Answer:
ΔF=125.22 %
Explanation:
We know that drag force on the car given as

=Drag coefficient
A=Projected area
v=Velocity
ρ=Density
All other quantity are constant so we can say that drag force and velocity can be given as

Now by putting the values



Percentage Change in the drag force



ΔF=125.22 %
Therefore force will increase by 125.22 %.
Answer:
W = 19.845 J
Explanation:
Work is defined as W = Fdcos
, where F is the force exerted and d is the distance. Because the direction the ball is falling is the same direction as the force itself,
= 0 deg, and since cos(0) = 1, this equation is equivalent to W = Fd. In this case, the force exerted is the weight force, which is equivalent to m * g. Substituting you get:
W = mgd = 0.810 kg * 9.8 m/s^2 * 2.5m
W = 19.845 J
Answer:
F = W + ma a> 0
Explanation:
For this exercise let's use Newton's second law
we assume the upward direction as positive
F - W = m a
F = W + ma
F = m (g + a)
In this case they indicate that the speed is less and less as it goes down, therefore the acceleration must be opposite to the speed, that is, the acceleration is upwards, consequently it is positive
We can see that since a> 0 the force F must have greater than the weight of the elevator
Answer:
7.82 s
Explanation:
Given:
Δy = 300 m
v₀ = 0 m/s
a = 9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
(300 m) = (0 m/s) t + ½ (9.8 m/s²) t²
t = 7.82 s