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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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A truck is traveling at 74.5 kilometers per hour away from you. The driver is blowing the horn which has a frequency of 415 Hz.
ArbitrLikvidat [17]
It would be a high pitched sound. Hope this helps.
8 0
3 years ago
Beings on spherical asteroid have observed that a large rock is approaching their asteroid in a collision course. At 7514 km fro
luda_lava [24]

Answer:

c. 4.582\times10^{21} kg

Explanation:

r_{i} = Initial distance between asteroid and rock = 7514 km = 7514000 m

r_{f} = Final distance between asteroid and rock = 2823 km = 2823000 m

v_{i} = Initial speed of rock = 136 ms⁻¹

v_{f} = Final speed of rock = 392 ms⁻¹

m = mass of the rock

M = mass of the asteroid

Using conservation of energy

Initial Kinetic energy of rock + Initial gravitational potential energy = Final Kinetic energy of rock + Final gravitational potential energy

(0.5) m v_{i}^{2} - \frac{GMm}{r_{i}} = (0.5) m v_{f}^{2} - \frac{GMm}{r_{f}} \\(0.5) v_{i}^{2} - \frac{GM}{r_{i}} = (0.5) v_{f}^{2} - \frac{GM}{r_{f}} \\(0.5) (136)^{2} - \frac{(6.67\times10^{-11}) M}{(7514000)} = (0.5) (392)^{2} - \frac{(6.67\times10^{-11}) M}{(2823000)} \\M = 4.582\times10^{21} kg

8 0
3 years ago
Each lunar cycle has one full moon, in which the relative positions of Earth, the sun, and the moon form in a straight line
aniked [119]

Complete Question:

Each lunar cycle has one full moon, in which the relative positions of Earth, the sun, and the moon form in a straight line. Which list represents the position of Earth, the sun, and the moon during a full moon?

Group of answer choices.

A. Earth, sun, moon

B. sun, moon, Earth

C. moon, sun, Earth

D. sun, Earth, moon

Answer:

D. sun, Earth, moon

Explanation:

A lunar eclipse is a phenomenon that occurs when the Earth comes between the Moon and the Sun thereby causing it to cover the Moon with its shadow.

Simply stated, lunar eclipse takes place when the Moon passes or moves through the Earth's shadow thereby blocking any ray of sunlight from reaching the Moon. Thus, the full moon appears deep red (blood moon).

Also, a lunar eclipse would occur only when the Sun, Earth, and Moon are closely aligned to form a straight line known as the syzygy.

There are three (3) types of lunar eclipse and these are;

1. Total lunar eclipse.

2. Partial lunar eclipse.

3. Penumbra lunar eclipse.

Each lunar cycle has one full moon, in which the relative positions of Earth, the sun, and the moon form in a straight line. Thus, the list which represents the position of Earth, the sun, and the moon during a full moon is sun, Earth, and moon

3 0
3 years ago
In a Broadway performance, an 84.5-kg actor swings from a R = 4.30-m-long cable that is horizontal when he starts. At the bottom
Arada [10]

Answer:

1.57772 m

Explanation:

M = Mass of actor = 84.5 kg

m = Mass of costar = 55 kg

v = Velocity of costar

V  = Velocity of actor

h_i = Intial height of actor = 4.3 m

g = Acceleration due to gravity = 9.81 m/s²

As the energy of the system is conserved

\frac{1}{2}MV^2=Mgh_i\\\Rightarrow V=\sqrt{2gh_i}\\\Rightarrow V=\sqrt{2\times 9.81\times 4.3}\\\Rightarrow V=9.18509\ m/s

As the linear momentum is conserved

MV=(m+M)v\\\Rightarrow v=\frac{MV}{m+M}\\\Rightarrow V=\frac{84.5\times 9.18509}{84.5+55}\\\Rightarrow v=5.56372\ m/s

Applying conservation of energy again

\frac{1}{2}(m+M)v^2=(m+M)gh_f\\\Rightarrow h_f=\frac{v^2}{2g}\\\Rightarrow h_f=\frac{5.56372^2}{2\times 9.81}\\\Rightarrow h_f=1.57772\ m

The maximum height they reach is 1.57772 m

3 0
3 years ago
A heat pump with a COP of 3.15 is used to heat an air-tight house. When running, the heat pump consumes 5 kW of power. If the te
Jet001 [13]

Answer: 1026s, 17.1m

Explanation:

Given

COP of heat pump = 3.15

Mass of air, m = 1500kg

Initial temperature, T1 = 7°C

Final temperature, T2 = 22°C

Power of the heat pump, W = 5kW

The amount of heat needed to increase temperature in the house,

Q = mcΔT

Q = 1500 * 0.718 * (22 - 7)

Q = 1077 * 15

Q = 16155

Rate at which heat is supplied to the house is

Q' = COP * W

Q' = 3.15 * 5

Q' = 15.75

Time required to raise the temperature is

Δt = Q/Q'

Δt = 16155 / 15.75

Δt = 1025.7 s

Δt ~ 1026 s

Δt ~ 17.1 min

5 0
3 years ago
Read 2 more answers
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