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WINSTONCH [101]
3 years ago
11

An electric heating element is connected to a 110 V circuit and a current of 3.2 A is flowing through the element. How much ener

gy is used up during a period of 5 hours by the element
Physics
2 answers:
Olin [163]3 years ago
4 0
Energy, E is given by =  IVt

Where E is Energy in Juoles.
I is Current in Amperes,
t is time in seconds,

I = 3.2 A,    V = 110 V,    t = 5 hours = 5 *3600 seconds.  1 hour = 3600s

E =  3.2 * 110 * 5 * 3600

E = 6 336 000 Juoles

E = 6.336 * 10^6 J  =  6.336 MJ
borishaifa [10]3 years ago
3 0
V=110 \ V \\ I=3,2 \ A \\ t=5 \ h=18 \ 000 \ s \\ \boxed{E-?} \\ \bold{Solving:} \\ \boxed{E=V \cdot t \cdot I} \\ E=110 \ V \cdot 3,2 \ A \cdot 18  \ 000 \ s \\ \Rightarrow \boxed{E=6 \ 336 \ 000 \ J}
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Answer:

orbitals

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Calculate kinetic energy of a planet using 5.97 x 10^24 kg mass and v at 30.29 km s-1
kiruha [24]

Answer:

2.74 × 10^33 J

Explanation:

the formula to calculate kinetic energy is:

1/2mv²

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v= velocity (m/s)

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4 0
2 years ago
Under what conditions will a moving 0.030 kg marble and a moving 2.43 kg rock have the same kinetic energy
Anestetic [448]

Answer:

To have the same kinetic energy the speed of the marble must be 9 times the speed of rock.

Explanation:

The general formula of kinetic energy is given as follows:

K.E = \frac{1}{2}mv^{2}

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m = mass of the object

v = speed of the object

So, for the marble and rock to have same kinetic energy, we can write:

K.E_{marble} = K.E_{rock}\\\\\frac{1}{2}m_{marble}v_{marble}^{2} =  \frac{1}{2}m_{rock}v_{rock}^{2}\\\\(0.03\ kg)v_{marble}^{2} = (2.43\ kg)v_{rock}^{2}\\\\taking\ square\ root\ on\ both\ sides:\\v_{marble} = \sqrt{\frac{2.43\ kg}{0.03\ kg}}v_{rock}\\\\v_{marble} = 9\ v_{rock}

<u>Hence, to have the same kinetic energy the speed of the marble must be 9 times the speed of rock.</u>

4 0
3 years ago
How to calculate the slope of position time graph?
Mashcka [7]

Answer:

The slope of a position-time graph can be calculated as:

m=\frac{\Delta y}{\Delta x}

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m=\frac{\Delta s}{\Delta t}

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4 years ago
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