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FinnZ [79.3K]
3 years ago
10

Earth has a surface area of 197 million square miles. Convert this area into each of the following units.

Physics
1 answer:
slamgirl [31]3 years ago
7 0

Answers:

a) 510,225,121.4 km^{2}

We know the Earth's surface area in square miles is:

197000000 mi^{2}

On the other hand, we know 1 mi=1.60934 km.

Then:

197000000 mi^{2} \frac{(1.60934 km)^{2}}{1 mi^{2}}=510,225,121.4 km^{2}

b) 510.225 Mm^{2}

In this part, we can work with the obtained value in part a:

510,225,121.4 km^{2}

And knowing 1 km=0.001 Mm

Hence:

510,225,121.4 km^{2} \frac{(0.001 Mm)^{2}}{1 km^{2}}=510.225 Mm^{2}

c) 5.1022(10)^{16} dm^{2}

Knowing 1 Mm=10^{7} dm:

510.225 Mm^{2} \frac{(10^{7} dm)^{2}}{1 Mm^{2}}=5.1022(10)^{16} dm^{2}

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Yuliya22 [10]

Answer:

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While,

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=> It is one of the condition of diffraction that the obstacle (coming in the way) must be comparable with the size of the wavelength.

=> This shows, that radio waves have a wavelength which is comparable with the size of buildings and can really easily diffract through it

=> While, X-rays are big enough to diffract through the wall.

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6 0
3 years ago
How far will it go, given that the coefficient of kinetic friction is 0.10 and the push imparts an initial speed of 3.9 m/s ?
Akimi4 [234]

Answer:19.5 m

Explanation:

Given

coefficient of kinetic Friction \mu _k=0.10

Initial speed u=3.9\ m/s

Friction is present so it tries to stop to the object and stops it completely after moving certain distance let say s

maximum deceleration provided by friction is

a_{max}=\mu _k\cdot g

a_{max}=0.1\times 3.9=0.39\ m/s^2

using equation of motion

v^2-u^2=2 as

where v=final\ velocity

u=initial velocity

a=acceleration

s=displacement

(0)-(3.9)^2=2\times (-0.39)\times s

s=19.5\ m

6 0
3 years ago
The layer of the earth where mantle convection occurs and on which the earth's crust resists is the ?
kow [346]
C. Lithosphere, I believe is correct
6 0
3 years ago
Read 2 more answers
your schools choir consisting of 10 singers gives a performance producing a sound intensity of 100 dB. At a point in the perform
laila [671]
There are two units of sound: intensity and in decibels. Decibels are not additive, you must convert it first to units of intensity (W/m²) using this formula:

dB = 10 log(I/10⁻¹²)

A.   100 dB = 10 log(I/10⁻¹²)
       Solving for I,
       I = 0.01 W/m²

      90 dB = 10 log(I/10⁻¹²)
      Solving for I,
       I = 0.001

Ratio = 0.01/0.001 = 10
<em>Thus,the choir is 10 times more intense than the soloist.</em>

B. Since there are 90 singers, there would be 9 groups of 10-person choir that produces 100 dB or 0.01 W/m². The total intensity would be

Total intensity = 0.01 W/m² (original choir) + 0.001 W/m² (soloist) + 10(0.01 W/m²) (additional 90 singers) = 0.111 W/m²
dB = 10 log(0.111/10⁻¹²) = <em>110.45 dB</em>

C. Rock concert:
    120 dB = 10 log(I/10⁻¹²)
    Solving for I,
    I = 1 W/m²

Ratio = 1/0.111 = 9
<em>Therefore, the rock concert is 9 times more intense than the choir concert.</em>


7 0
3 years ago
How much charge flows from a 6.0 v battery when it is connected across a completely discharged 15.7 μf capacitor?
Lilit [14]
The equation Q=CV (Charge = product of Capacitance and potential difference) tells us that the maximum charge that can be stored on a capacitor is equal to the product of it's capacitance and the potential difference across it. In this case the potential difference across the capacitor will be 6.0V (assuming circuit resistance is negligable) and it has a capacitance of 15.7<span>μf or </span> 15.7x10^-6f, therefore charge equals (15.7x10^-6)x6=9.42x10^-5C (Coulombs).


4 0
3 years ago
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