Complete Question
In an action movie, the villain is rescued from the ocean by grabbing onto the ladder hanging from a helicopter. He is so intent on gripping the ladder that he lets go of his briefcase of counterfeit money when he is 130 m above the water. If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?
Answer:
The speed of the helicopter is 
Explanation:
From the question we are told that
The height at which he let go of the brief case is h = 130 m
The time taken before the the brief case hits the water is t = 6 s
Generally the initial speed of the briefcase (Which also the speed of the helicopter )before the man let go of it is mathematically evaluated using kinematic equation as
Here s is the distance covered by the bag at sea level which is zero
=>
=> 
=>
Answer:
Explanation:
Impulse on an object is given by ![\mathrm{[impulse]}=F\Delta t](https://tex.z-dn.net/?f=%5Cmathrm%7B%5Bimpulse%5D%7D%3DF%5CDelta%20t)
.
However, it's also given as change in momentum (impulse-momentum theorem).
Therefore, we can set the change in momentum equal to the former formula for impulse:
.
Momentum is given by 
. Because the truck's mass is maintained, only it's velocity is changing. Since the truck is being slowed from 26.0 m/s to 18.0 m/s, it's change in velocity is 8.0 m/s. Therefore, it's change in momentum is:
.
Now we plug in our values and solve:
(two significant figures).
Answer:
Explanation:
reading of scale = reaction force of surface R
centripetal force = R - mg = m v² / R , m is mass , v is velocity and R is radius of the circular path .
R = mg + m v² / R
given ,
m v² / R = .80 mg
v² = .80 x g x R
= .8 x 9.8 x 9 = 70.56
v = 8.4 m /s
Solution:
<span>Starting with </span>g <span>equal to (1/0.6), solve for v.</span>
<span>v = 0.8c = 2.4E8 m/s</span>
