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)
2 is the answer have a nice day <3
Answer:
461.88 N
Explanation:
Let tension in each vine be T .
The vertical component of this tension of both the vine will add up and balance the weight .
2T cos 30 = mg
2 T x cos 30 = 800
T = 800 / (2 x cos30 )
= 461.88 N
First of all you have that the force that is in cable in its vertical posibion is:
Fy = m * g
Fy = (26) * (9.8) = 254.8 N
Then, the force in horizontal direction will be:
Tan (x) = Fy / Fx
Clearing:
Fx = Fy / tan (x)
Substituting values:
Fx = (254.8) / (0.17 / 4)
Fx = 5995.29 N
Answer:
The horizontal force that would be necessary to displace its position 0.17 m to one side is 5995.29 N
