b) intervalul de timp t3 in care prin fata sa trece vagonul urmator
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Answer:

The rule for kilometers is that every three seconds between a lightning flash and the following thunder gives the distance to the flash in kilometers.
Explanation:
In order to use the rule of thumb to find the speed of sound in meters per second, we need to use some conversion ratios. We know there is 1 mile per every 5 seconds after the lightning is seen. We also know that there are 5280ft in 1 mile and we also know that there are 0.3048m in 1ft. This is enough information to solve this problem. We set our conversion ratios like this:

notice how the ratios were written in such a way that the units got cancelled when calculating them. Notice that in one ratio the miles were on the numerator of the fraction while on the other they were on the denominator, which allows us to cancel them. The same happened with the feet.
The problem asks us to express the answer to one significant figure so the speed of sound rounds to 300m/s.
For the second part of the problem we need to use conversions again. This time we will write our ratios backwards and take into account that there are 1000m to 1 km, so we get:

This means that for every 3.11s there will be a distance of 1km from the place where the lightning stroke. Since this is a rule of thumb, we round to the nearest integer for the calculations to be made easily, so the rule goes like this:
The rule for kilometers is that every three seconds between a lightning flash and the following thunder gives the distance to the flash in kilometers.
The partial pressures of HBr when the system reaches equilibrium is 2.4 X 10⁻¹¹ atm
<u>Explanation:</u>
H₂ + Br₂ ⇒ 2HBr
PH₂ = 0.782atm
PBr₂ = 0.493atm
Kp = (PHBr)²/ (PH₂) (PBr₂) = 1.4 X 10⁻²¹
At equilibrium:
Let 2x = pressure of HBr
PH₂ = 0.782 -x
PBr₂ = 0.493 - x
Kp = (2x)^2 / (0.782-x)(0.493-x)
Now, because Kp is very small, x will be very small compared to 0.782 and 0.493.
Then,
Kp = 1.4X10⁻²¹ = (4x²) / (0.782)(0.493)
x = 1.2X10⁻¹¹
PHBr = 2x = 2.4 X 10⁻¹¹ atm
Therefore, the partial pressures of HBr when the system reaches equilibrium is 2.4 X 10⁻¹¹ atm
From the starting depth to the surface, the vertical distance is 35 ft.
From the surface to the peak of the jump, the vertical distance is 27 ft.
From the peak of the jump to the surface, the vertical distance is 27 ft.
From the surface to the ending depth, the vertical distance is 18 ft.
Then the total vertical distance is ...
35 ft + 27 ft + 27 ft + 18 ft = 107 ft
r = radius of the circle of the ride = 3.00 meters
v = linear speed of the person during the ride = 17.0 m/s
m = mass of the person in angular motion in the ride
L = angular momentum of the person in the ride = 3570 kg m²/s
Angular momentum is given as
L = m v r
inserting the values
3570 kg m²/s = m (17 m/s) (3.00 m)
m = 3570 kg m²/s/(51 m²/s)
m = 7 kg
hence the mass comes out to be 7 kg