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SVETLANKA909090 [29]
3 years ago
7

Technician A says that side post terminals need to be removed to inspect them for corrosion. Technician B says that side post te

rminals can be damaged by over tightening the terminal bolt. Who is correctA. Technician AB. Technician BC. Both A and BD. Neither A and B.
Physics
1 answer:
zlopas [31]3 years ago
4 0

Answer: C

Explanation: Side post terminals need to be removed to inspect them for corrosion.

Over tightening the terminal bolt can damage side post terminals.

The battery terminals and cable ends can corrode especially when the battery or car is not used for a long period of time. Corrosion limits a battery's lifespan and so should be prevented. To inspect the areas where corrosion occur on a side-post battery, you need to remove the terminals.

Also, it is true that over tightening the terminal bolt can damage the side post terminals. The covering on the battery can become twisted, and make the seals on the terminals leak.

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if the instantaneous current in the circuit is giveen by I=3 sin theta amperes, the rms value of the current will be
Kisachek [45]

Answer:

I_{rms}=2.12\ A

Explanation:

Given that,

The instantaneous current in the circuit is giveen by :

I=3\sin\theta\ A

We need to find the rms value of the current.

The general equation of current is given by :

I=I_o\sin\theta

It means, I_o=3\ A

We know that,

I_{rms}=\dfrac{I_o}{\sqrt2}\\\\=\dfrac{3}{\sqrt2}\\\\=2.12\ A

So, the rms value of current is 2.12 A.

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3 years ago
Multidimensional development means that:
Snezhnost [94]

Answer:

option D

Explanation:

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3 years ago
A small metal object is tied to the end of a string and whirled round a circular path of radius 30cm. The object makes 20 oscill
LUCKY_DIMON [66]

Answer:

(i) The angular speed of the small metal object is 25.133 rad/s

(ii) The linear speed of the small metal object is 7.54 m/s.

Explanation:

Given;

radius of the circular path, r = 30 cm = 0.3 m

number of revolutions, n = 20

time of motion, t = 5 s

(i) The angular speed of the small metal object is calculated as;

\omega = \frac{20 \ rev}{5 \ s} \times \frac{2 \pi \ rad}{1 \ rev} = \frac{40\pi \ rad}{5 \ s} = 8\pi \ rad/s = 25.133 \ rad/s

(ii) The linear speed of the small metal object is calculated as;

v = \omega r\\\\v = 25.133 \ rad/s \ \times \ 0.3 \ m\\\\v = 7.54 \ m/s

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Maybe the picture helps. The blue block represents the cart with a mass of 3 kg. The person(black block) is pulling the cart to the right with a force F so that the acceleration a is 2 m/s². According to Newton's 2nd law: F = m*a.

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