How long would it take for a 60 w lamp to consume 30.24 kilowatt-hours of power?

1 answer:

7 0

A kilowatt-hour measures energy, which is slightly different than defining power. The computation for this question comes in the following form: 30.24 kW-hour = 30240 watt-hour (because 1 kW = 1000 watts); 30240 watt-hour/60 watts = 504 hours. This is known as dimensional analysis.

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A 4.9-MeV (kinetic energy) proton enters a 0.28-T field, in a plane perpendicular to the field. Part APart complete What is the

BartSMP [9]

Answer:

$r=1.14m$

Explanation:

$\theta$ is the angle between the velocity and the magnetic field. So, the magnetic force on the proton is:

$F_m=qvBsen\theta\\F_m=qvBsen(90^\circ)\\F_m=qvB$

A charged particle describes a semicircle in a uniform magnetic field. Therefore, applying Newton's second law to uniform circular motion:

$F_m=F_c\\qvB=F_c(1)$

$F_c$ is the centripetal force and is defined as:

$F_c=m\frac{v^2}{r}$

Here $v$ is the proton's speed and $r$ is the radius of the circular motion. Replacing this in (1) and solving for r:

$qvB=\frac{mv^2}{r}\\r=\frac{mv^2}{qvB}\\r=\frac{mv}{qB}$

Recall that 1 J is equal to $6.242*10^{12}MeV$ , so:

$4.9MeV*\frac{1J}{6.242*10^{12}MeV}=7.85*10^{-13}J$

We can calculate $v$ from the kinetic energy of the proton:

$K=\frac{mv^2}{2}\\\\v=\sqrt{\frac{2K}{m}}\\v=\sqrt{\frac{2(7.85*10^{-13}J)}{1.67*10^{-27}kg}}\\v=3.06*10^{7}\frac{m}{s}$

Finally, we calculate the radius of the proton path:

$r=\frac{mv}{qB}\\r=\frac{1.67*10^{-27}kg(3.06*10^{7}\frac{m}{s})}{1.6*10^{-19}C(0.28T)}\\r=1.14m$

8 0

3 years ago

I need help if you can help me to get 10 points.

Alinara [238K]

Answer:

It allows the reader to feel more connected to her purpose

Explanation:

If Cooper is using personal examples for her argument, she is allowing the reader see more of her point of view, which can also lead them to feel connected to her purpose

6 0

3 years ago

Read 2 more answers

Energy captured during the ""photo"" part of photosynthesis is stored in _____ during the ""synthesis"" part of the process.

andre [41]

Energy captured during the ""photo"" part of photosynthesis is stored in covalent bond during the ""synthesis"" part of the process.

Explanation:

When carbon dioxide, water and sunlight are combindly processed by Plants, algae and a set of bacteria called cyanobacteria to become photoautotrophs, then the process goes is named as Photosynthesis. It generates oxygen, Glyceraldehyde-3-phosphate (G3P), common high-energy carbohydrate molecules which result into glucose, sucrose or other sugar molecules which comprises covalent energy-saving bonds.

Thus the species breakdown these molecules to exhibit energy for cellular functioning. In light-dependent processes, chlorophyll absorbs the radiation from the sunlight and converts it into chemical energy in the form of electron carrier derivatives such as ATP and NADPH. Carbohydrate molecules are constructed from carbon dioxide in light-independent processes i.e in the Calvin cycle, using the chemical energy obtained throughout the light-dependent processes.