A drag because the lift keeps it up and the drag brings it down
First you must convert Km/hr to m/s. 90 km/hr equals 25m/s (this can be done through a conversion table by plugging in the conversion values). Then you need to see what was given:
vi (initial velocity)= 0m/s
vf (final velocity= 25m/s (90km/hr)
t (time)= 10s
Next you should find an equation that requires only the values you know and gives you the value you're looking for. Sometimes that requires two equations to be used, but in this case you only need one. The best equation for this would be a=(vf-vi)/t. Finally, plug in your values (a=(25-0)/10) to get your answer which would be 2.5m/s^2. Hope this helped!
Answer:
a) 
, b) 
Explanation:
a) The acceleration due to gravity inside the planet is:
b) The acceleration at the surface of the planet is:
This is a Doppler effect. Generally, if you move to a frequency source, you would detect an increase in frequency and when you move away from a source you would detect a decrease.
For this question, before you pass them, you are actually approaching them, so you would hear a higher frequency than the constant 300 Hz they are playing at.
Using the condensed formula:
f ' = ((v <u>+</u> vd)/(v <u>+</u> vs)) * f
Where: vd = Velocity of the detector.
vs = Velocity of the frequency source.
v = Velocity of sound in air.
f ' = Apparent frequency.
f = Frequency of source.
v = 343 m/s, vd = detector = 27.8 m/s, vs = velocity of the source =0. (the flautists are not moving).
f = 300 Hz.
There would be an overall increase in frequency, so we maintain a plus at the numerator and a minus at the denominator.
f ' = ((v + vd)/(v - vs)) * f
f ' = ((343+ 27.8)/(343 - 0)) * 300
= (370.8/343)* 300 = 324.3
Therefore frequency before passing them = 324.3 Hz.
Cheers.
Answer:
c. in the absence of an unbalanced force, an object at rest will stay at rest and an object in motion will stay in motion.
Explanation:
First law: things keep doing what they are doing, unless force is applied.
