Answer:
i)-6.25m/s
ii)18 metres
iii)26.5 m/s or 95.4 km/hr
Explanation:
Firstly convert 90km/hr to m/s
90 × 1000/3600 = 25m/s
(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)
0 = (25)^2 + 2A(50)
0 = 625 + 100A....then moved the other value to one
-625 = 100A
Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)
(ii) Firstly convert 54km/hr to m/s
In which this is 54 × 1000/3600 = 15m/s
then apply the same formula as that in (i)
0 = (15)^2 + 2(-6.25)s
-225 = -12.5s
Hence the stopping distance = 18metres
(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question
0 = u^2 + 2(-6.25)(56)
u^2 = 700
Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s
In km/hr....26.5 × 3600/1000 = 95.4 km/hr
Answer:
You will fly forward in the bus until you hit something.
Explanation:
While standing there on the bus, you are traveling at the same speed as the bus. If the bus suddenly stops, you will still be traveling at the same speed you started with. That is until you hit something hard enough or big enough to stop you.
1. Resonance. Mechanical waves act on or through a medium, these waves can often have frequencies that are synchronized in a way that makes them act on the matter in the medium more "aggressively."
The molecules which evaporate presumably take heat away from the liquid. So, I'd disagree with the classmate. Whether the amount of cooling would differ from the usual case wherein the molecules have different speeds is another question.
I guess the argument goes something along the lines of that the faster moving and therefore most kinetically energetic molecues evaporate and take away most heat. But if there's no faster moving molecules, 'cos they all have the same speed well, then presumably they'd all take away the same amount of heat. So, maybe the cooling would be less. No cooling though ??? Hmmmm dunno .... i think not ....