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docker41 [41]
3 years ago
9

Que formas de energía utilizas habitualmente?

Physics
1 answer:
Luda [366]3 years ago
5 0
Esta energía<span> puede ser convertida en otras, como calor para calentar agua o edificios, invernaderos etc. o electricidad. Podemos convertir la </span>energía<span> solar en eléctrica de dos </span>formas<span>: Fotovoltáica (PV): La radiación solar se convierte directamente en electricidad

hope this help mark brainliest plz</span>
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The relationship among speed, distance, and time is
alexdok [17]
Distance= speed (multiplied by) time
3 0
3 years ago
Read 2 more answers
Un objeto de 1400 g de masa se mueve bajo la acción de una fuerza constante con una aceleración de 0,5 m/s2 , sobre una superfic
igomit [66]

Answer:

if you spoke this in english i can help you out

Explanation:

3 0
3 years ago
Small rockets are used to make small adjustments in the speed of satellites. One such rocket has a thrust of 42 N. If it is fire
Over [174]

To solve this problem we will apply the concepts related to Newton's second law that relates force as the product between acceleration and mass. From there, we will get the acceleration. Finally, through the cinematic equations of motion we will find the time required by the object.

If the Force (F) is 42N on an object of mass (m) of 83000kg we have that the acceleration would be by Newton's second law.

F = ma \rightarrow a = \frac{F}{m}

Replacing,

a =\frac{42N}{83000kg}

a =5.06*10^{-4}m/s^2

The total speed change

\Delta v = v_f -v_0 \rightarrow v_f =\text{Final velocity and } v_0 = \text{Initial velocity } we have that the value is 0.71m/s

If we know that acceleration is the change of speed in a fraction of time,

a= \frac{\Delta v}{t} \rightarrow t = \frac{\Delta v}{a}

We have that,

t= \frac{0.71m/s}{5.06*10^{-4}m/s^2 }

t = 1403.16s

Therefore the Rocket should be fired around to 1403.16s

7 0
2 years ago
How can we maximise the rate of energy transfer to keep things cool?
faltersainse [42]
To do this we may use things that are good conductors - are painted dull black -
Have a air flow around them Maximised.


6 0
3 years ago
Yamin is running 50 feet of No. 14 wire (with a cross section of 4,110 cmils) to a load that draws current of 11 amps. What appr
castortr0y [4]

Resistance per 1000 feet for gauge 14 wire is given as

R = 2.525 ohm

now if wire is of length 50 feet only then the resistance is given as

R = \frac{2.525}{1000}\times 50

R = 0.126 ohm

now if 11 A current flows through the wire then the voltage drop is given by ohm's law

V = iR

V = 11 \times 0.126

V = 1.4 Volts

so most appropriate answer in given options is

A. 1.8 Volts

8 0
3 years ago
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