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kipiarov [429]
3 years ago
5

Please help!!!!!! 26 points

Physics
2 answers:
Mashutka [201]3 years ago
8 0
It’s like a fusion. They all come together to make white
Dovator [93]3 years ago
7 0
The color of the sun is white. I believe the sun has allll the colors of the rainbow which combines to make white.
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Suppose a family replaces ten 60-watt incandescent bulbs with ten 30-watt
mixer [17]

Initially there were 10 bulbs of 60 Watt power

So total power of all bulbs = 60 * 10 = 600 W

now each bulb used for 4 hours daily

so total energy consumed daily

E = p*t

E = 600*4 = 2400Wh

E = 2.4kWH

now we have total power consumed in 1 year

E = 365 * 2.4 = 876kWh

cost of electricity = 10 cents/ kWh

so total cost of energy for one year

P1  = 876*10 = 8760cents = 87.60

Now if all 60 Watt bulbs are replaced by 30 Watt bulbs

So total power of all bulbs = 30 * 10 = 300 W

now each bulb used for 4 hours daily

so total energy consumed daily

(The next steps are in the pic.)

5 0
3 years ago
What is described as the flow of charged particles?
rodikova [14]

Answer:

An electric current

Explanation:

8 0
3 years ago
Read 2 more answers
What does charge does a balloon have when its rubbed on your hair?
emmainna [20.7K]

Answer:

im pretty sure thats called friction, or did you mean something else?

3 0
3 years ago
A cube with sides of length 2cm has a mass of 7.36g . calculate the density of the cube
kondor19780726 [428]
Density = 7.36 grams ÷ (2 cm × 2 cm × 2cm) = 0.92 g/cm^3

7 0
3 years ago
A 200. kg object is pushed 12.0 m to the top of an incline to a height of 6.0 m. If the force applied along the incline is 3000.
Nataliya [291]

Answer:

Approximately 1.2 \times 10^{4}\; {\rm J} (assuming that g = 9.81\; {\rm N \cdot kg^{-1}}.)

Explanation:

The strength of the gravitational field near the surface of the earth is approximately constant: g = 9.81\; {\rm N \cdot kg^{-1}}.

The change in the gravitational potential energy ({\rm GPE}) of an object near the surface of the earth is proportional to the change in the height of this object. If the height of an object of mass m increased by \Delta h, the {\rm GPE} of that object would have increased by m\, g\, \Delta h.

In this question, the height of this object increased by \Delta h = 6.0\; {\rm m}. The mass of this object is m = 200\; {\rm kg}. Thus, the {\rm GPE} of this object would have increased by:

\begin{aligned}& (\text{Change in GPE}) \\ =\; & m\, g\, \Delta h \\ =\; & 200\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times 6.0\; {\rm m} \\ \approx\; & 1.2 \times 10^{4}\; {\rm J}\end{aligned}.

(Note that 1\; {\rm N \cdot m} = 1\; {\rm J}.)

3 0
2 years ago
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