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uranmaximum [27]

4 years ago

9

An airplane accelerates from rest to its required takeoff speed with an acceleration 5.0 m/s^2 after travelling on the runway a

distance of 0.75 km. Find the takeoff speed of the airplane.

1 answer:

Alex Ar [27]4 years ago

5 0

Answer:

Explanation:

.75 km = 750 m

acceleration a = 5 m/s²

distance s = 750 m

initial velocity u = 0

Final velocity v = ?

v² = u² + 2as

v² = 0 + 2 x 5 x 750

v = 86.60 m /s.

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A 10cm long wire is pulled along a U-shaped conducting rail in a perpendicular magnetic field. The total resistance of the wire

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Explanation:

From the question we are told that:

Wire Length $l=10cm=>0.10m$

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Force $F=1.0N$

Power $P=4.0W$

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A fire hose held near the ground shoots water at a speed of 6.8 m/s. At what angle(s) should the nozzle point in order that the

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So, based on the angle values that have been found, the angle of elevation of the nozzle can be <u>16° or 74°</u>.

<h3>Introduction</h3>

Hi ! This question can be solved using the principle of parabolic motion. Remember ! When the object is moving parabolic, the object has two points, namely the highest point (where the resultant velocity is 0 m/s in a very short time) and the farthest point (has the resultant velocity equal to the initial velocity). At the farthest distance, the object will move with the following equation :

$\boxed{\sf{\bold{x_{max} = \frac{(v_0)^2 \cdot \sin(2 \theta)}{g}}}}$

With the following condition :

$\sf{x_{max}}$ = the farthest distance of the parabolic movement (m)
$\sf{v_0}$ = initial speed (m/s)
$\sf{\theta}$ = elevation angle (°)
g = acceleration due to gravity (m/s²)

<h3>Problem Solving :</h3>

We know that :

$\sf{x_{max}}$ = the farthest distance of the parabolic movement = 2.5 m
$\sf{v_0}$ = initial speed = 6.8 m/s
g = acceleration due to gravity = 9.8 m/s²

<h3>What was asked :</h3>

$\sf{\theta}$ = elevation angle = ... °

Step by Step :

Find the equation value $\sf{\bold{theta}}$ (elevation angle)

$\sf{x_{max} = \frac{(v_0)^2 \cdot \sin(2 \theta)}{g}}$

$\sf{x_{max} \cdot g = (v_0)^2 \cdot \sin(2 \theta)}$

$\sf{\frac{x_{max} \cdot g}{(v_0)^2} = \sin(2 \theta)}$

$\sf{\frac{2.5 \cdot 9.8}{(6.8)^2} = \sin(2 \theta)}$

$\sf{\frac{2.5 \cdot 9.8}{(6.8)^2} = \sin(2 \theta)}$

$\sf{\frac{24.5}{46.24} = \sin(2 \theta)}$

$\sf{\sin(2 \theta) \approx 0.53}$

$\sf{\cancel{\sin}(2 \theta) \approx \cancel{\sin}(32^o)}$

Find the angle value of the equation by using trigonometric equations. Provided that the parabolic motion has an angle of elevation 0° ≤ x ≤ 90°.

First Probability

$\sf{2 \theta = 32^o + k \cdot 360^o}$

$\sf{\theta = 16^o + k \cdot 180^o}$

→ $\sf{k = 0 \rightarrow 16^o + 0 = 16^o}$ (T)

→ $\sf{k = 1 \rightarrow 16^o + 180^o = 196^o}$ (F)

Second Probability

$\sf{2 \theta = (180^o - 32^o) + k \cdot 360^o}$

$\sf{2 \theta = 148^o + k \cdot 360^o}$

$\sf{\theta = 74^o + k \cdot 180^o}$

→ $\sf{k = 0 \rightarrow 74^o + 0 = 74^o}$ (T)

→ $\sf{k = 1 \rightarrow 74^o + 180^o = 254^o}$ (F)

$\boxed{\sf{\therefore \theta \{16^o , 74^o\} }}$

<h3>Conclusion</h3>

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