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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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AysviL [449]
Number one- numbers of items sold. Number two- Thursday and Friday. Number three- 1,200. Number four-150
3 0
4 years ago
Convert 400 mm to m using the method of dimensional analysis
s2008m [1.1K]

Answer:

To convert 400 mm to m you can apply the formula [m] = [mm] / 1000; use 400 for mm. Thus, the conversion 400 mm m is the result of dividing 400 by 1000. 0.4

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7 0
2 years ago
Cho hai điện tích q1=q2=8.10^-7 C đặt cách nhau 5cm. Xác định cường độ điện trường tại điểm:
Aleksandr [31]

Answer: b

Explanation:

5 0
3 years ago
Una barra de plata de 335.2 g con una temperatura de 100 ºC se introduce un calorímetro de aluminio de 60 g de masa que contiene
sdas [7]

Respuesta:

0,0560 cal / gºC.

Explicación:

Cantidad de calor; (Q)

Q = mcΔt; Δt = t2 - t1

m = masa, c = capacidad calorífica específica; Δt = cambio de temperatura

c de agua = 1 cal / gºC

c de aluminio = 0,22 cal / gºC

QTotal = Q de agua + Q de aluminio

Q de agua = 450 * 1 * (26 - 23) = 1350 cal

Q de aluminio = 60 * 0.22 * (26 - 23) = 39.6 cal

QTotal = 1350 + 39,6 = 1389,6 cal

Calor perdido = calor ganado

QTotal = calor perdido

- 1389,6 = 335,2 * c * (26 - 100)

-1389,6 = −24804,8 * c

c = 1389,6 / 24804,8

c = 0,056021 cal / gºC.

Capacidad calorífica específica de la plata = 0,0560 cal / gºC.

8 0
3 years ago
A basketball with a mass of 0.5 kilograms is accelerated at 2
Paul [167]

Answer: 1N

Explanation: its not 0N.

5 0
3 years ago
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