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pickupchik [31]
3 years ago
12

When performing an.experiment similar to Millikan's oil drop, a student measured the following load magnitudes: 3.26x10 ^-19 C 5

.09x10 ^-19 C 1.53x10 ^-19 C 6.39x10 ^-19 C 4.66x10 ^-19 C I used these measurements to find the charge on the electron
Physics
1 answer:
torisob [31]3 years ago
8 0

Answer:

Explanation:

We put the charges in the ascending order as follows

1.53 P

3.26 P

4.66 P

5.09 P

6.39 P

where P is equal to 10⁻¹⁹

we round off given charges as follows

1.53 P → 1.6 P

3.26 P → 3.2 P

4.66 P → 4.8 P

5.09 P → 4.8 P

6.39 P → 6.4 P

We see that 2 nd to 4 th charges are integral multiples of first charge . That means these charges are supposed to be made of combination of first charge . So first charge appears to be minimum possible charge .

Hence this charge may exist on single electron.

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1)Light of wavelength 588.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 55.5 cm from the slit
Talja [164]

Answer:

These are Diffraction Grating Questions.

Q1. To determine the width of the slit in micrometers (μm), we will need to use the expression for distance along the screen from the center maximum to the nth minimum on one side:  

Given as  

y = nDλ/w                                                       Eqn 1

where  

w = width of slit  

D = distance to screen  

λ = wavelength of light  

n = order number  

Making x the subject of the formula gives,  

w = nDλ/y  

Given  

y = 0.0149 m  

D = 0.555 m  

λ = 588 x 10-9 m  

and n = 3

w = 6.6x10⁻⁵m

Hence, the width of the slit w, in micrometers (μm) = 66μm

Q2. To determine the linear distance Δx, between the ninth order maximum and the fifth order maximum on the screen

i.e we have to find the difference between distance along the screen (y₉-y₅) = Δx

Recall Eqn 1,     y = nDλ/w  

given, D = 27cm = 0.27m  

λ = 632 x 10-9 m  

w = 0.1mm = 1.0x10⁻⁴m

For the 9th order, n = 9,

y₉ = 9 x 0.27 x 632 x 10-9/ 1.0x10⁻⁴m = 0.015m

Similarly, for n = 5,

y₅ = 5x 0.27 x 632 x 10-9/ 1.0x10⁻⁴m = 0.0085m

Recall,  Δx = (y₉-y₅) = 0.015 - 0.0085 = 0.0065m

Hence, the linear distance Δx between the ninth order maximum and the fifth order maximum on the screen = 6.5mm

8 0
3 years ago
a student drops an object from the top of a building which is 19.6 m high. How long does it take the object to fall to the groun
zubka84 [21]

Here's a formula that's simple and useful, and if you're really in
high school physics, I'd be surprised if you haven't see it before. 
This one is so simple and useful that I'd suggest memorizing it,
so it's always in your toolbox.

This formula tells how far an object travels in how much time,
when it's accelerating:

               Distance = (1/2 acceleration) x (Time²).

                           D = 1/2 A T²

For your student who dropped an object out of the window,

     Distance = 19.6 m
     Acceleration = gravity = 9.8 m/s²

                                              D = 1/2 G T²

                                          19.6 =   4.9   T²

Divide each side by 4.9 :       4  =           T²

Square root each side:           2  =          T

When an object is dropped in Earth gravity,
it takes  2  seconds to fall the first 19.6 meters.

8 0
3 years ago
Read 2 more answers
PLEASE HURRY Take 40 points.please just look at the picture.​
Fynjy0 [20]

Answer:

It you get a sushi for work, for a party, then it leads to people being happy, as an independant variable

It is changing the molecular structure of a protein

Explanation:

7 0
3 years ago
4. The value of the before and after-collision momentum of two colliding objects is shown in the
ExtremeBDS [4]

Answer:A) WHICH is 0kgm/s

Explanation:

3 0
3 years ago
2. Three blocks, A,B and C of mass 2kg. 3kg. 5kg respectively kept side by side with one another are accelerated at 2m/s2 across
gulaghasi [49]

Answer:

Total mass of combination = 2+3+5 = 10kg.

Acceleration produced = 2m/s^2

hence force =( total mass × acceleration)= (2×10)= 20 N.

Net force on 3kg block = acceleration × mass = (2 × 2 )= 4 N

applied force on 2 kg block = 20N

Force between 2 kg and 3 kg block = (20-4) = 16N. ans

Net force on 3 kg block = 3 × 2 =6N.

Applied force on 3 kg block due to 2 kg block = 16N.

hence, force between 3 kg and 5 kg block = (16-6) = 10N .

answers:-

(a) 20 N

(b) 16N

(c) 10 N

4 0
1 year ago
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