A gas has to become ionisied in order to become a conductor. It must have a chain reaction in which atoms in it became unstable, in which they loose stabile electronic configuration. In order for a gas to become a conductor, it must have free particles, and it can happen only in ionisied gas.
Answer: A 100-lb child stands on a scale while riding in an elevator. Then, the scale reading approaches to 100lb, while the elevator slows to stop at the lowest floor
Explanation: To find the correct answer, we need to know more about the apparent weight of a body in a lift.
<h3>What is the apparent weight of a body in a lift?</h3>
- Consider a body of mass m kept on a weighing machine in a lift.
- The readings on the machine is the force exerted by the body on the machine(action), which is equal to the force exerted by the machine on the body(reaction).
- The reaction we get as the weight recorded by the machine, and it is called the apparent weight.
<h3>How to solve the question?</h3>
- Here we have given with the actual weight of the body as 100lbs.
- This 100lb child is standing on the scale or the weighing machine, when it is riding .
- During this condition, the acceleration of the lift is towards downward, and thus, a force of ma .
- There is also<em> mg </em>downwards and a normal reaction in the upward direction.
- when we equate both the upward force and downward force, we get,
i.e. during riding the scale reads a weight less than that of actual weight.
- When the lift goes slow and stops the lowest floor, then the acceleration will be approaches to zero.
Thus, from the above explanation, it is clear that ,when the elevator moves to the lowest floor slowly and stops, then the apparent weight will become the actual weight.
Learn more about the apparent weight of the body in a lift here:
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Answer:
19.08 m/s
Explanation:
f = actual frequency emitted by the parked car's horn = 440 Hz
V = speed of sound = 342 m/s
f' = frequency of the horn observed by you = 466 Hz
v = speed of your car moving towards the parked car = ?
frequency of the horn observed by you is given as


v = 19.08 m/s
<span>Place a test charge in the middle. It is 2cm away from each charge.
The electric field E= F/Q where F is the force at the point and Q is the charge causing the force in this point.
The test charge will have zero net force on it. The left 30uC charge will push it to the right and the right 30uC charge will push it to the left. The left and right force will equal each other and cancel each other out.
THIS IS A TRICK QUESTION.
THe electric field exactly midway between them = 0/Q = 0.
But if the point moves even slightly you need the following formula
F= (1/4Piε)(Q1Q2/D^2)
Assume your test charge is positive and make sure you remember two positive charges repel, two unlike charges attract. Draw the forces on the test charge out as vectors and find the magnetude of the force, then divide by the total charge to to find the electric field strength:)</span>
The car’s velocity as a function of time is b + 2ct and the car’s average velocity during this interval is 0.9 m/s.
<h3>Average velocity of the car</h3>
The average velocity of the car is calculated as follows;
x(t) = a + bt + ct2
v = dx/dt
v(t) = b + 2ct
v(0) = -10.1 m/s + 2(1.1)(0) = -10.1 m/s
v(10) = -10.1 + 2(1.1)(10) = 11.9 m/s
<h3>Average velocity</h3>
V = ¹/₂[v(0) + v(10)]
V = ¹/₂ (-10.1 + 11.9 )
V = 0.9 m/s
Thus, the car’s velocity as a function of time is b + 2ct and the car’s average velocity during this interval is 0.9 m/s.
Learn more about velocity here: brainly.com/question/4931057
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