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Svetradugi [14.3K]
3 years ago
9

PROJECTILE MOTION-Why isn't there any accelertion in the x direction while there is an acceleration of -g in the y direction? ..

.?
Physics
1 answer:
Ahat [919]3 years ago
5 0
There is no acceleration of g in the x direction because the gravitational acceleration points downward. Also, on most studies we ignore the tidal forces since we are dealing with small bodies compared to the size of the earth.

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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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}
 
7 0
3 years ago
The amplitude of a sound is the A. frequency of the sound. B. magnitude of displacement of a sound pressure wave. C. psychologic
AlladinOne [14]

Answer:

Option B. magnitude of displacement of a sound pressure wave

Explanation:

Amplitude is simply the maximum displacement of a wave from its mean position.

6 0
3 years ago
Read 2 more answers
An archer shoots an arrow at a 75.0 m distant target; the bull’s-eye of the target is at same height as the release height of th
jeyben [28]

Answer:

the shooting angle ia 18.4º

Explanation:

For resolution of this exercise we use projectile launch expressions, let's see the scope

      R = Vo² sin (2θ) / g

      sin 2θ = g R / Vo²

      sin 2θ = 9.8 75/35²

      2θ = sin⁻¹ (0.6)

      θ = 18.4º

To know how for the arrow the tree branch we calculate the height of the arrow at this point

       X2 = 75/2 = 37.5 m

We calculate the time to reach this point since the speed is constant on the X axis

       X = Vox t

       t2 = X2 / Vox = X2 / (Vo cosθ)

        t2 = 37.5 / (35 cos 18.4)

        t2 = 1.13 s

With this time we calculate the height at this point

        Y = Voy t - ½ g t²

        Y = 35 sin 18.4   1.13 - ½ 9.8 1,13²

        Y = 6.23 m

With the height of the branch is 3.5 m and the arrow passes to 6.23, it passes over the branch

8 0
3 years ago
Read 2 more answers
Pretty cool, the atoms that exist in your body are some of the same atoms that have existed since the beginning of the universe.
marusya05 [52]

Answer:

was is carl sagan?

Explanation:

please forgive me if im wrong :(

8 0
2 years ago
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30 POINTS!!! CAN U AWNSER IT?? :)
solniwko [45]

Answer:

5235.84 kg

Explanation:

There is one theorem - whose proof I will never remember without having to drag calculus in there - that says that the variation of momentum is equal to the force applied times the time the application last.

F\Delta t = m \Delta v As long as the engine isn't ejecting mass - at this point it's a whole new can of worm - we know the force, we know the variation in speed, time to find the mass. But first, let's convert the variation of speed in meters per second. The ship gains 250 kmh, \Delta v = 69.4 m/s;

45 450 \cdot 8 = 69.4 m \rightarrow m = \frac{45450\cdot 8}{69.4} = 5235.84 kg

7 0
2 years ago
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