Answer:
float
Explanation:
The mass of the cube with volume 125 ml = 125 cm³= 20g.
The Mass of 125 ml of water is about 125g.
The same volume of water is heavier than the cube.
First of all, the question is vague as you haven't mentioned the medium of propagation of the wave, which is extremely crucial.
For example light travels at <span>299,792</span> km/second in vacuum, but in certain semiconductors, it travels as slow as 9 km/second. Sound, ocean and seismic waves don't exist in vacuum at all.
If you mean the maximum possible speed any of these options can attain in any medium of choice for the different options, then the answer would be
b. radio waves, which travel exactly at the speed of light in vacuum (299,792 km/second) and with an almost similar(slightly less) speed in air. (Radio waves are nothing but electromagnetic waves with low frequency)
solution:
As Given plane is flying in east direction.
It throws back some supplies to designated target.
Time taken by the supply to reach the target =10 seconds
g = Acceleration due to gravity = - 9.8 m/s²[Taken negative as object is falling Downwards]
As we have to find distance from the ground to plane which is given by d.
d =
= meters
Distance from the ground where supplies has to be land to plane = Option B =490 meters
<span>As a wave moves past a point in the ocean, water molecules move up and down as the wave goes by. They also move in another way. The answer which describes this second type of movement of water molecules is B. they move in a circular motion. </span>
Answer:
Explanation:
a )
in the regions r < R₁
charge q inside sphere of radius R₁ = 0
Applying gauss's law for electric field E at distance r <R₁
electric flux through Gaussian surface of radius r = 4π r² E
4π r² E = q / ε₀ = 0 / ε₀
E = 0
Applying gauss's law for electric field E at distance R₁ < r < R₂ .
charge q inside sphere of radius R₁ = q₁
Applying gauss's law for electric field E at distance R₁ < r < R₂
electric flux through Gaussian surface of radius r = 4π r² E
4π r² E = q₁ / ε₀
E = q₁ / 4πε₀
in the regions r> R₂
charge q inside sphere of radius R₂ = (q₁ + q₂)
Applying gauss's law for electric field E at distance r > R₂
electric flux through Gaussian surface of radius r = 4π r² E
4π r² E = (q₁ + q₂) / ε₀
E = (q₁ + q₂) /4π ε₀
b )
For electric flux to be zero at r > R₂
(q₁ + q₂) /4π ε₀ = 0
q₁ + q₂ = 0
q₁ / q₂ = - 1 .