The minimum stopping distance when the car is moving at 32.0 m/s is 348.3 m.
<h3>
Acceleration of the car </h3>
The acceleration of the car before stopping at the given distance is calculated as follows;
v² = u² + 2as
when the car stops, v = 0
0 = u² + 2as
0 = 15² + 2(76.5)a
0 = 225 + 153a
-a = 225/153
a = - 1.47 m/s²
<h3>Distance traveled when the speed is 32 m/s</h3>
If the same force is applied, then acceleration is constant.
v² = u² + 2as
0 = 32² + 2(-1.47)s
2.94s = 1024
s = 348.3 m
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i would say, are sour and corrode metals
Answer:
v= 3.18 m/s
Explanation:
Given that
m= 150 g = 0.15 kg
M= 240 g = 0.24 kg
Angular speed ,ω = 150 rpm
The speed in rad/s
ω = 15.7 rad/s
The distance of center of mass from 150 g
r= 20.30 cm
The speed of the mass 150 g
v= ω r
v= 20.30 x 15.7 cm/s
v= 318.71 cm/s
v= 3.18 m/s
North-East
If the wins makes it to east, then the motor goes north Then it would go North-East
