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irga5000 [103]
3 years ago
11

Please help me - i beg u

Physics
1 answer:
oksano4ka [1.4K]3 years ago
8 0

Answer:

516526.863 m

Explanation:

From the question given above, the following data were obtained:

Acceleration due to gravity (g) = 2.24 m/s²

Mass (M) = 8.96×10²¹ Kg

Gravitational constant (G) = 6.67×10¯¹¹Nm²/Kg²

Radius (r) =?

The radius of the planet can be obtained as follow:

g = GM/r²

2.24 = 6.67×10¯¹¹ × 8.96×10²¹ / r²

2.24 = 5.97632×10¹¹ / r²

Cross multiply

2.24 × r² = 5.97632×10¹¹

Divide both side by 2.24

r² = 5.97632×10¹¹ / 2.24

r² = 2.668×10¹¹

Take the square root of both side

r = √2.668×10¹¹

r = 516526.863 m

Thus, the radius of the planet is 516526.863 m

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Suppose that two point charges, each with a charge of +1.00 C, are separated by a distance of 1.0 m. If the distance between the
Aleksandr-060686 [28]

Given:

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Explanation:

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\begin{gathered} r^{\prime}=\text{ 2r} \\ =2\times1 \\ =2\text{ }m \end{gathered}

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A sound is recorded at 19 decibels. What is the intensity of the sound?
sp2606 [1]

1 \times 10^{-10.1} \mathrm{Wm}^{-2} is the intensity of the sound.

Answer: Option B

<u>Explanation:</u>

The range of sound intensity that people can recognize is so large (including 13 magnitude levels). The intensity of the weakest audible noise is called the hearing threshold. (intensity about 1 \times 10^{-12} \mathrm{Wm}^{-2}). Because it is difficult to imagine numbers in such a large range, it is advisable to use a scale from 0 to 100.

This is the goal of the decibel scale (dB).  Because logarithm has the property of recording a large number and returning a small number, the dB scale is based on a logarithmic scale. The scale is defined so that the hearing threshold has intensity level of sound as 0.

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Where,

I = Intensity of the sound produced

I_{0} = Standard Intensity of sound of 60 decibels = 1 \times 10^{-12} \mathrm{Wm}^{-2}

So for 19 decibels, determine I as follows,

                   19 d B=(10 d B) \times \log _{10}\left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)

                  \log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=\frac{19}{10}

                  \log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=1.9

When log goes to other side, express in 10 to the power of that side value,

                  \left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)=10^{1.9}

                  I=1 \times 10^{-12} \mathrm{Wm}^{-2} \times 10^{1.9}=1 \times 10^{-12-1.9}=1 \times 10^{-10.1} \mathrm{Wm}^{-2}

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