Physics, 26.08.2020 06:01, blackeyes7659

the acceleration due to gravity at the surface of the moon is 1.6m/s. if the radius of the moon is 1.7*10^6n, calculate the mass of the moon.

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answered: Guest

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

so, refractive index of medium = 300000/200000 = 1.5

1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

∞ refractive index is usually for opaque mediums. so third medium is opaque medium.

answered: Guest

Initial velocity, u = 4.9 m/sfinal velocity = v m/sinitial height, h₁ = 245 m final height, h₂ = 0a = -9.8 m/s²time taken = t secondss = ut + 0.5at²⇒ (h₂-h₁) = ut + 0.5at²⇒ 0-245 = 4.9t + 0.5×(-9.8)×t²⇒ -245 = 4.9t - 4.9t²⇒ 4.9t² -4.9t -245 =0solving it, we get t = 7.59sv = u + at = 4.9 -9.8×7.59 = 4.9 - 74.38 = -69.48 m/svelocity is 69.48 m/s downward

answered: Guest

Sounds propagates through some kind of medium. usually it's air, but it goes just as well through solid objects like walls, windows etc. a soundwave will compress the (air)molecules in front of it, and if this increased density wave hits your ears, you register it as sound. solid objects like walls are a lot denser than air, this means that they will damp a wave, but also carry it easier.

answered: Guest

The figure is missing but we can answer anyway.the correct answer is "releasing the string". in fact, when the person is holding the bow and the arrow, the kinetic energy of the arrow is still zero (because the arrow is not moving). the moment the arrow is released, it acquires a velocity v, so its kinetic energy will become different from zero: $k= \frac{1}{2}mv^2$ where m is the mass of the arrow and v its velocity. all the other options, instead, do not increase the kinetic energy of the system: - pulling farther back on the string : this will just increase the elastic potential energy of the system- holding the arrow higher: this will increase the gravitational potential energy of the system (which is given by u=mgh, where h is the height above the ground)- decreasing the mass of the arrow: this will decrease the gravitational potential energy of the system (which is given by u=mgh, where m is the mass)