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Anastaziya [24]
3 years ago
10

WHOEVER ANSWERS RIGHT GETS BRAINLY!!

Physics
1 answer:
Mademuasel [1]3 years ago
6 0

Answer:

I think the answer is A

Explanation:

I need this brainliest answer please

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An 85,000 kg stunt plane performs a loop-the-loop, flying in a 260-m-diameter vertical circle. at the point where the plane is f
konstantin123 [22]
A) When the plane is flying straight down, there are three forces acting on it:
- the centripetal force  F=m \frac{v^2}{r}, directed toward the center of the circle (so, horizontally)
- the weight of the plane: W=mg, downward, so vertically
- a third force, given by the propulsion of the plane, which is accelerating it towards the ground (because the problem says that the plane has an acceleration of a=12 m/s^2 towards the ground)

The radius of the circle is r= \frac{260 m}{2} = 130 m, so the centripetal force acting on the plane is
F_c=m \frac{v^2}{r} = \frac{(85000 kg)(55 m/s)^2}{130 m}=1.98 \cdot 10^6 N
On the vertical axis, we have two forces: the weight
W=mg=(85000 kg)(9.81 m/s^2)=8.34 \cdot 10^5 N
and the other force F given by the propulsion. Since we know that their sum should generate an acceleration equal to a=12 m/s^2, we can find the magnitude of this other force F by using Newton's second law:
F+mg=ma
F=m(a-g)=(85000kg)(12 m/s^2-9.81 m/s^2)=1.86 \cdot 10^5 N

So, the net force acting on the plane will be the resultant of the centripetal force (acting in the horizontal direction) and the two forces W and F (acting in the vertical direction):
R= \sqrt{(F_c^2+(W+F)^2}=
= \sqrt{(1.98\cdot 10^6N)^2+(8.34 \cdot 10^5N+1.86 \cdot 10^5 N)^2}  =2.23 \cdot 10^6 N

(b) The tangent of the angle with respect to the horizontal is the ratio between the sum of the forces in the vertical direction (taken with negative sign, since they are directed downward) and the forces acting in the horizontal direction, so:
\tan \theta =  \frac{-(W+F)}{F_c}= -0.5
And so, the angle is
\theta = \arctan (-0.5)=-26.8 ^{\circ}
 
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