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Troyanec [42]
3 years ago
13

Can we write names while writing conversation in board exam​

Physics
1 answer:
lora16 [44]3 years ago
5 0

Answer:

ya we can write the imaginary character's name .

So that we  can identify these imaginary people, as we cannot simply write the conversation and leave it .

Or maybe sometimes the reader will get confused as there is no name for the two people .

So, i suggest that you should write the names

Explanation:

You can even ask to your class teacher for further clarification

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A 0.22kg mousetrap car has 3.1 J of potential energy due to the spring in the mouse
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Answer:

r56

Explanation:

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When dolphin swims close to its prey, it transmits sound waves to figure out details (like direction where the fish is moving) o
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The frequency of the waves depends on the distance between wave fronts - considering a front as a maximum disturbance of the wave

(Consider the waves emitted by an organ pipe: condensation and rarefactions)

The waves themselves are a fixed distance apart -

as one moves towards the source the waves received will be closer together (higher frequency)

So if the frequency received increases, the distance between the source and the observer must be decreasing

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3 years ago
Examine the lightbulbs in the circuit below. Write a sentence explaining what would happen if lightbulb A burned out. Repeat thi
tatiyna

If the lightbulb A in the circuit shown in the image burned out, the path for the  current to flow is disrupted because one of its terminals is connected direct to the source. So, there will be no current through the lightbulbs B, C, and D, and they will turn off. Similarly it will happen, if the lightbulb D burned out.

If the lightbulb B burned out the current will continue circulating through the lightbulbs A, C, and D, because lightbulb B is connected in parallel. Similarly it will happen, if the lightbulb C burned out.

8 0
3 years ago
During the hammer throw at a track meet, an 8.0 kg hammer is accidentally thrown straight up. If 784.0 J of
jeka57 [31]

The height risen by the hammer when the work were done is 10 m.

The given parameters;

  • <em>Mass of the hammer, m = 8.0 kg</em>
  • <em>Work done on the hammer, W = 780 J</em>

<em />

Apply work energy - theorem to determine the height risen by the hammer when the work is done on it;

E = mgh\\\\h = \frac{E}{mg} \\\\h = \frac{784}{8 \times 9.8} \\\\h = 10 \ m

Thus, the height risen by the hammer when the work were done is 10 m.

Learn more about work-energy theorem here: brainly.com/question/22236101

4 0
2 years ago
The allowed energies of a simple atom are 0.0 eV, 3.0 eV, and 4.0 eV. An electron traveling at a speed of 1.3*10^6 m/s collision
ss7ja [257]

A) 5.34\cdot 10^5 m/s

The minimum speed of the electron occurs when the electron loses the maximum energy: this occurs when the electron excites the atom from 0.0 eV to 4.0 eV, because in this case the energy given to the atom is maximum.

The energy given by the electron to the atom is equal to the difference between the two energy levels:

\Delta E= 0.0 eV - 4.0 eV =-4.0 eV = -6.4\cdot 10^{-19}J

This is equal to the kinetic energy lost by the electron:

K_f - K_ i = \Delta E\\\frac{1}{2}m(v^2-u^2) = \Delta E

where

m is the electron's mass

v is the final speed of the electron after the collision

u=1.3\cdot 10^6 m/s is the speed of the electron before the collision

Solving for v, we find

v=\sqrt{ \frac{2\Delta E}{m}+u^2}=\sqrt{\frac{2 (-6.4\cdot 10^{-19} J)}{9.11\cdot 10^{-31} kg}+(1.3\cdot 10^6 m/s)^2}=5.34\cdot 10^5 m/s

B) 1.16\cdot 10^6 m/s

The maximum speed of the electron occurs when the electron loses the minimum amount of energy: this occurs when the electron excites the atom from 3.0 eV to 4.0 eV, because in this case the energy given to the atom is minimum.

The energy given by the electron to the atom is equal to the difference between the two energy levels, so in this case we have:

\Delta E= 3.0 eV - 4.0 eV =-1.0 eV = -1.6\cdot 10^{-19}J

And so, this time the final speed of the electron after the collision will be given by:

v=\sqrt{ \frac{2\Delta E}{m}+u^2}=\sqrt{\frac{2 (-1.6\cdot 10^{-19} J)}{9.11\cdot 10^{-31} kg}+(1.3\cdot 10^6 m/s)^2}=1.16\cdot 10^6 m/s

3 0
4 years ago
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