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
A change in momentum is called impulse.
The kinetic energy of an object is given by
KE = 0.5mv²
where m is the mass and v is the velocity.
To calculate the change in kinetic energy...
Initial KE:
KEi = 0.5mVi²
where Vi is the initial velocity.
Final KE:
KEf = 0.5mVf²
where Vf is the final velocity.
ΔKE = KEf - KEi
ΔKE = 0.5mVi² - 0.5mVf²
ΔKE = 0.5m(Vf²-Vi²)
Given values:
m = 16kg
Vi = 25m/s
Vf = 20m/s
Plug in the given values and solve for ΔKE:
ΔKE = 0.5×16×(20²-25²)
ΔKE = -1800J
Answer:
An object appears white when it reflects all wavelengths and black when it absorbs them all. Red, green and blue are the additive primary colors of the color spectrum. Combining balanced amounts of red, green and blue lights also produces pure white.
Explanation:
