2. Is right..... if not then go with 1
Answer:
The shortest distance in which you can stop the automobile by locking the brakes is 53.64 m
Explanation:
Given;
coefficient of kinetic friction, μ = 0.84
speed of the automobile, u = 29.0 m/s
To determine the the shortest distance in which you can stop an automobile by locking the brakes, we apply the following equation;
v² = u² + 2ax
where;
v is the final velocity
u is the initial velocity
a is the acceleration
x is the shortest distance
First we determine a;
From Newton's second law of motion
∑F = ma
F is the kinetic friction that opposes the motion of the car
-Fk = ma
but, -Fk = -μN
-μN = ma
-μmg = ma
-μg = a
- 0.8 x 9.8 = a
-7.84 m/s² = a
Now, substitute in the value of a in the equation above
v² = u² + 2ax
when the automobile stops, the final velocity, v = 0
0 = 29² + 2(-7.84)x
0 = 841 - 15.68x
15.68x = 841
x = 841 / 15.68
x = 53.64 m
Thus, the shortest distance in which you can stop the automobile by locking the brakes is 53.64 m
Answer:
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Answer:
10 cm/sec
Explanation:
using formula velocity = frequency * wave length
where: f=1/T
1/1=1
so frequency will be 1HZ ( due to 1 oscillation in 1 sec)
wave length will be 10 cm because while sloshing cup hold (1/2) half full length.
velocity = 1*10 = 10cm/sec.
velocity = 10 cm/sec
A) Using:
2as = v² - u², where v will be 0 at max height
s = -(160)² / 2 x -32.174
s = 397.8 ft
b) Using:
s = ut + 1/2 at²
256 = 160t - 16.1t²
solving for t,
t = 2.0, t = 7.9
Now, v = u + at, for both times:
v(2) = 160 - 32.174(2)
v(2) = 95.7 ft/sec on the way up
v(7.9) = 160 - 32.174(7.9)
v(7.9) = -94.3 ft/sec; 94.3 ft/sec on the way down
c) -32.174 ft/s², which is the acceleration due to gravity.
d) s = 0
0 = 160t - 1/2 x 32.174t²
t = 9.94 seconds
