Answer:
0.705 m/s²
Explanation:
a) The sprinter accelerates uniformly from rest and reaches a top speed of 35 km/h at the 67-m mark.
Using newton's law of motion:
v² = u² + 2as
v = final velocity = 35 km/h = 9.72 m/s, u = initial velocity = 0 km/h, s = distance = 67 m
9.72² = 0² + 2a(67)
134a = 94.484
a = 0.705 m/s²
b) The sprinter maintains this speed of 35 km/h for the next 88 meters. Therefore:
v = 35 km/h = 9.72 m/s, u = 35 km/h = 9.72 m/s, s = 88 m
v² = u² + 2as
9.72² = 9.72² + 2a(88)
176a = 9.72² - 9.72²
a = 0
c) During the last distance, the speed slows down from 35 km/h to 32 km/h.
u = 35 km/h = 9.72 m/s, v = 32 km/h = 8.89 m/s, s = 200 - (67 + 88) = 45 m
v² = u² + 2as
8.89² = 9.72² + 2a(45)
90a = 8.89² - 9.72²
90a = -15.4463
a = -0.1716 m/s²
The maximum acceleration is 0.705 m/s² which is from 0 to 67 m mark.
Answer:
<h3>The power of headlight in series connection is 29.64 W</h3>
Explanation:
Given :
Power of headlight 
W
Power of starter 
W
Voltage of headlight and starter 
V
From equation of power,
For finding the resistance of headlight and starter,
⇒ For headlight,
Ω
⇒ For starter,
Ω
Since equivalent resistance,
Ω
So power in series is given by,
W
Answer:
6 m/s²
Explanation:
From the question given above, the following data were obtained:
Velocity (v) = 30 m/s
Time (t) = 5 s
Acceleration (a) =..?
Acceleration is defined mathematically as:
Acceleration (a) = Velocity (v) /time (t)
a = v /t
With the above formula, we can obtain the acceleration of the object as follow:
Velocity (v) = 30 m/s
Time (t) = 5 s
Acceleration (a) =..?
a= v/t
a= 30/5
a = 6 m/s²
Therefore, the acceleration of the object is 6 m/s² due East.
I think its c sorry if I’m wrong
