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asambeis [7]
3 years ago
9

A 12 volt battery in a motor vehicle is capable of supplying the starter motor with 150 A. It is noticed that the terminal volta

ge of the battery drops to 10 V when the engine is cranked over with the starter motor. Determine the internal resistance of the 12 volt battery.
Physics
1 answer:
qwelly [4]3 years ago
4 0

Answer:0.0133 \Omega

Explanation:

Given

Voltage=12 V

Current(I)=150 A

V_{terminal}=10 V

r_{internal}=\frac{\Delta V}{I}

r_{internal}=\frac{12-10}{150}

r_{internal}=\frac{2}{150}=0.0133 \Omega

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Answer:

The required  total area is 1.48 m²

Explanation:

Given that,

Latitude = 44+° N

New Mexico,

Latitude= 35+° N

Heat capacity = 4200 J/Kg°C

Temperature = 60°C

Let us assume the input temperature 22°C

Estimate volume of water 100 ltr for 4 person.

We need to calculate the heat

Using formula of heat

H=mc_{p}\Delta T

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Put the value into the formula

H=100\times4200\times(60-22)

H=15960\ KJ...(I)

Let solar radiation for 6 hours/day.

We need to calculate the total energy per unit area

Using formula of energy

E=1000\times6\times3600\ J/m^2

E=21600\ KJ/m^2

Let the efficiency of collector is 50 %

Then,  the total energy per unit area will be

E=21600\times\dfrac{50}{100}

E=10800\ KJ/m^2....(II)

We need to calculate the required total area

Using equation (I) and (II)

A=\dfrac{H}{E}

Where, H = heat

E = total energy

Put the value into the formula

A=\dfrac{15960}{10800}

A=1.48\ m^2

Hence, The required  total area is 1.48 m²

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3 years ago
Mass of water= 357g density water= 1.0g/cm3
aliya0001 [1]

m=357g\\\\\rho=1.0\ \dfrac{g}{cm^3}\\\\\rho=\dfrac{m}{V}\to V=\dfrac{m}{\rho}\\\\\text{substitute}\\\\V=\dfrac{357g}{1.0\frac{g}{cm^3}}=375g\cdot1.0\dfrac{cm^3}{g}=375\ cm^3

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What is alpha, beta and gamma radiation made of
algol13

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Explanation:

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When is a white dwarf formed?
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A small block with mass 0.0400 kg is moving in the xy-plane. The net force on the block is described by the potentialenergy func
OlgaM077 [116]

Answer:

A= 148.92  m/s²

Explanation:

Given that

U(x,y) = (6.00  )x²  - (3.75  )y ³

m= 0.04 kg

Now force in the x-direction

Fx= - dU/dx

U(x,y) = (6.00  )x²  - (3.75  )y ³

dU/dx= 12 x

When x=0.4 m

dU/dx= 12 x 0.4 = 4.8

So we can say that

Fx= - 4.8 N

From Newtons law

F= m a

- 4.8 = 0.04 x a

a = -120 m/s²

Acceleration in x direction ,a = -120 m/s²

In y -direction

F= - dU/dy

U(x,y) = (6.00  )x²  - (3.75  )y ³

dU/dy = 0 - 3.75 x 3 y²

When y = 0.56 m

dU/dy = - 3.75 x 3 x 0.56 x 0.56

dU/dy = - 3.52

So we can say that force in y -direction

F= 3.52 N

F= m a'

3.52 = 0.04 x a'

a'=88.2 m/s²

acceleration in y direction is 88.2 m/s²

The resultant acceleration

A=\sqrt{a^2+a'^2}

A=\sqrt{120^2+88.2^2}

A= 148.92  m/s²

7 0
3 years ago
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