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3 years ago

A fully charged battery with 6 cells should indicate

Engineering

1 answer:

finlep [7]3 years ago

7 0

Answer:

Battery Voltage

Voltage refers to the amount of electrical potential your battery holds. The standard automotive battery in today's vehicles is a 12-volt battery. Each battery has six cells, each with 2.1 volts at full charge. A car battery is considered fully charged at 12.6 volts or higher.

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Steam enters a well-insulated turbine operating at steady state at 4 MPa with a specific enthalpy of 3015.4 kJ/kg and a velocity

Sedbober [7]

Answer:

attached below

Explanation:

6 0

4 years ago

The equation for the velocity V in a pipe with diameter d and length L, under laminar condition is given by the equation V=Δpdsq

Citrus2011 [14]

Answer:

Given that

$V=\dfrac{\Delta Pd^2}{32\mu L}$

LHS of above given equation have dimension $[M^oL^{1}T^{-1}]$ .

Now find the dimension of RHS

Dimension of P = $[ML^{-1}T^{-2}]$ .

Dimension of d= $[M^{0}L^{1}T^{0}]$ .

Dimension of μ = $[ML^{-1}T^{-1}]$ .

Dimension of L= $[M^{0}L^{1}T^{0}]$ .

$\dfrac{\Delta Pd^2}{32\mu L}=\dfrac{[ML^{-1}T^{-2}].[M^{0}L^{1}T^{0}]^2}{[ML^{-1}T^{-1}].[M^{0}L^{1}T^{0}]}$

$\dfrac{\Delta Pd^2}{32\mu L}=[M^0L^{1}T^{-1}]$

It means that both sides have same dimensions.

5 0

4 years ago

The atmospheric pressure at the top and the bottom of a building is read by a barometer to be 98.5 kPa and 100 kPa, respectively

Fynjy0 [20]

Answer:

127.42m

Explanation:

The air pressure can be understood as the weight exerted by the air column on a body, for this case we must remember that the pressure is calculated by the formula P=αgh, Where P=pressure, h=gravity, h= height,α=density

So what we must do to solve this problem is to find the length of the air column above and below the building and then subtract them to find the height of the building, taking into account the above the following equation is inferred

h2-h1= building height=H

$H=\frac{P1-P2}{g\alpha }$

P1=100kPa=100.000Pa

P2=98.5kPa=98.500Pa

α=1.2 kg/m^3

g=9.81m/s^2

$H=\frac{100000-98500}{(9.81)(1.2) }=127.42m$

4 0

4 years ago

A motorist is driving his car at 60km/hr when he observes that a traffic light 250m ahead turns red. The traffic light is

Alecsey [184]

Explanation:

Okay soo-

Given-

u = 60 km/hr = 60×1000/3600=50/3 m/s

t = 20 s

s = 250 m

a = ?

v = ?

Solution -

Here, acceleration is uniform.

(a) According to 2nd kinematics equation,

s = ut + ½at^2

250 = 50/3 ×20 + 0.5×a×20×20

250-1000/3=200a

(750-1000)/3=200a

a = -250/(3×200)

a = -5/12

a = 0.4167 m/s^2

The required uniform acceleration of the car is 0.4167 m/s^2.

(b) According to 1st kinematics equation

v = u + at

v = 50/3 + (-5/12)×20

v = 50/3-25/3

v = 25/3

v = 8.33 m/s

The speed of the car as it passes the traffic light is 8.33 m/s.

Good luck!

5 0

3 years ago

A. Name the major strengthening mechanisms in metals and explain the working principle under each mechanism.Give the relevant eq

Sever21 [200]

Answer:

a) Solid solution strengthening and alloying, Precipitation hardening, work hardening

b) Absence of enough crystallographic misalignment in the grain boundary region for a small-angle

Explanation:

A) strengthening mechanism

i) Solid solution strengthening and alloying:

In solid solution strengthening and alloying mechanism there is an addition of one atom of solute to another during this process, there might be substitution of interstitial point defect in crystal

also the shear stress required can be represented as: Δz = Gb√Ce^3/2

where : C = solute concentration , e = strain on material

ii) Precipitation hardening:

During precipitation hardening the alloying above the concentrate will lead to the formation of a second phase also under precipitation hardening a second phase can also be created via thermal treatments

particle bowing cab be written as : Δz = Gb / L-2x

iii) work hardening :

Dislocation caused by stress fields been generated hardens metals under the work hardening mechanism

dislocation can be represented as ; Gb √ p

where : G = shear modulus , b = Burgess vector, p = dislocation density

B) The small angle grain boundaries are not effective enough because there is less crystallographic misalignment in the grain boundary region for a small-angle

3 0

3 years ago