Answer:
he factor for the temporal part 1.296 107 s² = h²
m / s² = 12960 km / h²
Explanation:
This is a unit conversion exercise.
In the unit conversion, the size of the object is not changed, only the value with respect to which it is measured is changed, for this reason in the conversion the amount that is in parentheses must be worth one.
In this case, it is requested to convert a measure km/h²
Unfortunately, it is not clearly indicated what measure it is, but the most used unit in physics is m / s² , which is a measure of acceleration. Let's cut this down
the factor for the distance is 1000 m = 1 km
the factor for time is 3600 s = 1 h
let's make the conversion
m / s² (1km / 1000 m) (3600 s / 1h)²
note that as time is squared the conversion factor is also squared
m / s² = 12960 km / h²
the factor for the temporal part 1.29 107 s² = h²
For a wave:
v = fλ
v is the velocity, f is the frequency, and λ is the wavelength.
Assuming the velocity of the wave doesn't change...
If you increase its frequency, its wavelength will shorten.
Answer:
yes it doesn't matter
Explanation:
it doesn't matter because troughs and crests are the same and either can be used
Answer:
a)
Y0 = 0 m
Vy0 = 15 m/s
ay = -9.81 m/s^2
b) 7.71 m
c) 3.06 s
Explanation:
The knowns are that the initial vertical speed (at t = 0 s) is 15 m/s upwards. Also at that time the dolphin is coming out of the water, so its initial position is 0 m. And since we can safely assume this happens in Earth, the acceleration is the acceleration of gravity, which is 9.81 m/s^2 pointing downwards
Y(0) = 0 m
Vy(0) = 15 m/s
ay = -9.81 m/s^2 (negative because it points down)
Since acceleration is constant we can use the equation for uniformly accelerated movement:
Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2
To find the highest point we do the first time derivative (this is the speed:
V(t) = Vy0 + a * t
We equate this to zero
0 = Vy0 + a * t
0 = 15 - 9.81 * t
15 = 9.81 * t
t = 0.654 s
At this time it will have a height of:
Y(0.654) = 0 + 15 * 0.654 - 1/2 * 9.81 * 0.654^2 = 7.71 m
The doplhin jumps and falls back into the water, when it falls again it position will be 0 again. So we can equate the position to zero to find how long it was in the air knowing that it started the jump at t = 0s.
0 = Y0 + Vy0 * t + 1/2 * a * t^2
0 = 0 + 15 * t - 1/2 * 9.81 t^2
0 = 15 * t - 4.9 * t^2
0 = t * (15 - 4.9 * t)
t1 = 0 This is the moment it jumped into the air
0 = 15 - 4.9 * t2
15 = 4.9 * t2
t2 = 3.06 s This is the moment when it falls again.
3.06 - 0 = 3.06 s
