Answer:
4. 10.0 m/s²
Explanation:
I) if initial velocity is 'v₀', the final velocity is 'v', the accelaration is 'a', the distance is 'L' and elapsed time if 't', then:
II) using these two equations after substitution v₀=0; v=30 and L=45:
The correct answers are as follows:
<span>1) hydrogenous sediment
2)sand and gravel
3) They rapidly break down at surface temperatures and pressures.</span>
Answer:
1.) 4m
2.) 37 m
3.) 62m
4.) 2.5 s
Explanation:
1.) Given that the
Thinking distance = 1m
Breaking distance = 3m
Stopping distance = breaking distance + thinking distance
Stopping distance = 1 + 3 = 4m
2.) Given that the
Stopping distance = 52 m
Thinking distance = 15m
Breaking distance = 52 - 15 = 37m
3.) The stopping distance = 76m
Thinking distance = 14m
Breaking distance = 76 - 14 = 62m
It take the brakes 62m to slow the car down to a stop.
4.) Given that a lorry travels 28m when stopping from a speed of 4m/s. If its braking distance was 18m, what was the driver’s reaction time?
Thinking = stopping distance - braking distance
Thinking distance = 28 - 18 = 10m
Speed = distance/time
4 = 10/reaction time
Reaction time = 10/4
Reaction time = 2.5 s
5.) Question incomplete
Answer:
b) 16 cm
Magnification, m = v/u
3 = v/u
⇒ v = 3u
Lens formula : 1/v – 1/u = 1/f
1/3u = 1/u = 1/12
-2/3u = 1/12
⇒ u = -8 cm
V = 3 × (-8) = -24
Distance between object and image = u – v = -8 – (-24) = -8 + 24 = 16 cm
Explanation:
