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seraphim [82]
3 years ago
7

Suppose that the collector is held at a small negative voltage with respect to the grid. Will the accelerated electrons reach th

e collecting plate?
A) Yes, but only those electrons with energy less than the potential difference established between the grid and the collector will reach the collector.
B) Yes, but only those electrons with energy greater than the potential difference established between the grid and the collector will reach the collector.
C) Yes; all the accelerated electrons will reach the collector.
D) No, none of the electrons will reach the collector
Physics
1 answer:
Leona [35]3 years ago
3 0

Answer:

B) Yes, but only those electrons with energy greater than the potential difference established between the grid and the collector will reach the collector.

Explanation:

In the case when the collector would held at a negative voltage i.e. small with regard to grid So yes the accelerated electrons would be reach to the collecting plate as the kinetic energy would be more than the potential energy that because of negative potential

so according to the given situation, the option b is correct

And, the rest of the options are wrong

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The graph shows two runners participating in a race.
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Answer:

'Daniela had a 5-meter head start, and Leonard caught up to her at 25 meters.'

Explanation:

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Explain Newton's third law in your own words.
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Newton's third law states "for every action, there is an equal and opposite reaction."

What this is pretty much saying is that for every action, there is a consequence. One force connects and triggers another.
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A 20kg mass approaches a spring at a speed of 30 m/s. The mass compresses the spring 12cm before coming to a stop. Calculate the
Oksana_A [137]

Answer:

625000 N/ m

Explanation:

m= 20 kg

v= 30 m/s

x= 12 cm

k = ?

Here when the mass when hits at spring its speed is

Vi= 30 m/s

Finally it comes to rest after compressing for 12 cm

i-e Vf = 0 m/s

Distance= S= 12 cm = 0.12 m

using

2aS= Vf2 - Vi2

==> 2a ×0.12 = o- 30 × 30

==> a = 900 ÷ 0.24 = 3750 m/sec2

Now we know;

F = ma

F= -Kx

==> ma= -kx

==> 20 × 3750 = -K × 0.12

==> k = 625000 N/ m

5 0
3 years ago
What amount of heat is required to raise the temperature of 200 g of water by 15°C (the specific heat of water is 1 cal/g°C)
Sergeeva-Olga [200]

Answer:

Heat energy required (Q) = 3,000 J

Explanation:

Find:

Mass of water (M) = 200 g

Change in temperature (ΔT) = 15°C

Specific heat of water (C) = 1 cal/g°C

Find:

Heat energy required (Q) = ?

Computation:

Q = M × ΔT × C

Heat energy required (Q) = Mass of water (M) × Change in temperature (ΔT) × Specific heat of water (C)

Heat energy required (Q) = 200 g × 15°C × 1 cal/g°C

Heat energy required (Q) = 3,000 J

4 0
3 years ago
A surveyor is using a magnetic compass 5.6 m below a power line in which there is a steady current of 140 A. (a) What is the mag
ArbitrLikvidat [17]

Answer:

(a) B = 5.6 micro Tesla

Explanation:

Current in the wire, i = 140 A

distance, r = 5 m

The formula for the magnetic field at a distance r due to the current carrying wire

B=\frac{\mu _{0}}{4\pi }\times \frac{2i}{r}

B=10^{-7}\times \frac{2\times140}{5}

B = 5.6 x 10^-6 Tesla

B = 5.6 micro Tesla

(b) As the magnetic field of earth at this site is 20 micro tesla so the magnetic field due to current carrying wire interfere the magnetic compass.

4 0
3 years ago
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