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alexus6339
14.09.2019 •
Physics
Over a time interval of 1.71 years, the velocity of a planet orbiting a distant star reverses direction, changing from +17.3 km/s to -22.8 km/s. find (a) the total change in the planet's velocity (in m/s) and (b) its average acceleration (in m/s2) during this interval. include the correct algebraic sign with your answers to convey the directions of the velocity and the acceleration.
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Ответ:
a)40100m/s
b)-4.348x10^- m/s^2
Explanation:
to calculate the change in the planet's velocity we have to rest the speeds
ΔV=-22.8-17.3=-40.1km/s=40100m/s
A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.
When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.
Vf=Vo+a.t (1)
{Vf^{2}-Vo^2}/{2.a} =X(2)
X=Xo+ VoT+0.5at^{2} (3)\\
Where
Vf = final speed
Vo = Initial speed
T = time
A = acceleration
X = displacement
In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve
for this problem we have to convert the time interval ins seconds, we know that a year has 53926560s
t=1.71years=53926560*1.71=92214417.6
then we can use the ecuation number 1 to calculate the aceleration
Vf=-22.8km/s
Vo=17.3km/s
Vf=Vo+at
a=(vf-vo)/t
a=(-22.8-17.3)/92214417.6
a=-4.348x10^-7 km/s^2=-4.348x10^- m/s^2
Ответ:
4.23 beats / seconds
Explanation:
Frequency heard = f = f₁ - f₂
f₂ = f₁ √ ( F₂ / F₁)
f = f₁ - f₁ √ ( F₂ / F₁) = f₁ ( 1 - √ ( F₂ / F₁) = 523 Hz ( 1 - 0.9919 ) = 4.23 beats / seconds