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2 years ago

Which si unit whould be most appropriate for describing the length of a shoe?

Physics

1 answer:

aniked [119]2 years ago

4 0

Answer:

Milimeters

Explanation:

This kind of question must be checked with everyday elements, in this way we will analyze each of the options until we reach the appropriate:

kilograms can not be because this unit serves to measure the mass not the length.

kilometers (km) is a unit of length, but however it does not serve for this measurement since a km is equivalent to 1000 meters, therefore measuring the length of a shoe with a ruler measuring 1000 meters is very impractical.

milliliters is a unit of volume, and can often be found in the measurement of volume content of liquids, for example 700 milliliters of milk or 1000 milliliters of water. Therefore it is impractical to measure a shoe with a unit of volume.

In this way the way to measure a shoe correctly is using a ruler or measuring instrument graduated in millimeters.

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Which planet is most likely to have acid rain? A. Mercury B. Venus C. Mars D. Uranus

timofeeve [1]

Its VENUS because the planet is basicily a hot planet

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2 years ago

An air bubble at the bottom of a lake 36.0 m deep has a volume of 1.22 cm^3. If the temperature at the bottom is 5.9°C and at th

AlexFokin [52]

Answer:

volume of the bubble just before it reaches the surface is 5.71 cm³

Explanation:

given data

depth h = 36 m

volume v2 = 1.22 cm³ = 1.22 × $10^{-6}$ m³

temperature bottom t2 = 5.9°C = 278.9 K

temperature top t1 = 16.0°C = 289 K

to find out

what is the volume of the bubble just before it reaches the surface

solution

we know at top atmospheric pressure is about P1 = $10^{5}$ Pa

so pressure at bottom P2 = pressure at top + ρ×g×h

here ρ is density and h is height and g is 9.8 m/s²

pressure at bottom P2 = $10^{5}$ + 1000 × 9.8 ×36

pressure at bottom P2 =4.52 × $10^{5}$ Pa

so from gas law

$\frac{P1*V1}{t1} = \frac{P2*V2}{t2}$

here p is pressure and v is volume and t is temperature

so put here value and find v1

$\frac{10^{5}*V1}{289} = \frac{4.52*10^{5}*1.22}{278.9}$

V1 = 5.71 cm³

volume of the bubble just before it reaches the surface is 5.71 cm³

6 0

2 years ago

Sound level B in decibels is defined as

Lelechka [254]

Answer:

The approximate combined sound intensity is $I_{T}=1.1\times10^{-4}W/m^{2}$

Explanation:

The decibel scale intensity for busy traffic is 80 dB. so intensity will be

$10log(\frac{I_{1}}{I_{0}} )=80$ , therefore $I_{1}=1\times10^{8}I_{0}=1\times10^{8} * 1\times10^{-12}W/m^{2}=1\times10^{-4}W/m^{2}$

In the same way for the loud conversation having a decibel intensity of 70 dB.

$10log(\frac{I_{2}}{I_{0}} )=70$ , therefore $I_{2}=1\times10^{7}I_{0}=1\times10^{7} * 1\times10^{-12}W/m^{2}=1\times10^{-5}W/m^{2}$

Finally we add both of them $I_{T}=I_{1}+I_{2}=1\times10^{-4}W/m^{2}+1\times10^{-5}W/m^{2}=1.1\times10^{-4}W/m^{2}$ , is the approximate combined sound intensity.

3 0

2 years ago

An astronaut floating in space throws a wrench forward with the force of 10 N.

Jobisdone [24]

Answer:

10N

Explanation:

1. Every Action has an equal and opposite reAction.

2. If 10N of force is acted upon an wrench, then the wrench will react with an equal amount of force, but in the opposite direction.

7 0

2 years ago

Read 2 more answers

On a hot day, the deck of a small ship reaches a temperature of 48

AlekseyPX

The final temperature of the seawater-deck system is 990°C.

The increment in temperature adds up the thermal energy into the object. This energy is Heat energy.

The deck of a small ship reaches a temperature Ti= 48.17°C seawater on the deck to cool it down. During the cooling, heat Q =3,710,000 J are transferred to the seawater from the deck. Specific heat of seawater= 3,930 J/kg°C.

Suppose for 1 kg of sea water, the heat transferred from the system is given by

3,710,000 = 1 x 3,930 x (T - 48.17)

T = 990°C to the nearest tenth.

The final temperature of the seawater-deck system is 990°C.

Learn more about heat.

brainly.com/question/13860901

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6 0

1 year ago