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OleMash [197]
3 years ago
7

A diffraction grating with 1000 lines per mm is used in a spectrometer to measure the wavelengths of light emitted from a gas di

scharge tube. You measure the diffraction angle to be at 20° for one emission line. What would happen to this emission line’s diffraction angle if you replaced the diffraction grating with one having 500 lines per mm instead
Physics
1 answer:
frez [133]3 years ago
4 0

Answer:

= 9.8°

Explanation:

Width of one slit (a₁ ) = 1 / 1000 mm=0.001 mm = 10⁻⁶ m.

width of one slit in case 2 (a₂ ) = 1/500 =2 x 10⁻⁶ m

angular position of fringe, Sinθ  = n λ /a

n is order of fringe , λ is wave length of light and a  is slit aperture

So Sinθ  ∝ 1 / a

Sin θ₁ /Sin θ₂ = a₂/a₁ ;

Sin20°/sinθ₂ = 2 / 1

sinθ₂ = Sin 20° / 2 = .342/2 = .171

θ₂ = 9.8 °

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A cannon sends a projectile towards a target a distance 1420 m away. The initial velocity makes an angle 35◦ with the horizontal
ss7ja [257]

Answer:

v_{o}=141.51m/s

Explanation:

From the exercise we know the final x distance, the angle which the projectile is being released and acceleration of gravity

x=1420m\\g=-9.8m/s^{2}

From the equation of x-position we know that

x=v_{ox}t=v_{o}cos(35)t

Solving for v_{o}

v_{o}=\frac{x}{tcos(35)} =\frac{1420m}{tcos(35)} (1)

Now, if we analyze the equation of y-position we got

y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}

At the end of the motion y=0

0=v_{o}sin(35)t+\frac{1}{2}gt^{2}

Knowing the equation for v_{o} in (1)

0=\frac{1420}{tcos(35)}tsin(35)-\frac{1}{2}(9.8)t^{2}

\frac{1}{2}(9.8)t^{2}=1420tan(35)

Solving for t

t=\sqrt{\frac{2(1420tan(35))}{9.8} } =14.25s

Now, we can solve (1)

v_{o}=\frac{1420m}{(14.25s)cos(35)}=141.51m/s

6 0
3 years ago
Read 2 more answers
If the magnitude of a positive charge is tripled, what is the ratio of the original value of the electric field at a point to th
ipn [44]

Answer:

b)1 :3

Explanation:

Lets that

The value of a positive charge = q

As we know that electric filed on a point charge given as

E=\dfrac{Kq}{r^2}

Where ,K=Constant

q=Charge ,r=Distance

If the value of the charge gets tripled ,q'= 3 q

Then electric filed E'

E'=\dfrac{Kq'}{r^2}

E'=\dfrac{3Kq}{r^2}

E' = 3 E

Therefore we can say that

\dfrac{E}{E'}==\dfrac{1}{3}

therefore the answer will be --

b)1 :3

3 0
3 years ago
Have you ever visited an amusement park and taken a ride on a parachute drop ride? These types of rides take the passengers to a
Triss [41]

Answer & Explanation:

a)

Lenz's law states that the direction of induced electric current is always such that, it opposes the change in magnetic flux.

In a drop ride, the hub on which we sit and are hung to is an electromagnet and there are many such magnets mounted on the columns of the support. what happens is these electromagnets (in support) generate a repulsive magnetic field with respect to the field generated by the hub solenoids. this results in lift generation till the top of ride. reaching the top, the bar solenoids are at their maximum repulsive force. Then the solenoids in column are set current less means electric supply is cut off. this makes you fall under the effect of gravity. by the time you are half way down, column  solenoids are turned on again. As the hub solenoid approaches every single electromagnet in supporting columns. Due to change in magnetic field (with respect to lenz's law) an opposing current induces further providing resistance to the fall, this continues until the ride comes to rest completely. This is how it works.  

c) In addition, highly compressive springs, dampers, viscous dampers, etc. could be used in its place.

but the above listed cannot provide a differential braking,

have a limited lifecycle,

will provide resistance during lift also,

require higher maintenance

3 0
3 years ago
In your own words describe how matter is defined
gizmo_the_mogwai [7]

Matter is any substance that has mass and takes up space.

5 0
3 years ago
Which of these letters is the symbol for current in equations A: c B: i C: r D: t
True [87]
  Answer:  [B]:  The letter, "<em /> I " ; for current;  in units of "Amps" .
__________________________________________________
4 0
3 years ago
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