All categories

3 years ago

How does a car batteray NOT die?

Engineering

1 answer:

erik [133]3 years ago

6 0

Answer:

bye hooking plugs up to it to amp it up

You might be interested in

Why would an aerospace engineer limit the maximum angle of deflection of the control surfaces?

WITCHER [35]

Answer:

You can create high drag which allows a steeper angle without increasing your air speed on landing. you can reduce the length of landing role. Flaps are also used to increase the drag they are retracted when they are not needed. it is adviseable to down he flaps during the time of take off.

4 0

3 years ago

Ball Throws

professor190 [17]

Jake because the more speed, the more kinetic energy which = force.

3 0

3 years ago

A customer executive from Southeast Asia will visit your HQ facility and meet with your

Rufina [12.5K]

<u>Answer</u>:

Say no!

Explanation:

<u>Remember</u>, this scenario is an ethical dilemma. Been ethical implies doing what's professionally right, and as such saying no to that request is the right need to do. In the workplace a company's policies should be obeyed always by individual employees even though some may resent it.

7 0

3 years ago

The minimum size for overhead service conductors shall not be smaller than ____ AWG copper. A. 4 B. 6 C. 8 D. 10

ivolga24 [154]

Answer: C. 8AWG

Explanation: According to the NEC system

4 0

3 years ago

A 4,000-km^2 watershed receives 102cm of precipitation in one

Svetradugi [14.3K]

Answer:

1) The change in storage of the catchment is 707676800 cubic meters.

2) The runoff coefficient of the catchment is 0.83.

Explanation:

The water budget equation of the catchment can be written as

$P+Q_{in}=ET+\Delta Storage+Q_{out}+I$

where

'P' is volume of precipitation in the catchment = $Area\times Precipitation$

$Q_{in}$ Is the water inflow

ET is loss of water due to evapo-transpiration

$\Delta Storage$ is the change in storage of the catchment

$Q_{out}$ is the outflow from the catchment

I is losses due to infiltration

Applying the values in the above equation and using the values on yearly basis (Time scale is taken as 1 year) we get

$4000\times 10^{6}\times 1.02+0=0.40\times 4000\times 10^{6}+\Delta Storage+34.2\times 3600\times 24\times 365\times 5.5\times 10^{-9}\times 4000\times 10^{6}\times 3600\times 24\times 365$

$\therefore \Delta Storage=707676800m^3$

Part b)

The runoff coefficient C is determined as

$C=\frac{P-I}{P}$

where symbols have the usual meaning as explained earlier

$\therefore C=\frac{102-5.5\times 10^{-7}\times 3600\times 24\times 365}{102}=0.83$

5 0

3 years ago