Scientists conduct crash tests on cars to determine their safety. Suppose the maximum allowed deceleration during a crash is –23
m/s². The crash tests on the latest model sedan show a deceleration from 10 m/s to 0 m/s in just 0.4 s during a particular type of collision. Is the deceleration within the acceptable range?
A. Yes, the deceleration is 10 m/s².
B. Yes, the deceleration is 20 m/s².
C. No, the deceleration is 24 m/s².
D. No, the deceleration is 25 m/s².
A. Yes, the deceleration is 10 m/s².
B. Yes, the deceleration is 20 m/s².
C. No, the deceleration is 24 m/s².
D. No, the deceleration is 25 m/s².
1 answer:
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Answer:
Explanation:
The amplitude of the oscillation under SHM will be .5 m and the equation of
SHM can be written as follows
x = .5 sin(ωt + π/2) , here the initial phase is π/2 because when t = 0 , x = A ( amplitude) , ω is angular frequency.
x = .5 cosωt
given , when t = .2 s , x = .35 m
.35 = .5 cos ωt
ωt = .79
ω = .79 / .20
= 3.95 rad /s
period of oscillation
T = 2π / ω
= 2 x 3.14 / 3.95
= 1.6 s
b )
ω =
ω² = k / m
k = ω² x m
= 3.95² x .6
= 9.36 N/s
c )
v = ω
At t = .2 , x = .35
v = 3.95
= 3.95 x .357
= 1.41 m/ s
d )
Acceleration at x
a = ω² x
= 3.95 x .35
= 1.3825 m s⁻²
Answer:
The normal force the seat exerted on the driver is 125 N.
Explanation:
Given;
mass of the car, m = 2000 kg
speed of the car, u = 100 km/h = 27.78 m/s
radius of curvature of the hill, r = 100 m
mass of the driver, = 60 kg
The centripetal force of the driver at top of the hill is given as;
where;
Fc is the centripetal force
is downward force due to weight of the driver
is upward or normal force on the drive
Therefore, the normal force the seat exerted on the driver is 125 N.