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3 years ago

10

________________________involves connecting a voltmeter set to read DC volts to the suspected high-resistance cable ends and cra

nking the engine.

1 answer:

AnnyKZ [126]3 years ago

4 0

Answer:

Voltage Drop Testing

Explanation:

The motives behind conducting Voltage Drop Testing is to recognize or identify the presence of unwanted resistance in the wire. or any other electrical appliance. The appliance used to calculate the voltage drop is digital volt ohmmeters.

voltage drop is the amount of loss of voltage due to resistance when electric current flow through the circuit. Thus measuring the voltage drop in respect to the amount of current flow, resistance can be measured.

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What acceleration will you give to a 24.5 kg

Ludmilka [50]

Heya!!

For calculate aceleration, lets applicate second law of Newton:

$\boxed{F=ma}$

<u>Δ Being Δ</u>

F = Force = 78,3 N

m = Mass = 24,5 kg

a = Aceleration = ?

⇒ Let's replace according the formula and clear "a":

$\boxed{a=78,3\ N / 24,5\ kg}$

⇒ Resolving

$\boxed{a=3.19\ m/s^{2}}$

Result:

The aceleration is <u>3,19 meters per second squared (m/s²)</u>

Good Luck!!

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What is the mechanical advantage of an inclined plane that has a length of 9 feet and a height of 3 feet?

Rainbow [258]

Answer:

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Explanation:

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The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me

Sav [38]

Answer:

(a): $\rm meter/ second^2.$

(b): $\rm meter/ second^3.$

(c): $\rm 2ct-3bt^2.$

(d): $\rm 2c-6bt.$

(e): $\rm t=\dfrac{2c}{3b}.$

Explanation:

Given, the position of the particle along the x axis is

$\rm x=ct^2-bt^3.$

The units of terms $\rm ct^2$ and $\rm bt^3$ should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of $\rm ct^2=meter$

Therefore, unit of $\rm c= meter/ second^2.$

(b):

Unit of $\rm bt^3=meter$

Therefore, unit of $\rm b= meter/ second^3.$

(c):

The velocity v and the position x of a particle are related as

$\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.$

(d):

The acceleration a and the velocity v of the particle is related as

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.$

(e):

The particle attains maximum x at, let's say, $\rm t_o$ , when the following two conditions are fulfilled:

$\rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.$
$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Applying both these conditions,

$\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.$

For $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c$

Since, c is a positive constant therefore, for $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0$

Thus, particle does not reach its maximum value at $\rm t = 0\ s.$

For $\rm t_o = \dfrac{2c}{3b}$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.$

Here,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Thus, the particle reach its maximum x value at time $\rm t_o = \dfrac{2c}{3b}.$

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3 years ago