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Sidana [21]
3 years ago
10

Here is the picture of an aircraft, with the forces working on it shown. Identify the force that corresponds to the question mar

k.
A. Buoyancy
B. Atmospheric pressure
C. Gravity
D. Density

Physics
2 answers:
mel-nik [20]3 years ago
8 0
Your answer is option c
Alisiya [41]3 years ago
3 0
Answer- c: gravity
explanation- it just usb
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The prop blades of an airplane spin with a linear velocity of 875 m/s and have a centripetal acceleration on the farthest edge o
storchak [24]

The radius of the prop blade of an airplane is determined as 4.25 m.

<h3>Radius of the prop blade</h3>

The radius of the prop blade of an airplane is calculated as follows;

a = v²/r

where;

  • v is the linear speed
  • r is the radius of the prop blade
  • a is the centripetal acceleration

r = v²/a

r = (875²)/(180,000)

r = 4.25 m

Thus, the radius of the prop blade of an airplane is determined as 4.25 m.

Learn more about centripetal acceleration here: brainly.com/question/79801

#SPJ1

4 0
1 year ago
A paperboy rode his bike 3m/s. After being chased by a dog for 8 seconds he was traveling 6m/s. What is his accleration
aev [14]
I believe the answer is 0.375 m/s²
8 0
3 years ago
What are many scientific laws represented with?
salantis [7]

hypothesises i think

3 0
3 years ago
Consider a particle on which several forces act, one of which is known to be constant in time: Fi = 3.00 i +4.00 ) N. As a resul
Leviafan [203]

Given that.

F=3•i+4•j

And it from point (0,0)m to (5,6)m

dx=final position - initial position

dx=(5,6)-(0,0)

dx=(5,6)m

dx=5•i +6•j

Work done by the force is give by

W = F•dx

W=F•dx

Note that i•i=j•j=1 and i•j=j•i=0

Then,

W=(3i+4j)•(5i+6j)

Therefore,

W=3i•(5i+6j)+4j•(5i+6j)

W=15i•i+18i•j+20j•i+24j•j

W=15+0+0+24

W=39J

Then the work done by the force is 39 Joules

4 0
3 years ago
A person who weighs 685 N steps onto a spring scale in the bathroom, and the spring compresses by 0.88 cm. (a) What is the sprin
kotykmax [81]

To solve this problem it is necessary to use the concepts of Force of a spring through Hooke's law, therefore,

F = kx

Where,

k = Spring constant

x = Displacement

Initially our values are given,

F = 685N

x = 0.88 cm

PART A ) With this values we can calculate the spring constant rearranging the previous equation,

k = \frac{F}{x}

k = \frac{685}{0.88*10^-2}

k = 77840.9N/m

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F = kx

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