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3 years ago

Compare the speed of light in water to the speed of sound in water

1 answer:

6 0

Answer:As a rule sound travels slowest through gases,faster through liquids,and fastest through solids.The speed of light as it travels through air and space is much faster than that of sound; it travels at 300 million meters per second or 273,400 miles per hour.Speed of light in water = 226 million m/s or 205,600 mph.

Explanation:dont have one

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3 years ago

An unbalanced force of 500 N is applied to a 75 kg object. What is the acceleration of the object?

NemiM [27]

1. Define Newtons second law of motion (this will help put things into perspective)

2.Get the mass of the object (in this case 75 kg)

3.The net force acting on the object...find it (in this case 500 N)

4.Change the equation to F=ma (500=75a)

5.Divide both sides by 75 and that is the acceleration.

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3 years ago

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What is cosmic background radiation?

Kobotan [32]

Cosmic background radiation is electromagnetic radiation from the sky with no discernible source. The origin of this radiation depends on the region of the spectrum that is observed.

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3 years ago

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A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 4.444 N

Dmitriy789 [7]

Answer:

The value of third charge is 0.8μC.

Explanation:

Given that.

Magnitude of net force=4.444 N

According to figure,

Suppose, First charge = 2.4 μC

Second charge = 6.2 μC

Distance r₁ = 9.8 cm

Distance r₂ = 2.1 cm

We need to calculate the value of r

Using Pythagorean theorem

$r=\sqrt{(r_{1})^2+(r_{2})^2}$

Put the value into the formula

$r=\sqrt{(9.8)^2+(2.1)^2}$

$r=10.02\ cm$

We need to calculate the force

Using formula of force

$F_{12}=\dfrac{kq_{1}q_{2}}{(r)^2}$

Force F₁₂,

$F_{12}=\dfrac{9\times10^{9}\times2.4\times10^{-6}\times6.2\times10^{-6}}{(10.02\times10^{-2})^2}$

$F_{12}=13.33\ N$

$F_{21}=-13.33\ N$

Force F₂₃,

$F_{23}=\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(10.02)^2}$

We need to calculate the value of third charge

$F_{net}=F_{21}+F_{23}$

$4.444=-13.33+\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(5.01)^2}$

$q_{3}=\dfrac{(4.444+13.33)\times(5.01\times10^{-2})^2}{9\times10^{9}\times6.2\times10^{-6}}$

$q_{3}=7.99\times10^{-7}\ C$

$q_{3}=0.8\times10^{-6}\ C$

Hence, The value of third charge is 0.8μC.

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3 years ago

The diameter of an atom is 1.1×10−10m and the diameter of its nucleus is 1.0×10−14m. Part A What percent of the atom's volume is

77julia77 [94]

To solve this problem we will use the basic concept given by the Volume of a sphere with which the atom approaches. The fraction in percentage terms would be given by the division of the total volume of the nucleus by that of the volume of the atom, that is,

$\% Percent = \frac{V_{nucleus}}{V_{atom}}*100$