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schepotkina
2 days ago
13

The rotational speeds of four generators are listed in RPM (revolutions per minute). Arrange the generators in order based on th

e electric fields they create. Start with the weakest electric field and end with the strongest. Assume that the generators are identical except for their speed.
Tiles
3,600 RPM
3,200 RPM
3,400 RPM
3,000 RPM
Sequence


<

<

<
Physics
2 answers:
Keith_Richards [2.2K]2 days ago
4 0

Utilizing the same generator, the only variable affecting the electric field is the speed. The higher the generator's rotation speed, the more extensive the electric field it generates. Hence, the order is 3000 rpm<3200 rpm<3400 rpm<3600 rpm

Keith_Richards [2.2K]2 days ago
3 0

Sequence: 3000

3200

3400

3600

Reasoning: I completed the test with a score of 5 out of 5

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Laser beam with wavelength 632.8 nm is aimed perpendicularly at opaque screen with two identical slits on it, positioned horizon
Softa [2029]

Response:

1 slit width = 0.158 mm, slit separation = 0.633 mm, distance between diffraction maxima = 12.7 mm

Explanation:

This involves defining several terms:

λ, the wavelength of the beam

D, the distance from the plane to the slit

x, the distance between minima in the diffraction pattern (in single slit setups)

w, the fringe width for double slit setups

1. In a double slit experiment, the fringe width is also recognized as the distance between Maxima.

Thus, w=λD/d, leading to d=λD/w when rearranged.

So, d= (632.8 x 10^-9 x 1 x 10^3)/1

which equals d= 0.633mm.

For single slit diffraction, minima are defined by a*sinΘ= m*x

where a is the slit width and m is an integer.

For small angles, it simplifies to Θ= (x/D) = (λ/a), allowing us to solve for a:

a = λD/a

a = (632.8x10^-9 x 1)/4

yielding a= 0.158mm.

2. A grating with 20 slits/mm gives d= 1/20mm= 0.05x10^-3.

To find y (the distance between maxima), apply y= λL/d:

Finally, y= (632.8 x 10^-9 x 1)/0.05x10^-3, which results in y= 12.7mm.

6 0
15 days ago
When you apply the torque equation ∑τ = 0 to an object in equilibrium, the axis about which torques are calculated:
Softa [2029]

Answer:

option D.

Explanation:

The correct choice is option D.

For an object in equilibrium, the torque measured at any point will be zero.

An object is deemed to be in equilibrium when the net moment acting on it equals zero.

If the object experiences a net moment not equal to zero, it will rotate and will not remain stable.

3 0
1 month ago
What is the minimum frequency of light necessary to emit electrons from titanium via the photoelectric effect?
kicyunya [2264]

<span>E = h x f </span>

<span>Thus: </span>

<span>f = E / h </span>
<span>f = 4.41•10^-19 / 6.62•10^-34 </span>
<span>f = 6.66•10^14 Hz (s^-1) </span>


<span>b/ What is the wavelength of this light? </span>
<span>------------------------------ </span>

<span>λ = c / f </span>
<span>λ = 3•10^8 / 6.66•10^14 </span>
<span>λ = 4.50•10^-7 m </span>
4 0
28 days ago
Read 2 more answers
Assume you are driving 20 mph on a straight road. Also, assume that at a speed of 20 miles per hour, it takes 100 feet to stop.
Maru [2337]

Answer:900 feet

Explanation:

Given

Velocity \left ( V_1\right )=20 mph\approx 29.334 ft/s

It takes 100 feet to come to a stop.

Utilizing the equation of motion

v^2-u^2=2as

Where

v,u=Final and initial velocities

a=acceleration

s=distance traveled

0-\left ( 29.334\right )^2=2\left (-a\right )\left ( 100\right )

a=\frac{29.334^2}{2\times 100}=4.302 ft/s^2

When the speed is 60 mph \approx 88.002 ft/s

v^2-u^2=2as

0-\left ( 88.002\right )^2=2\left ( -4.302\right )\left ( s\right )

s=900.08 feet

8 0
29 days ago
Read 2 more answers
The angle θ is slowly increased. Write an expression for the angle at which the block begins to move in terms of μs.
kicyunya [2264]

Answer:

tan \theta = \mu_s

Explanation:

Aby obiekt był w spoczynku na nachyleniu, wynikowa siła działająca na niego musi wynosić zero. Równanie sił działających w kierunku równoległym do nachylenia jest następujące:

mg sin \theta - \mu_s R =0 (1)

gdzie

mg sin \theta to składowa ciężaru równoległa do nachylenia, przy czym m oznacza masę obiektu, g oznacza przyspieszenie grawitacyjne, a \theta to kąt nachylenia

\mu_s R to siła tarcia, z \mu_s jako współczynnikiem tarcia oraz R jako reakcją normalną nachylenia

Równanie sił w kierunku prostopadłym do nachylenia to

R-mg cos \theta = 0

gdzie

R to reakcja normalna

mg cos \theta to składowa ciężaru prostopadła do nachylenia

Obliczając R,

R=mg cos \theta

I podstawiając do (1)

mg sin \theta - \mu_s mg cos \theta = 0

Rearanżując równanie,

sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s

To jest warunek, przy którym równowaga jest zachowana: gdy tangens kąta staje się większy niż wartość \mu_s, siła tarcia nie jest w stanie zrównoważyć składowej ciężaru równoległej do nachylenia, dlatego obiekt zaczyna zsuwać się w dół.

4 0
16 days ago
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