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3 years ago

When the electrical connection to the alternator from the battery is not tight, it can cause?

Engineering

1 answer:

Oksi-84 [34.3K]3 years ago

3 0

Answer:

affects the flow of electricity

Explanation:

A loose battery terminal affects the flow of electricity. There is less power going to the electrical systems and the vehicle will not start or start sluggishly. Also, a loose battery terminal causes the car's electrical components like navigation, car lights, and audio among others to dim or fail completely.

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As Becky was driving "Old Betsy," the family station wagon, the engine finally quit, being worn out after 171,000 miles. It can

andriy [413]

Answer:

1) 4.361 x 10 raised to power 8 revolutions

2) 1.744 x 10 raised to power 9 firings

3) 2.18 x 10 raised to power 8 intake strokes

Explanation:

The step by step explanation is as shown in the attachment

8 0

4 years ago

Which of the following is NOT a line used on blueprints?

jonny [76]

Answer: Photo lines

Explanation: made more sense

4 0

3 years ago

A surveyor knows an elevation at Catch Basin to be elev=2156.77 ft. The surveyor takes a BS=2.67 ft on a rod at BM Catch Basin a

fenix001 [56]

Answer:

the elevation at point X is 2152.72 ft

Explanation:

given data

elev = 2156.77 ft

BS = 2.67 ft

FS = 6.72 ft

solution

first we get here height of instrument that is

H.I = elev + BS ..............1

put here value

H.I = 2156.77 ft + 2.67 ft

H.I = 2159.44 ft

and

Elevation at point (x) will be

point (x) = H.I - FS .............2

point (x) = 2159.44 ft - 6.72 ft

point (x) = 2152.72 ft

3 0

3 years ago

A well insulated rigid tank contains 4 kg of argon gas at 450 kPa and 30 C. A valve is opened, allowing the argon to escape unti

natima [27]

Answer:

Final mass of Argon= 2.46 kg

Explanation:

Initial mass of Argon gas ( M1 ) = 4 kg

P1 = 450 kPa

T1 = 30°C = 303 K

P2 = 200 kPa

k ( specific heat ratio of Argon ) = 1.667

assuming a reversible adiabatic process

<u>Calculate the value of the M2 </u>

Applying ideal gas equation ( PV = mRT )

P₁V / P₂V = m₁ RT₁ / m₂ RT₂

hence : m2 = P₂T₁ / P₁T₂ * m₁

= (200 * 303 ) / (450 * 219 ) * 4

= 2.46 kg

<em>Note: Calculation for T2 is attached below</em>

5 0

3 years ago

A program is seeded with 30 faults. During testing, 21 faults are detected, 15 of which are seeded faults and 6 of which are ind

Vesna [10]

Answer:

Estimated number of indigenous faults remaining undetected is 6

Explanation:

The maximum likelihood estimate of indigenous faults is given by,

$N_F=n_F\times \frac{N_S}{n_S}$ here,

$n_F$ = the number of unseeded faults = 6

$N_S$ = number of seeded faults = 30

$n_s$ = number of seeded faults found = 15

So NF will be calculated as,

$N_F=6\times \frac{30}{15}=12$

And the estimate of faults remaining is $N_F-n_F$ = 12 - 6 = 6

8 0

3 years ago