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3 years ago

F a light bulb produces 60.0 J of electrical energy and is 45% efficient, how much energy does it use?

2 answers:

7 0

The answer is A
60 J = 45%
So meaning it wants to know all 100%
So it will be a little over 120 and the closest one is 133
(please don’t do what do!!)

liberstina [14]3 years ago

6 0

Answer:

133 j

Explanation:

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Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 4 MPa, 500◦C, and 80 m/s, and the exit

Cerrena [4.2K]

Answer:

a) ΔEC=-23.4kW

b)W=12106.2kW

c) $A=0.01297m^2$

Explanation:

The kinetic energy is defined as:

$\frac{m*vel^2}{2}$ (vel is the velocity, to differentiate with v, specific volume).

The kinetic energy change will be: Δ ( $\frac{mvel^2}{2}$ )= $\frac{m*vel_2^2}{2}-\frac{m*vel_1^2}{2}$

Δ ( $\frac{mvel^2}{2}$ )= $\frac{m}{2}*(vel_2^2-vel_1^2)$

Where 1 and 2 subscripts mean initial and final state respectively.

Δ( $\frac{mvel^2}{2}$ )= $\frac{12\frac{kg}{s}}{2}*(50^2-80^2)\frac{m^2}{s^2}=-23400W=-23.4kW$

This amount is negative because the steam is losing that energy.

Consider the energy balance, with a neglective height difference: The energy that enters to the turbine (which is in the steam) is the same that goes out (which is in the steam and in the work done).

$H_1+\frac{m*vel_1^2}{2}=H_2+\frac{m*vel_2^2}{2}+W\\W=m*(h_1-h_2)+\frac{m}{2} *(vel_1^2-vel_2^2)$

We already know the last quantity: $\frac{m}{2} *(vel_1^2-vel_2^2)$ = $-$ Δ ( $\frac{mvel^2}{2}$ )=23400W

For the steam enthalpies, review the steam tables (I attach the ones that I used); according to that, $h_1=h(T=500C,P=4MPa)=3445.3\frac{kJ}{kg}$

The exit state is a liquid-vapor mixture, so its enthalpy is:

$h_2=h_f+xh_{fg}=289.23+0.92*2366.1=2483.4\frac{kJ}{kg}$

Finally, the work can be obtained:

$W=12\frac{kg}{s}*(3445.3-2438.4)\frac{kJ}{kg} +23.400kW)=12106.2kW$

C) For the area, consider the equation of mass flow:

$m=p*vel*A$ where $p$ is the density, and A the area. The density is the inverse of the specific volume, so $m=\frac{vel*A}{v}$

The specific volume of the inlet steam can be read also from the steam tables, and its value is: $0.08643\frac{m^3}{kg}$ , so:

$A=\frac{m*v}{vel}=\frac{12\frac{kg}{s}*0.08643\frac{m^3}{kg}}{80\frac{m}{s}}=0.01297m^2$

Download pdf

7 0

3 years ago

Three forces are applied to a solid cylinder of mass 12 kg (see the drawing). The magnitudes of the forces are F1 = 15 N, F2 = 2

crimeas [40]

Answer:

α = 13.7 rad / s²

Explanation:

Let's use Newton's second law for rotational motion

∑ τ = I α

we will assume that the counterclockwise turns are positive

F₁ 0 + F₂ R₂ - F₃ R₃ = I α

give us the cylinder moment of inertia

I = ½ M R₂²

α = (F₂ R₂ - F₃ R₃) $\frac{2}{M R_2^2}$

let's calculate

α = (24 0.22 - 13 0.10) $\frac{2}{12 \ 0.22^2}$ 2/12 0.22²

α = 13.7 rad / s²

6 0

3 years ago

Will mark as brainliest if correct!!!!!

irakobra [83]

Refraction refers to C. the bending of light rays when they pass from one medium into another

Explanation:

Refraction is a phenomenon typical of wave. Refraction occurs when a wave travels through the boundary between two different mediums. When this occurs, the wave changes speed, wavelength and direction (but the frequency remains the same).

In particular, the direction of the refracted ray is determined by Snell's Law:

$n_1 sin \theta_1 = n_2 sin \theta_2$