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AlexFokin [52]
3 years ago
7

Which uses more electricity a 1700 w hair dryer or a 1200 w microwave: both running for 5 minutes?

Physics
2 answers:
Umnica [9.8K]3 years ago
4 0
Hair dryer as you can tell from the difference, 500w
Artist 52 [7]3 years ago
3 0
The hair dryer dissipates 1700 joules of energy every second.

The microwave dissipates 1200 joules of energy every second.

In 5 minutes (300 seconds), the hair dryer uses up 150,000 more joules of energy than the microwave does.
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The manufacturer of a 1.5 V D flashlight battery says that the battery will deliver 9 mA for 42 continuous hours. During that ti
Sladkaya [172]

Answer:

The manufacturer of a 9V dry-cell flashlight battery says that the battery will deliver 20 mA for 80 continuous hours. During that time the voltage will drop from 9V to 6V. Assume the drop in voltage is linear with time. How much energy does the battery deliver in this 80 h interval?

Explanation:

4 0
4 years ago
How are longitudinal and transverse waves different?
Natalija [7]

Answer:

A longitudinal wave is a wave where the movement of the medium is in the same direction as the wave. On the other hand, a transverse wave is a wave where the movement of the medium is at a right angle to the wave direction.

Explanation:i got this right on a quiz so i know its right

3 0
4 years ago
Use the information and diagram to answer the following question.
Mumz [18]

Answer:b

Explanation:

Because I don’t know

8 0
3 years ago
A manufacturer selected a metal to use in producing a lightweight button for clothing. A metal that has a density of 2.71 g/cm3
Natali5045456 [20]

Just find the density of every metal and select the one with a density of 2.71 g/cm³ . This is:

Metal 1

ρ = m/V

ρ = 22.1 g / 3 cm³

ρ = 7.367 g / cm³

Metal 2

ρ = m/V

ρ = 42 g / 4 cm³

ρ = 10.5 g / cm³

Metal 3

ρ = m/V

ρ = 9.32 g / 5 cm³

ρ = 1.864 g / cm³

Metal 4

ρ = m/V

ρ = 8.13 g / 3 cm³

ρ = 2.71 g / cm³

<h2>R / Metal 4 was selected.</h2>
4 0
3 years ago
An element has the following natural abundances and isotopic masses: 90.92% abundance with 19.99 amu, 0.26% abundance with 20.99
sashaice [31]

<u>Answer:</u> The average atomic mass of the given element is 20.169 amu.

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of the isotopes each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i     .....(1)

We are given:

  • For isotope 1:

Mass of isotope 1 = 19.99 amu

Percentage abundance of isotope 1 = 90.92 %

Fractional abundance of isotope 1 = 0.9092

  • For isotope 2:

Mass of isotope 2 = 20.99 amu

Percentage abundance of isotope 2 = 0.26%

Fractional abundance of isotope 2 = 0.0026

  • For isotope 3:

Mass of isotope 3 = 21.99 amu

Percentage abundance of isotope 3 = 8.82%

Fractional abundance of isotope 3 = 0.0882  

Putting values in equation 1, we get:

\text{Average atomic mass}=[(19.99\times 0.9092)+(20.99\times 0.0026)+(21.99\times 0.0882)]

\text{Average atomic mass}=20.169amu

Hence, the average atomic mass of the given element is 20.169 amu.

4 0
4 years ago
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