Answer: 3.49 s
Explanation:
We can solve this problem with the following equation of motion:
(1)
Where:
is the final height of the ball
is the initial height of the ball
is the initial velocity (the ball was dropped)
is the acceleratio due gravity
is the time
Isolating :
(2)
(3)
Finally we find the time the ball is in the air:
(4)
Answer: 0.258 N
Explanation:
As the density of the object is much less than the density of water, it’s clear that the buoyant force, is greater than the weight of the object, which means that in normal conditions, it would float in water.
So, in order to get the ball submerged in water, we need to add a downward force, that add to the weight, in order to compensate the buoyant force, as follows:
F = Fb – Fg
Fb= δH20* 4/3*π*(d/2)³ * g
Fg = δb* 4/3*π*(d/2)³ *g
F= (δH20- δb) * 4/3*π*(d/2)³*g
Replacing by the values of the densities, and the ball diameter, we finally get:
F= 0.258 N
Answer:
n = 1,875
Explanation:
The speed of light in vacuum is constant (c) and in a material medium it is
v = d / t
The refractive index of a material is defined by
n = c / v
Let's look for the speed of light in the material, in general the length that light travels is known, this value is high, x = 1, when we place a block on the road, a small amount is lengthened by the length of the block, which in general is despised
These measurements are made on a digital oscilloscope that allows to stop the signals and measure their differences, that is, the zero is taken when the first ray arrives and the time for the second ray is measured,
v = d / t
v = 1 / 6.25 10⁻⁹
v = 1.6 10⁸ m / s
we calculate the refractive index
n = 3 10⁸ / 1.6 10⁸
n = 1,875
Answer:
The answer is number 2 :)
•THAT THE PROPAGATION OF SOUND WAVES NEED MEDIUM TO TRAVEL
•THE MEDIUM SHOULD POSSES ELASTICITY
•FOR THE FASTER PROPAGATION OF SOUND THE PARTICLES SHOULD BE VERY CLOSE TO EACH OTHER
