Answer:
Explanation:
The difference in time will be due to travel through atmosphere where speed of light slows down. If t be the thickness of atmosphere and c be the speed of light in space and μ be the refractive index of atmosphere difference in travel time will be as follows .
difference = \frac{2t\mu }{c}-\frac{2t }{c}
=\frac{2t}{c }\left ( 1-\mu \right )
Now t = 40 x 10³m ,μ = 1.000293 , c = 3 x 10⁸.
difference =\frac{2t\mu }{c}-\frac{2t }{c}
=\frac{2t}{c }\left ( \mu -1 \right )\\
=\frac{ 2\times 40\times 10^3}{3\times10^3 }\left ( 1.000293-1 \right )\\
=7.81\times 10^{-3}
s
Answer:
62.601kJ
Explanation:
We have a force with a horizontal component given by the angle of 15°. So wue can take this component to calculate the total work done, that is,

Our values are,


,
Substituting in Work equation,



Buffers neutralize the acid and the bases
Answer:
x = 0.176 m
Explanation:
For this exercise we will take the condition of rotational equilibrium, where the reference system is located on the far left and the wire on the far right. We assume that counterclockwise turns are positive.
Let's use trigonometry to decompose the tension
sin 60 =
/ T
T_{y} = T sin 60
cos 60 = Tₓ / T
Tₓ = T cos 60
we apply the equation
∑ τ = 0
-W L / 2 - w x + T_{y} L = 0
the length of the bar is L = 6m
-Mg 6/2 - m g x + T sin 60 6 = 0
x = (6 T sin 60 - 3 M g) / mg
let's calculate
let's use the maximum tension that resists the cable T = 900 N
x = (6 900 sin 60 - 3 200 9.8) / (700 9.8)
x = (4676 - 5880) / 6860
x = - 0.176 m
Therefore the block can be up to 0.176m to keep the system in balance.