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Soloha48 [4]
3 years ago
12

ASAP please!! The force vectors on an aircraft are as shown. Find the net (resultant force).

Physics
1 answer:
ioda3 years ago
6 0

Answer:

The magnitude of the net force is 5430N

Explanation:

I suggest to define the axes as aligned to the axis of the plane. This will require you to decompose only one vector, namely the Weight. We need two components of the W force: one in horizontal direction of the plane, the other perpendicular to it. Through a simple triangle argument you will se that the plane-horizontal component of W is

W_D=3600 N\cdot\sin 27^\circ=1634N

acting in the direction of the Drag, and the plane-perpendicular component is:

W_L=-3600N\cdot\cos 27^\circ=-3208N

with negative sign since it counteracts the Lift.

So the components of the netforce F are:

F_h=T-D-W_D=(8000-1000-1634)N=5366N\\F_v=L+W_L=(4100-3208)N=829N

The magnitude of the  net force is:

|F|=\sqrt{5366^2+829^2}N = 5430N


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Answer:

T = 540 N   (to two significant digits)

Explanation:

Let the crate dimension L be from strap attachment to floor contact

Let T be the strap tension

sum moments about the floor contact point to zero

mg[½Lcos25] - Tsin61[Lcos25] + Tcos61[Lsin25] = 0

L is common to all terms, so divides out.

½(71)(9.8)cos25 = T(sin61cos25 - cos61sin25)

T = (71)(9.8)cos25 / (2(sin61cos25 - cos61sin25))

T = 536.428020...

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A needle 5cm long can just rest on the surface of the water without wetting. What us it's weight? Surface tension is 0.007N/m​
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c

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3 years ago
sledge (including load) weighs 5000 N. It is pulled on level snow by a dog team exerting ahorizontal force on it. The coefficient
lara31 [8.8K]

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2.5 x 10^{5} J

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3 years ago
a bicycle travels at a velocity of -2.33 m/s and has a displacement of -58.3 m. how much time did it take?
Alexandra [31]
We know that velocity is equal to the total displacement of an object over time.
velocity =  \frac{displacement}{time}
Deriving from that equation, we can say that:
t =  \frac{d}{v}
Okay, so here it goes:
t =  \frac{58.3m}{2.33 \frac{m}{s} } \\ t = 25.02s
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5 0
3 years ago
You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headligh
spayn [35]

To solve this problem we will apply the concepts related to Reyleigh's criteria. Here the resolution of the eye is defined as 1.22 times the wavelength over the diameter of the eye. Mathematically this is,

\theta = \frac{1.22 \lambda }{D}

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D is diameter of the eye

D = \frac{1.22 (539nm)}{5.11 mm}

D= 1.287*10^{-4}m

The angle that relates the distance between the lights and the distance to the lamp is given by,

Sin\theta = \frac{d}{L}

For small angle, sin\theta = \theta

sin \theta = \frac{d}{L}

Here,

d = Distance between lights

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Therefore the distance is 5.367km.

4 0
3 years ago
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