All categories

2 years ago

If a 9V battery produces a current of 3 A through a load, what is the resistance of the load

Engineering

2 answers:

Elden [556K]2 years ago

7 0

3 ohms hope this helps :D ❤

Andrej [43]2 years ago

5 0

Answer:

3 ohms

Explanation:

You might be interested in

g Consider a thin opaque, horizontal plate with an electrical heater on its backside. The front end is exposed to ambient air th

xxTIMURxx [149]

Answer:

The electrical power is 96.5 W/m^2

Explanation:

The energy balance is:

Ein-Eout=0

$qe+\alpha sGs+\alpha skyGsky-EEb(Ts)-qc=0$

if:

Gsky=oTsky^4

Eb=oTs^4

qc=h(Ts-Tα)

$\alpha s=\frac{\int\limits^\alpha _0 {\alpha l Gl} \, dl }{\int\limits^\alpha _0 {Gl} \, dl }$

$\alpha s=\frac{\int\limits^\alpha _0 {\alpha lEl(l,5800 } \, dl }{\int\limits^\alpha _0 {El(l,5800)} \, dl }$

if Gl≈El(l,5800)

$\alpha s=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))$

lt= 2*5800=11600 um-K, at this value, F=0.941

$\alpha s=(0.8*0.941)+0.3(1-0.941)=0.77$

The hemispherical emissivity is equal to:

$E=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))$

lt=2*333=666 K, at this value, F=0

$E=0+(1-0.7)(1)=0.3$

The hemispherical absorptivity is equal to:

$qe=EoTs^{4}+h(Ts-T\alpha )-\alpha sGs-\alpha oTsky^{4}=(0.3*5.67x10^{-8}*333^{4})+10(60-20)-(0.77-600)-(0.3*5.67x10^{-8}*233^{4})=96.5 W/m^{2}$

3 0

3 years ago

The input and output signals of a system is related by the following equation: fraction numerator d squared y over denominator d

Colt1911 [192]

Answer:

Explanation:

The given equation is :

$\frac{d^{2}y }{dx^{2} } + sin(3y) \frac{dy}{dt} + y = t\frac{df}{dt} + f$

5 0

2 years ago

What are two factors that determine the thermal energy of a substance

Sav [38]

Mass and chemical composition

6 0

2 years ago

Fully developed conditions are known to exist for water flowing through a 25-mm-diameter tube at 0.01 kg/s and 27 C. What is the

Irina18 [472]

Answer:

0.0406 m/s

Explanation:

Given:

Diameter of the tube, D = 25 mm = 0.025 m

cross-sectional area of the tube = (π/4)D² = (π/4)(0.025)² = 4.9 × 10⁻⁴ m²

Mass flow rate = 0.01 kg/s

Now,

the mass flow rate is given as:

mass flow rate = ρAV

where,

ρ is the density of the water = 1000 kg/m³

A is the area of cross-section of the pipe

V is the average velocity through the pipe

thus,

0.01 = 1000 × 4.9 × 10⁻⁴ × V

V = 0.0203 m/s

also,

Reynold's number, Re = $\frac{VD}{\nu}$

where,

ν is the kinematic viscosity of the water = 0.833 × 10⁻⁶ m²/s

thus,

Re = $\frac{0.0203\times0.025}{0.833\times10^{-6}}$

Re = 611.39 < 2000

thus,

the flow is laminar

hence,

the maximum velocity = 2 × average velocity = 2 × 0.0203 m/s

maximum velocity = 0.0406 m/s

5 0

2 years ago

A rocket is launched from rest with a constant upwards acceleration of 18 m/s2. Determine its velocity after 25 seconds

lisabon 2012 [21]

Answer:

The final velocity of the rocket is 450 m/s.

Explanation:

Given;

initial velocity of the rocket, u = 0

constant upward acceleration of the rocket, a = 18 m/s²

time of motion of the rocket, t = 25 s

The final velocity of the rocket is calculated with the following kinematic equation;

v = u + at

where;

v is the final velocity of the rocket after 25 s

Substitute the given values in the equation above;

v = 0 + 18 x 25

v = 450 m/s

Therefore, the final velocity of the rocket is 450 m/s.

5 0

2 years ago