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
He noticed that the outlines of the continents looked like they could fit together similar to puzzle pieces
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A) Net horizontal force:
Fx= -35 + 25 + -10N
Net vertical force
Fy= 50 - 35 = 15N
Net force
F^2 = fX^2 +Fy^2
F= 18N
B.) Acceleration of system
a=18/5=3.6m/s^2
C). Equilibriant force must be equal to the net force above
F1= 18N
D.) Acceleration becomes half
a1= a/2 =1.8m/s^2
E.) Equilibrant force also doubles
F" = 2F' = 36N
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False, atoms with the the same number of protons but a differing number of neutrons is called an isotope.
An isomer is a compound that has the same atomic makeup as another, but in a different arrangement.
60.9 - 4.9= 56
56 is the net force
using the formula F = ma
56= 8a
a = 7
acceleration is 7ms^-2