.11 mm wide slit on a screen. You measure on the screen

that the 11th dark fringe is 9.85 cm away from the center of the central maximum. How far is the screen located from the slit
Physics
1 answer:
velikii [3]3 years ago
5 0

Answer:

13.8 m

Explanation:

Dark fringes are formed in a single slit experiment due to destructive interference that occurs due to interference.

The position of these dark fringes formed on a screen is given by:

y = \frac{\lambda }{d} (m + 1/2)D

where y = position of  mth minimum

m = order of the minimum

D = distance of the slit from the screen

d = width of the slit

λ = wavelength of the light used

We need to find D:

D = \frac{yd} {\lambda (m + 1/2)} \\

From the question:

m = 11

y = 9.85 cm = 0.0985 m

λ  = 6.83 * 10^{-7} m

d = 1.11 mm = 0.0011 m

Therefore:

D = \frac{0.0985 *0.0011} {6.83 * 10^{-7} *(11 + 1/2)} \\\\D = \frac{0.00010835} {6.83 * 10^{-7}  * (23/2)} \\\\D = 13.8 m

The slit is 13.8 m far from the screen

You might be interested in
on a very muddy football field, a 120 kg linebacker tackles an 75 kg halfback. immediately before the collision, the linebacker
DIA [1.3K]
<u>Momentum</u> 
- a vector quantity; has both magnitude and direction
- has the same direction as object's velocity
- can be represented by components x & y.

Find linebacker momentum given m₁ = 120kg, v₁ = 8.6 m/s north
P₁ = m₁v₁
P₁ = (120)(8.6)
[ P₁ = 1032 kg·m/s ] = y-component, linebacker momentum

Find halfback momentum given m₂ = 75kg, v₂ = 7.4 m/s east
P₂ = m₂v₂
P₂ = (75)(7.4)
[ P₂ = 555 kg·m/s ] = x-component, halfback momentum

Find total momentum using x and y components.
P = √(P₁)² + (P₂)²
P = √(1032)² + (555)²
[[ P = 1171.77 kg·m/s ]] = magnitude 

!! Finally, to find the magnitude of velocity, take the divide magnitude of momentum by the total mass of the players.
P = mv
P = (m₁ + m₂)v
1171.77 = (120 + 75)v      <em>[solve for v]</em>
<em />v = 1171.77/195
v = 6.0091 ≈ 6.0 m/s

If asked to find direction, take inverse tan of x and y components.
tanθ = (y/x)
θ = tan⁻¹(1032/555)
[ θ = 61.73° north of east. ]

The magnitude of the velocity at which the two players move together immediately after the collision is approximately 6.0 m/s.
6 0
3 years ago
How much heat must be added to make a 5g substance with a specific heat of 2 J/gC that has its temperature go up 10 degrees? Q =
zlopas [31]

Answer:

100 Joule

Explanation:

Amount of heat in agiven body is given by Q = m•C•ΔT

where m is the mass of the body

c is the specific heat capacity of body. It is the amount of heat stored in 1 unit weight of body which raises raises the temperature of body by 1 unit of temperature.

ΔT is the change in the temperature of body

___________________________________________

coming back to problem

m = 5g

C = 2J/gC

since, it is given that temperature of body increases by 10 degrees, thus

ΔT = 10 degrees

Using the formula for heat as given

Q = m•C•ΔT

Q = 5* 2 * 10  Joule= 100 Joule

Thus, 100 joule heat must be added to  a  5g substance with a specific heat of 2 J/gC to raise its temperature go up by 10 degrees.

8 0
3 years ago
A car goes round a curve of radius 48m, the road is banked at an angle of 15 with the horizontal,at what maximum speed may the c
Marizza181 [45]

Answer:

11 m/s

Explanation:

Draw a free body diagram.  There are two forces acting on the car:

Weigh force mg pulling down

Normal force N pushing perpendicular to the incline

Sum the forces in the +y direction:

∑F = ma

N cos θ − mg = 0

N = mg / cos θ

Sum the forces in the radial (+x) direction:

∑F = ma

N sin θ = m v² / r

Substitute and solve for v:

(mg / cos θ) sin θ = m v² / r

g tan θ = v² / r

v = √(gr tan θ)

Plug in values:

v = √(9.8 m/s² × 48 m × tan 15°)

v = 11.2 m/s

Rounded to 2 significant figures, the maximum speed is 11 m/s.

