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cestrela7 [59]
3 years ago
15

The wing of an airplane experiences the forces as depicted in the vector diagram to the right. Using both one and two dimensiona

l vector addition, find the result force acting on the wing. Give the magnitude and direction of the vector and draw the vector. (Show work)

Physics
1 answer:
Vedmedyk [2.9K]3 years ago
6 0

Answer:

A.) 3605.6 N

B.) 33.7 degree

Explanation:

To find the result force acting on the wing of the airplane, we need to resolve the forces into x and y components

Resolving into x component :

Sum of forces = 3500 - 500 = 3000N

Resolving into y component:

Sum of forces = 2000N

Resultant force Fr = sqrt ( Fx^2 + Fy^2)

Fr = sqrt ( 3000^2 + 2000^2 )

Fr = sqrt ( 9000000 + 4000000 )

Fr = sqrt ( 13000000)

Fr = 3605.6 N

Therefore, resultant force acting on the wing is 3605.6 N

The direction of the vector will be:

Tan Ø = Fy / Fx

Substitute Fx and Fy into the formula

Tan Ø = 2000 / 3000

Tan Ø = 0.66666

Ø = tan^-1(0. 66666)

Ø = 33.7 degree.

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Which is a scalar quantity displacement distance force acceleration​
cestrela7 [59]

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I hope this helps you.

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3 years ago
Shanti is riding on a train that is moving at a speed of 90 km/h. He is carrying a power cord for his phone that is 1.2 m long.
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3 0
3 years ago
Read 2 more answers
A high powered rifle can shoot a bullet at a speed of 1500 mi/hr. On the moon, with almost no atmosphere and an acceleration due
Amiraneli [1.4K]

Answer:

Explanation:

On the Moon :----

1500 x 1.6 = 2400 m /s is initial velocity of bullet .

g = 1.6 m /s²

v = u - gt

0 = 2400 - 1.6 t

t = 1500 s

This is time of ascent

Time of decent will also be the same

Total time of flight = 2 x 1500 = 3000 s

On the Earth : ---

v = u - a₁ t

0 = u - a₁ x 18

u = 18a₁

v² = u² - 2 x a₁ x 2743.2

0 = (18a₁ )² - 2 x a₁ x 2743.2

a₁ = 16.93

For downward return

s = ut + 1/2 a₂ x t²

2743.2 = 0 + .5 x a₂ x 31²

a₂ = 5.7 m /s²

If d be the deceleration produced by air

g + d = 16.93 ( during upward journey )

g - d = 5.7

g = (16.93 + 5.7) / 2  

= 11.315 m / s

d = 5.6 m /s²

So air is creating a deceleration of 5.6 m /s².

6 0
3 years ago
In which space, outdoors or in your classroom, would it be easier to hear a musician? Explain
bixtya [17]

Answer:

It is easier to hear a musician in the classroom than outdoors

Explanation:

It is easier to hear a musician in the classroom due to the improved acoustics provided by the walls of the classroom whereby along with the direct sound of the musician, which is the lead source of the sounds, there is an increased number of indirect sound reaching the ear in the classroom than outdoors and due to precedence effect, all the sound appear to come from the musician

In music played outside, along side the direct sound from the musician, the indirect sound that reach the ear is echoed from maybe by only the ground while the majority of the sound from the music wanders away with the wind and in other directions as well as being absorbed such that speakers will be required to improve the sound of the music outdoors.

7 0
3 years ago
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