All categories

3 years ago

If you operate a 2.0 kilowatt for 2 hours how much energy do you use

Physics

1 answer:

kap26 [50]3 years ago

3 0

Answer:

4 kWh/day

Explanation:

hope this helps!

~mina-san

You might be interested in

Resistance in wires causes electrical energy to be converted to what form of energy? sound chemical energy nuclear energy therma

ella [17]

Resistance in wires is transformed into thermal energy, you can observe this fact when you are charging your phone and feel the wire is quite hot. That is a result of the resistance.

4 0

3 years ago

An object has a mass of 0.250 kg. What is the gravitational force of on the object by the earth?

sesenic [268]

Answer:

2.4525 N

Explanation:

The earths gravity is 9.81 N/Kg

And so to work this out you would multiply 9.81 by 0.250 which equals to 2.4525N

7 0

3 years ago

Because she had serious traffic accident on Friday the 13th of last month, Felicia is conceived that all Friday the 13th will br

elena-s [515]

Answer:

Illusory correlation

Explanation:

Illusory correlation is a psychological phenomenon of sensing a correlation or relationship between a particular quantity and other variables however unrelated the two variables may be. A person may establish such connections, which are indeed false connections, as a result of over emphasis on novel incidents that are easily noticed due to them being salient

For example, when a waiter at a restaurant is being rude to a person after the person was made to walk under a ladder due to ongoing construction.

4 0

3 years ago

In the ground state of hydrogen, according to the Bohr model, an electron orbits 5.3 x 10-11 m from the nucleus. It undergoes a

Readme [11.4K]

Answer:

Explanation:

Given

radius of electron $(r)=5.3\times 10^{-11} m$

centripetal acceleration $(a_c)=9\times 10^22 m/s^2$

we know

$a_c=\frac{v^2}{r}$

$v=\sqrt{r\times a_c}$

$v=\sqrt{5.3\times 10^{-11}\times 9\times 10^{22}}$

$v=\sqrt{47.7\times 10^11}$

$v=21.84\times 10^5 m/s$

(b)For n=10

$r=100\times 5.3\times 10^{-11} m\approx 5.3\times 10^{-9} m$

$a_c=10^4\times 9\times 10^{22} m/s^2$

$a_c=9\times 10^{26} m/s^2$

$v=\sqrt{r\times a_c}$

$v=\sqrt{9\times 10^{26}\times 5.3\times 10^{-9}}$

$v=21.84\times 10^8 m/s$

8 0

3 years ago

In the following figure, electric field at y axis will be maximum at y=?

Strike441 [17]

Because of symmetry electric field component in the x axis cancels out. Now just use electric field formula and slap that sine of theta cause you want the vertical component of electric field and multiply that by two since there’s two charges. I’ve shown my work. Hope it helps✌

5 0

3 years ago

Read 2 more answers