Answer: 6s
Explanation:
Vs=32m/s speed at beginning of slowing down
Vf=0m/s stop speed
a= -6 m/s² acceleration
----------------
Use equation for acceleration :
a=(Vf-Vs)/t
a*t=Vf-Vs
t=(Vf-Vs)/a
t=(0-36)/-6
t=-36/-6
t=6 s
Answer:
μ = 0.18
Explanation:
Let's use Newton's second Law, the coordinate system is horizontal and vertical
Before starting to move the box
Y axis
N-W = 0
N = W = mg
X axis
F -fr = 0
F = fr
The friction force has the formula
fr = μ N
fr = μ m g
At the limit point just before starting the movement
F = μ m g
μ = F / m g
calculate
μ = 34.8 / (19.8 9.8)
μ = 0.18
Answers:
a) -171.402 m/s
b) 17.49 s
c) 1700.99 m
Explanation:
We can solve this problem with the following equations:
(1)
(2)
(3)
Where:
is the bomb's final jeight
is the bomb'e initial height
is the bomb's initial vertical velocity, since the airplane was moving horizontally
is the time
is the acceleration due gravity
is the bomb's range
is the bomb's initial horizontal velocity
is the bomb's fina velocity
Knowing this, let's begin with the answers:
<h3>b) Time</h3>
With the conditions given above, equation (1) is now written as:
(4)
Isolating
:
(5)
(6)
(7)
<h3>a) Final velocity</h3>
Since
, equation (3) is written as:
(8)
(9)
(10) The negative sign ony indicates the direction is downwards
<h3>c) Range</h3>
Substituting (7) in (2):
(11)
(12)