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s344n2d4d5 [400]
3 years ago
8

A small, single engine airplane is about to take off. The airplane becomes airborne, when its speed reaches 161.0 kmph. The cond

itions at the airport are ideal, there is no wind. When the engine is running at its full power, the acceleration of the airplane is 2.60 m/s2. What is the minimum required length of the runway?
Physics
1 answer:
lukranit [14]3 years ago
5 0

Answer:

384.6 m

Explanation:

The length of the runway airport should be long enough to accommodate the aircraft during its acceleration from rest to 161 km/h at rate of 2.6 m/s. We can use the following equation of motion to solve for this:

v^2 - v_0^2 = 2a\Delta s

where v0 = 0 m/s is the initial velocity of the airplane when it start accelerating, v = 161 km/h = 161*1000*(1/60)(1/60) = 44.72 m/s is the airborn speed, a = 2.6 m/s2 is the acceleration, and \Delta s is the distance of the runway, which we care looking for

44.72^2 - 0 = 2*2.6\Delta s

\Delta s = 44.72^2/(2*2.6) = 384.6 m

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<h3>It takes 60 seconds to do the work</h3>

<em><u>Solution:</u></em>

Given that,

Force = 100 newtons

Distance = 15 meters

Power = 25 watts

To find: time it takes to do the work

<em><u>Find the work done:</u></em>

work = force \times displacement\\\\work = 100\ newtons \times 15\ meters\\\\work = 1500\ joule

<em><u>Find the time taken</u></em>

power = \frac{work}{time}\\\\25\ watts = \frac{1500\ joule}{time}\\\\time = \frac{1500\ joule}{25\ watt}\\\\time = 60\ second

Thus it takes 60 seconds to do the work

3 0
3 years ago
What is the magnitude of the electric field at a point midway between a −5.0μC and a +5.8μC charge 8.4cm apart? Assume no other
Alex73 [517]

Answer:

Electric Field = E = 36.848 N/C

Explanation:

In accordance with Columb's law

E = k Q1 Q2 / r.r = 8.99 x 10^9 x 5.0 x 10^-6 x 5.8 x 10^-6 / 0.084 x 0.084

= 36948.6961 x 10^-3 = 36.848 N/C

4 0
3 years ago
If mechanical energy is conserved, then a pendulum that has a potential energy of 20 J at its highest point and 0.5 J at its low
liraira [26]
At the highest point: kinetic energy is 0 due to the speed  is 0

So the total mechanical energy is 20

Assume no frictions present, then the mechanical energy is conserved

So at the lowest point, kinetic energy = mechanical energy - potential energy

Answer will be 20 - 0.5 = 19.5 J
4 0
3 years ago
Why are some consolation visible to New York State observers at midnight during April , but not visible at midnight during Octob
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3 years ago
A particle (charge = +0.8 mC) moving in a region where only electric forces act on it has a kinetic energy of 6.7 J at point A.
Maksim231197 [3]

Answer:

The kinetic energy of the particle as it moves through point B is 7.9 J.

Explanation:

The kinetic energy of the particle is:

\Delta K = \Delta E_{p} = q\Delta V

<u>Where</u>:

K: is the kinetic energy

E_{p}: is the potential energy

q: is the particle's charge = 0.8 mC

ΔV: is the electric potential = 1.5 kV                                    

\Delta K = q \Delta V= 0.8 \cdot 10^{-3} C*1.5 \cdot 10^{3} V = 1.2 J

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\Delta K = K_{f} - K_{i}

K_{f} = \Delta K + K_{i} = 1.2 J + 6.7 J = 7.9 J

Therefore, the kinetic energy of the particle as it moves through point B is 7.9 J.

I hope it helps you!      

8 0
3 years ago
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