It’s gonna have to b since it’s decreasing
Answer:
The sport utility vehicle was traveling at V2= 11.5 m/s.
Explanation:
m1= 1090 kg
V1= 30.4 m/s
m2= 2880 kg
V2= ?
m1*V1 = m2*V2
V2= (m1*V1)/m2
V2= 11.5 m/s
Answer:
a) 
b) 
Explanation:
Given:
- upward acceleration of the helicopter,

- time after the takeoff after which the engine is shut off,

a)
<u>Maximum height reached by the helicopter:</u>
using the equation of motion,

where:
u = initial velocity of the helicopter = 0 (took-off from ground)
t = time of observation


b)
- time after which Austin Powers deploys parachute(time of free fall),

- acceleration after deploying the parachute,

<u>height fallen freely by Austin:</u>

where:
initial velocity of fall at the top = 0 (begins from the max height where the system is momentarily at rest)
time of free fall


<u>Velocity just before opening the parachute:</u>



<u>Time taken by the helicopter to fall:</u>

where:
initial velocity of the helicopter just before it begins falling freely = 0
time taken by the helicopter to fall on ground
height from where it falls = 250 m
now,


From the above time 7 seconds are taken for free fall and the remaining time to fall with parachute.
<u>remaining time,</u>



<u>Now the height fallen in the remaining time using parachute:</u>



<u>Now the height of Austin above the ground when the helicopter crashed on the ground:</u>



Answer:
COMPLETE QUESTION
A spring stretches by 0.018 m when a 2.8-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 3.0 Hz?
Explanation:
Given that,
Extension of spring
x = 0.0208m
Mass attached m = 3.39kg
Additional mass to have a frequency f
Let the additional mass be m
Using Hooke's law
F= kx
Where F = W = mg = 3.39 ×9.81
F = 33.26N
Then,
F = kx
k = F/x
k = 33.26/0.0208
k = 1598.84 N/m
The frequency is given as
f = ½π√k/m
Make m subject of formula
f² = ¼π² •(k/m
4π²f² = k/m
Then, m4π²f² = k
So, m = k/(4π²f²)
So, this is the general formula,
Then let use the frequency above
f = 3Hz
m = 1598.84/(4×π²×3²)
m = 4.5 kg
Answer:
body position 4 is (-1,133, -1.83)
Explanation:
The concept of center of gravity is of great importance since in this all external forces are considered applied, it is defined by
x_cm = 1 /M ∑
m_{i}
y_cm = 1 /M ∑ y_{i} mi
Where M is the total mass of the body, mi is the mass of each element
give us the mass and position of this masses
body 1
m1 = 2.00 ka
x1 = 0 me
y1 = 0 me
body 2
m2 = 2.20 kg
x2 = 0m
y2 = 5 m
body 3
m3 = 3.4 kg
x3 = 2.00 m
y3 = 0
body 4
m4 = 6 kg
x4=?
y4=?
mass center position
x_cm = 0
y_cm = 0
let's apply to the equations of the initial part
X axis
M = 2.00 + 2.20 + 3.40
M = 7.6 kg
0 = 1 / 7.6 (2 0 + 2.2 0 + 3.4 2 + 6 x4)
x4 = -6.8 / 6
x4 = -1,133 m
Axis y
0 = 1 / 7.6 (2 0 + 2.20 5 +3.4 0 + 6 y4)
y4 = -11/6
y4 = -1.83 m
body position 4 is (-1,133, -1.83)