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ser-zykov [4K]
3 years ago
8

The average lifespan of an incandescent lightbulb (at 60 W) is 1,200 hours. How much energy does the incandescent lightbulb use

during its lifetime? (Answer in (60W = 0.06kW)
Physics
1 answer:
grigory [225]3 years ago
3 0

Answer:

0072.00kW

Explanation:

1,200 hours multiply by 60W will give you 72000kW. Convert it to kW. That will give you 0072.00kW.

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a student calculates experimentally the value of density of an iron as 4.4 gcm³. if the actual density of an iron is 7.6 gcm³, c
ser-zykov [4K]
Hola!

Percentage Error is a measurement of the discrepancy between an observed and a true, or accepted value.

[ refer the attachment. ]

According to Question,

% error = \frac{7.4 - 7.6}{7.6} × 100

= 2.631 % = 2.7 % (approximately.)

hope it helps!

5 0
3 years ago
What is the displacement of the object from 2 seconds to 6 seconds?
masya89 [10]

the answer would be 4

8 0
3 years ago
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The table below shows the right ascensions of two stars at a location.
noname [10]
Star 1 - 4 hours right ascension
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Thus,
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7 0
3 years ago
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Explain why average velocity in one dimension can be positive or negative.
Softa [21]
Imagine an object is moving in one dimension on a number line, and for this we'll say that the numbers on the line are a metre apart. If the object moves from 2 m to 7 m, the change in position is 7-2=+5 metres. But if the object moves back from 7 m to 2 m, the change in position is 2-7=-5 metres. since velocity =  \frac{change in position}{time}, and time is always positive, velocity will be positive in one direction and negative in the other direction.
3 0
3 years ago
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A child (approximately 4 years old) takes her metal “slinky Toy” (a flexible coiled metal spring) and does various tests to dete
anzhelika [568]

Answer:

248

Explanation:

L = Inductance of the slinky = 130 μH = 130 x 10⁻⁶ H

l = length of the slinky = 3 m

N = number of turns in the slinky

r = radius of slinky = 4 cm = 0.04 m

Area of slinky is given as

A = πr²

A = (3.14) (0.04)²

A = 0.005024 m²

Inductance is given as

L = \frac{\mu _{o}N^{2}A}{l}

130\times 10^{-6} = \frac{(12.56\times 10^{-7})N^{2}(0.005024)}{3}

N = 248

8 0
3 years ago
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