3 0
3 years ago
A rigid, insulated tank whose volume is 10 L is initially evacuated. A pinhole leak develops and air from the surroundings at 1
balandron [24]

Answer:

The answer is "143.74^{\circ} \ C , 8.36\ g, and \ 2.77\ \frac{K}{J}"

Explanation:

For point a:

Energy balance equation:

\frac{dU}{dt}= Q-Wm_ih_i-m_eh_e\\\\

W=0\\\\Q=0\\\\m_e=0

From the above equation:

\frac{dU}{dt}=0-0+m_ih_i-0\\\\\Delta U=\int^{2}_{1}m_ih_idt\\\\

because the rate of air entering the tank that is h_i constant.

\Delta U = h_i \int^{2}_{1} m_i dt \\\\= h_i(m_2 -m_1)\\\\m_2u_2-m_1u_2=h_i(M_2-m_1)\\\\

Since the tank was initially empty and the inlet is constant hence, m_2u-0=h_1(m_2-0)\\\\m_2u_2=h_1m_2\\\\u_2=h_1\\\\

Interpolate the enthalpy between T = 300 \ K \ and\ T=295\ K. The surrounding air  

temperature:

T_1= 25^{\circ}\ C\ (298.15 \ K)\\\\\frac{h_{300 \ K}-h_{295\ K}}{300-295}= \frac{h_{300 \ K}-h_{1}}{300-295.15}

Substituting the value from ideal gas:

\frac{300.19-295.17}{300-295}=\frac{300.19-h_{i}}{300-298.15}\\\\h_i= 298.332 \ \frac{kJ}{kg}\\\\Now,\\\\h_i=u_2\\\\u_2=h_i=298.33\ \frac{kJ}{kg}

Follow the ideal gas table.

The u_2= 298.33\ \frac{kJ}{kg} and between temperature T =410 \ K \ and\  T=240\ K.

Interpolate

\frac{420-410}{u_{240\ k} -u_{410\ k}}=\frac{420-T_2}{u_{420 k}-u_2}

Substitute values from the table.

 \frac{420-410}{300.69-293.43}=\frac{420-T_2}{{u_{420 k}-u_2}}\\\\T_2=416.74\ K\\\\=143.74^{\circ} \ C\\\\

For point b:

Consider the ideal gas equation.  therefore, p is pressure, V is the volume, m is mass of gas. \bar{R} \ is\  \frac{R}{M} (M is the molar mass of the  gas that is 28.97 \ \frac{kg}{mol} and R is gas constant), and T is the temperature.

n=\frac{pV}{TR}\\\\

=\frac{(1.01 \times 10^5 \ Pa) \times (10\ L) (\frac{10^{-3} \ m^3}{1\ L})}{(416.74 K) (\frac{8.314 \frac{J}{mol.k} }{2897\ \frac{kg}{mol})}}\\\\=8.36\ g\\\\

For point c:

 Entropy is given by the following formula:

\Delta S = mC_v \In \frac{T_2}{T_1}\\\\=0.00836 \ kg \times 1.005 \times 10^{3} \In (\frac{416.74\ K}{298.15\ K})\\\\=2.77 \ \frac{J}{K}

5 0
3 years ago
How many electrons are in 204 C of charge?
zubka84 [21]

Answer:

The mass number 204 – 82 protons = 122 neutrons

Explanation:

Hope this helps!

5 0
3 years ago
Other questions:
  • How many gallons of water per day does the average American family of 4 use?
    13·2 answers
  • Two objects are attracted to each other by electromagnetic forces. Suppose I double the charge of one object. How can I return t
    7·1 answer
  • (Double points) A worker applied 27 newtons to a lever with a length of 4 meters that rotated around a hinge. What was the torqu
    15·2 answers
  • How does energy from the sun affect the motion of molecules in a solid compared to molecules in a gas
    9·1 answer
  • Mitchell records the amount of time a toy car takes to travel down a ramp one time for each of four slope angles. His lab
    8·1 answer
  • A 90N force towards postive x-axis, A 50N force towards negative x-axis applied on 5kg object. Find:- A) the net force on the ob
    10·1 answer
  • Pentru a scoate apă dintr‑o fântână, Ionel folosește dispozitivul reprezentat în figura 3, în care d = 20 cm și ℓ = 40 cm. Gălea
    15·1 answer
  • SHORT ANSWER TYPE QUESTIONS<br>1 1. Why school bags have broad shoulder straps ?​
    6·1 answer
  • When do i know if the acceleration (a) in newtons second law is 0? ​
    14·1 answer
  • ¿Qué pueden decirnos las capas de hielo del Ártico sobre nuestros cambios atmosféricos?
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!