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Arturiano [62]
3 years ago
7

the grid in a triode is kept negatively charged to prevent… a. the variations in voltage from getting too large. b. electrons be

ing repelled back to the cathode. c. electrons from escaping from the tube containing the triode. d. the electrons from being attracted to the grid instead of the anode.
Physics
2 answers:
Nezavi [6.7K]3 years ago
5 0
D:the electrons from being attracted to the grid instead of the anode
VLD [36.1K]3 years ago
5 0

Answer:

D the electrons from being attracted to the grid instead of the anode.

Explanation:

just because that's the correct answer

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Answer:They stop because jet streams follow boundaries between hot and cold air.

Explanation:

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Did Kepler propose the modern model of the solar system?
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Yes Kepler did propose the modern model of the solar system.
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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
a train starting from rest moving with uniform acceleration attains a speed of 36 km/hr in 10 seconds. Find its acceleration​
Savatey [412]

Answer:

\boxed{\sf Acceleration \ of \ train = 1 \ m/s^{2}}

Given:

Initial speed (u) = 0 km/hr = 0 m/s

Final speed (v) = 36 km/hr

Time taken (t) = 10 sec

To Find:

Acceleration (a) of the train

Explanation:

\sf 1 \ km/hr = \frac{5}{18} \ m/s \\ \therefore \\ \sf 36 \ km/hr = 36 \times \frac{5}{18} \ m/s \\  \\  \sf \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:    = 2 \times  \cancel{18} \times  \frac{5}{ \cancel{18}}  \ m/s \\  \\  \sf \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \:  \:  \:  \:  \: = 2 \times 5 \ m/s \\  \\  \sf \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  = 10  \ m/s

So,

Final speed (v) = 10 m/s

\sf From \ equation \ of \ motion: \\ \sf \implies \bold{v = u + at} \\ \\ \sf Substituting \ value \ of \ v, \ u \ and \ t: \\ \sf \implies 10 = 0 + a(10) \\ \sf \implies 10 = a(10) \\ \sf \implies a \times 10 = 10 \\ \\ \sf Dividing \ both \ sides \ by \ 10: \\ \sf \implies \frac{a \times 10}{\boxed{\sf 10}} = \frac{10}{\boxed{\sf 10}} \\ \\ \sf \frac{\cancel{10}}{\cancel{10}} = 1: \\ \sf \implies a = 1 \:  m/s^2

Acceleration of train = 1 m/s²

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25,000N of force will be ouputed

Explanation:

Detailed explanation and calculation is shown in the image below

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