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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.
Physics
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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Answer:

a)  v=4.0m/s

b)  B=2.958T

Explanation:

From the question we are told that:

Wire Length l=10cm=>0.10m

Resistance R=0.35

Force F=1.0N

Power P=4.0W

a)

Generally the equation for Power is mathematically given by

P=Fv

Therefore

v=\frac{P}{F}

v=\frac{4.0}{1.0}

v=4.0m/s

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Generally the equation for Magnetic Field is mathematically given by

B=\frac{\sqrt{PR}}{vl}

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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>

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

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