<h3><u>Question</u><u>:</u></h3>
A racing car is travelling at 70 m/s and accelerates at -14 m/s^2. What would the car’s speed be after 3 s?
<h3><u>Statement:</u></h3>
A racing car is travelling at 70 m/s and accelerates at -14 m/s^2.
<h3><u>Solution</u><u>:</u></h3>
- Initial velocity (u) = 70 m/s
- Acceleration (a) = -14 m/s^2
- Time (t) = 3 s
- Let the velocity of the car after 3 s be v m/s
- By using the formula,
v = u + at, we have
- So, the velocity of the car after 3 s is 28 m/s.
<h3><u>Answer:</u></h3>
The car's speed after 3 s is 28 m/s.
Hope it helps
Answer with Explanation:
We are given that
Distance,r=0.27 m
Tangential speed=v=0.49 m/s
a.Angular speed ,
Using the formula
Time period,
b.Amplitude,A=Distance of small eraser from the center of a turnable =0.27 m
c.Maximum speed,
d.Maximum acceleration=
Let
V the speed of the girl per step
v the speed of the escalator per step
L the length between ground and floor in number of steps
V*60+v*60=L => V+v=L/60
V*90-v*90=L => V-v=L/90
By adding the equations V+v=L/60 and V-v=L/90 we get:
2V=L(3+2)/180
V=5L/360
V = L/72
Time to climb (in steps) with v=0 (escalator standing still) is:
V*t=L
t = L/V
t = L/(L/72)
t =72 steps
The girl will count 72 steps.
Answer:
well the answers are 180 and 240 obviously
Explanation:
