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Nadya [2.5K]
3 years ago
13

The radius of earth is 6,370,000 m. Express this measurement in km in scientific notation with the correct number of significant

digits.
Physics
1 answer:
AlexFokin [52]3 years ago
7 0

Answer:

6.37 x 10³ Km

Explanation:

given,

Radius of earth = 6,370,000 m

we know,

1 km = 1000 m

1 m = 0.001 Km

6,370,000 m =  6,370,000 x 0.001

                       = 6,370 Km

The number 6,370 has 3 significant figure.

To transform this to an exponential number, it is necessary to move the decimal to the left so there is only one digit in front of the decimal point.

Representing the given number in scientific notation

      = 6.37 x 10³ Km

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During an investigation, a scientist heated 123.6 g of copper carbonate till it decomposed to form a black residue. The total ma
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Answer:

See below explanation

Explanation:

The correspondent chemical reaction for copper carbonate decomposed by heat is:

CuCO₃ (s) → CuO (s) + CO₂ (g)

Considering all molar mass (MM) for each element ( we consider rounded numbers) :

MM CuCO₃ = 123 g/mol

MM CuO = 79 g/mol

MM CO₂ = 44 g/mol

Statement mentions that scientis heated 123.6 g of CuCO₃ (almost a MM), until a black residue is obtained, which weights 79.6 g : this solid residue is formed by CuO, and the remaining mass (approximatelly 44 g) belongs to teh second product, this is, CO₂; as it is a gas compund, it is not certainly included on the solid residue.

So, law of conservation mass is true for this case, since: 123.6 g = 79.6 g + 44 g. As explained, on the solid residue, we don not include the 44 g, which  "escaped" from our system, since it is a gas compound (CO₂)

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3 years ago
You push your friend, whose mass is 54kg, down a hill so she can go sledding. Her acceleration is 3m/s2. Calculate the amount of
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Answer:

18 newtons

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3 years ago
A silver wire has a cross sectional area a = 2.0 mm2. a total of 9.4 × 1018 electrons pass through the wire in 3.0 s. the conduc
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This problem uses the relationships among current I, current density J, and drift speed vd. We are given the total of electrons that pass through the wire in t = 3s and the area A, so we use the following equation to to find vd, from J and the known electron density n, so: 

v_{d} =  \frac{J}{n\left | q \right |}

<span>The current I is any motion of charge from one region to another, so this is given by:

</span>I = \frac{\Delta Q}{\Delta t} = \frac{9.4x1018electrons}{3s} = 3189.73(A)

The magnitude of the current density is:

J = \frac{I}{A} = \frac{3189.73}{2x10^{-6}} = 1594.86(A/m^{2})

Being:

A=2mm^{2} = 2x10^{-6}m^{2}
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Finally, for the drift velocity magnitude vd, we find:

</span>v_{d} = \frac{1594.86}{5.8x1028\left |1.60x10^{-19}|\right } = 1.67x10^{18}(m/s)

Notice: The current I is very high for this wire. The given values of the variables are a little bit odd
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Wind in the atmosphere and currents in the ocean are caused by uneven heating that produces______. A. Convection B. The Coriolan
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A 477 g portion of soup is heated in a microwave oven from 25°C to 90°C, using radiation with a wavelength of 1.55 × 10⁻² m. Ass
zheka24 [161]

To solve this problem we will use the heat transfer equations, to determine the amount of heat added to the body. Subsequently, through the energy ratio given by Plank, we will calculate the energy of each of the photons. The relationship between total energy and unit energy will allow us to determine the number of photons

The mass of water in the soup is 477g

The change in temperate is

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Use the following equation to calculate the heat required to raise the temperature:

q = mc\Delta T

Here,

m = Mass

c = Specific Heat

q = (477)(4.184)(65)

q = 129724.92J

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The number of photons required is the rate between the total energy and the energy of each proton, then

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This energy of the photon is given by the Planck's equation which say:

E = \frac{hc}{\lambda}

Here,

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Replacing,

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Now replacing we have,

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Therefore the number of photons required for heating is 6.41*10^{27}

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