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miv72 [106K]
3 years ago
12

Building up electric charges on a Van de Graaff generator is an example of static

Physics
2 answers:
KonstantinChe [14]3 years ago
8 0
Yes, this is true. Van de Graaff generators are devices that produce static this electricity right now.
sdas [7]3 years ago
5 0
Yes, that is true. Van de Graaff generators are devices that produce static electricity. Static does not involve acceleration, meaning it is motionless which applies to the building of electric charges since the building up of these electric charges are created in a way to not move. Therefore, the sentence you have stated is an accurate statement. 

Let me know if you need anything else. :)

               ~ Dotz
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Why was Earth very hot at the beginning of its formation? Select all that apply.
Sladkaya [172]

Answer:

heat came from the supernova that led to the formation of solar system

4 0
3 years ago
An Amtrak going 250m/s comes to a stop in 12s. What is the<br> acceleration?
astraxan [27]

Answer:

a=\frac{v-u}{t}  \\a = \frac{0-250}{12} = -20.83 m/s

Explanation:

you mean deceleration right ? because the acceleration is 250m/s

7 0
3 years ago
If 745-nm and 660-nm light passes through two slits 0.54 mm apart, how far apart are the second-order fringes for these two wave
kotegsom [21]

Answer:

0.82 mm

Explanation:

The formula for calculation an n^{th} bright fringe from the central maxima is given as:

y_n=\frac{n \lambda D}{d}

so for the distance of the second-order fringe when wavelength \lambda_1 = 745-nm can be calculated as:

y_2 = \frac{n \lambda_1 D}{d}

where;

n = 2

\lambda_1 = 745-nm

D = 1.0 m

d = 0.54 mm

substituting the parameters in the above equation; we have:

y_2 = \frac{2(745nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}

y_2 = 0.00276 m

y_2 = 2.76 × 10 ⁻³ m

The distance of the second order fringe when the wavelength \lambda_2 = 660-nm is as follows:

y^'}_2 = \frac{2(660nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}

y^'}_2 = 1.94 × 10 ⁻³ m

So, the distance apart the two fringe can now be calculated as:

\delta y = y_2-y^{'}_2

\delta y = 2.76 × 10 ⁻³ m - 1.94 × 10 ⁻³ m

\delta y = 10 ⁻³ (2.76 - 1.94)

\delta y = 10 ⁻³ (0.82)

\delta y = 0.82 × 10 ⁻³ m

\delta y =  0.82 × 10 ⁻³ m (\frac{1.0mm}{10^{-3}m} )

\delta y = 0.82 mm

Thus, the distance apart the second-order fringes for these two wavelengths = 0.82 mm

6 0
4 years ago
QUESTION 9
Alik [6]

Answer:

The answer is option a.

Hope this helps

4 0
3 years ago
An 85 kg man and his 35 kg daughter are sitting on opposite ends of a 3.00 m see-saw. The see-saw is anchored in the center. If
wolverine [178]

Answer:

0.54m

Explanation:

Step one:

given data

length of seesaw= 3m

mass of man m1= 85kg

weight = mg

W1= 85*10= 850N

mass of daughter m2= 35kg

W2= 35*10= 350N

distance from the center= (1.5-0.2)= 1.3m

Step two:

we know that the sum of clockwise moment equals the anticlockwise moment

let the distance the must sit to balance the system be x

taking moment about the center of the system

350*1.3=850*x

455=850x

divide both sides by  850

x=455/850

x=0.54

Hence the man must sit 0.54m from the right to balance the system

3 0
3 years ago
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