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faust18 [17]
3 years ago
5

ITS EASY TRUST ME!!!!

Physics
2 answers:
VLD [36.1K]3 years ago
7 0
It’s most definitely b
prisoha [69]3 years ago
6 0

Answer:

it appears to be B. acceleration

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A model airplane with a mass of 0.741kg is tethered by a wire so that it flies in a circle 30.9 m in radius. The airplane engine
Tju [1.3M]

Answer:

_T}=24.57Nm

ω = 0.0347 rad/s²

a ≅ 1.07 m/s²

Explanation:

Given that:

mass of the model airplane = 0.741 kg

radius of the wire = 30.9 m

Force = 0.795 N

The torque produced by the net thrust about the center of the circle can be calculated as:

_T } = Fr

where;

F represent the magnitude of the thrust

r represent the radius of the wire

Since we have our parameters in set, the next thing to do is to replace it into the above formula;

So;

_T}=(0.795)*(30.9)

_T}=24.57Nm

(b)

Find the angular acceleration of the airplane when it is in level flight rad/s²

_T}=I \omega

where;

I = moment of inertia

ω = angular acceleration

The moment of inertia (I) can also be illustrated as:

I = mr^2

I = ( 0.741) × (30.9)²

I = 0.741 × 954.81

I = 707.51 Kg.m²

_T}=I \omega

Making angular acceleration the subject of the formula; we have;

\omega = \frac{_T}{I}

ω = \frac{24.57}{707.51}

ω = 0.0347 rad/s²

(c)

Find the linear acceleration of the airplane tangent to its flight path.m/s²

the linear acceleration (a) can be given as:

a =  ωr

a = 0.0347 × 30.9

a = 1.07223 m/s²

a ≅ 1.07 m/s²

5 0
3 years ago
Describe how these amendments and law expanded the political process:
Katarina [22]
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A 0.25 kg mass is placed on a vertically oriented spring that is stretched 0.56 meters from its equilibrium position. If the spr
Vitek1552 [10]

Answer:

The speed of the ball when it reaches equilibrium position is 3.31 m/s

Explanation:

Given;

mass of the object, m = 0.25 kg

initial displacement of the object, h₁ = 0.56 m

spring constant, k = 105 N/m

displacement at equilibrium position, h₂ = 0

initial velocity of the object, v₁ = 0

velocity of the object at equilibrium position = v₂

The change in gravitational potential energy at the equilibrium position is given as;

ΔP.E = mg(h₂ - h₁)

The change in kinetic energy of the object at the equilibrium position is given as;

ΔK.E = ¹/₂m(v₂² - v₁²)  

Apply the principle of conservation of mechanical energy;

ΔK.E  +  ΔP.E = 0

¹/₂m(v₂² - v₁²)  +  mg(h₂ - h₁) = 0

¹/₂m(v₂² - 0)  +  mg(0 - h₁) = 0

¹/₂mv₂²  -  mgh₁  =  0

¹/₂mv₂²  = mgh

¹/₂v₂² = gh

v₂² = 2gh

v₂ = √2gh

v₂ = √(2 x 9.8 x 0.56)

v₂ = 3.31 m/s

Therefore, the speed of the ball when it reaches equilibrium position is 3.31 m/s

8 0
3 years ago
What is the result of convection currents in the atmosphere?
Degger [83]
Wind is the result of convection currents. :)
4 0
3 years ago
PLEASE HELP THIS IS TIMED I WILL PICK YOU THE BRAINLIEST!!!!!!!!!!!How are secondary colors of light related to the primary colo
Kipish [7]

Explanation:

Because combining primary colors od light like red, blue, and green created secondary colors like yellow, cyan, and magenta

4 0
3 years ago
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