If an electric water heater uses 40 kW of power and runs for 8 hours, what is the total amount of energy used?

Joe first focuses his attention (and his eyes) on the tree. The focal length of the cornea-lens system in his eye must be ______

kotykmax [81]

Answer: The focal length of the cornea-lens system in his eye must be LESS THAN the distance between the front and back of his eye.

Explanation:

The human eye the front part of the eye is the CORNEA. This is the tough white transparent part of the eye that helps in the refraction of light rays. While the backside of the eye is the RETINA. This is the part of the eye when images are focused.

When a normal eye is at rest, parallel rays from a distant object are focused on the retina. The ability of the eye - lens to focus points at different distances on the retina is known as accomodation. The adjustment of the eye lens to focus objects of varying distances is brought about by the ciliary muscles. The have the ability to change the shape of the eye which leads to change in focal length.

When a person with normal vision looks at a distant object at infinity, the lens brings parallel rays to focus on the retina. Thus, the furthest point which the eye can see distinctly is called the far point of the eye and it's infinity for a normal eye. But Joe was able to focus his eye on the tree, meaning that the tree was within his near point. This is the nearest point at which an object is clearly seen. Therefore, when the effective focal length of the cornea-lens system changes, it changes the location of the image of any object in one's field of view.

5 0

3 years ago

What is the result of (305.120 + 267.443) x 0.50? How many answers can be written based on the principle of significant digits?

NNADVOKAT [17]

Answer:

The answer is 286.2815.

6 0

3 years ago

A student was walking with a mass of 40 kilograms and an acceleration of 1 m/s/s. What is the force exerted on her?

ValentinkaMS [17]

Answer:

B

Explanation:40x1=40 40=b

brainlyest plz

3 0

3 years ago

A stereo speaker is rated at P1000 = 52 W of output at 1000 Hz. At 20 Hz, the sound intensity level LaTeX: \betaβ decreases by 1

baherus [9]

Answer:

The value of the power is $P_c = 38.55 \ W$

Explanation:

From the question we are told that

The power rating $P_{1000} =P_b= 52 \ W$

The frequency is $f = 1000 \ Hz$

The frequency at which the sound intensity decreases $f_k = 20 \ Hz$

The decrease in intensity is by $\beta = 1.3 dB$

Generally the initial intensity of the speaker is mathematically represented as

$\beta_1 = 10 log_{10} [\frac{P_b}{P_a} ]$

Generally the intensity of the speaker after it has been decreased is

$\beta_2 = 10 log_{10} [\frac{P_c}{P_a} ]$

So

$\beta_1-\beta_2 = 10 log_{10} [\frac{P_c}{P_a} ]- 10 log_{10} [\frac{P_b}{P_a} ]$

=> $\beta = 10 log_{10} [\frac{P_c}{P_a} ]- 10 log_{10} [\frac{P_b}{P_a} ]= 1.3$

=> $\beta =10log_{10} [\frac{\frac{P_b}{P_a}}{\frac{P_c}{P_a}} ] = 1.3$

=> $\beta =10log_{10} [\frac{P_b}{P_c} ] = 1.3$

=> $10log_{10} [\frac{P_b}{P_c} ] = 1.3$

=> $log_{10} [\frac{P_b}{P_c} ] = 0.13$

taking atilog of both sides

$[\frac{P_b}{P_c} ] = 10^{0.13}$

=> $[\frac{52}{P_c} ] = 10^{0.13}$

=> $P_c = \frac{52}{1.34896}$

=> $P_c = 38.55 \ W$

3 0

3 years ago

refrigerant 134a enters a compressor operating at steady state as saturated vapor at 0.12 MPa and exits at 1.2 MPa and 70 C at a

Afina-wow [57]

Answer:

the power input to the compressor is 7.19Kw

Explanation:

Hello!

To solve this problem follow the steps below.

1. We will call 1 the refrigerant state at the compressor inlet and 2 at the outlet.

2. We use thermodynamic tables to determine enthalpies in states 1 and 2.

(note: Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)

through prior knowledge of two other properties such as pressure and temperature. )

h1[quality=1, P=0.12Mpa)=237KJ/Kg

h2(P=1.2Mpa, t=70C)=300.6KJ/kg

3. uses the first law of thermodynamics in the compressor that states that the energy that enters a system is the same that must come out

Q=heat=0.32kJ/s

W=power input to the compressor

m=mass flow=0.108kg/S

m(h1)+W=Q+m(h2)

solving for W

W=Q+m(h2-h1)

W=0.32+0.108(300.6-237)=7.19Kw

the power input to the compressor is 7.19Kw

7 0

3 years ago