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ollegr [7]
3 years ago
13

A jumbo jet has a mass of 100,000 kg. The thrust of each of its four engines is 50,000 N. What is the jet's acceleration in mete

rs per second squared right before taking off? Neglect air resistance and friction.
Physics
1 answer:
lorasvet [3.4K]3 years ago
5 0

Answer:

The acceleration is   a =2\  m/s^2

Explanation:

From the question we are told that

       The  mass of the jumbo jet is  m_j  =  100000\ kg

        The thrust is  F_k =  50000 \ N

Generally given that the jet has four engines the total thrust is  

        F_t =  4 * F_k

substituting values

       F_t  =  4 * 50000

      F_t  =  200000 \ N

Generally the acceleration of the is mathematically represented as

         a = \frac{F_t}{m}

substituting values

       a =2 \frac{N}{kg}

Now  

        N  =  kg  \cdot m/s^2

Hence

         a =2 \frac{kg * \cdot m/s^2}{kg}

        a =2\  m/s^2

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Assume that a pendulum used to drive a grandfather clock has a length L0=1.00m and a mass M at temperature T=20.00°C. It can be
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The period will change a 0,036 % relative to its initial state

Explanation:

When the rod expands by heat its moment of inertia increases, but since there was no applied rotational force to the pendulum , the angular momentum remains constant. In other words:

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since there was no torque ( no rotational force applied)

ζ=0 → Δ(Iω)=0 → I₂ω₂ -I₁ω₁ = 0 → I₁ω₁ = I₂ω₂

thus

I₂/I₁ =ω₁/ω₂ , (2) represents final state and (1) initial state

we know also that ω=2π/T , where T is the period of the pendulum

I₂/I₁ =ω₁/ω₂ = (2π/T₁)/(2π/T₂)= T₂/T₁

Therefore to calculate the change in the period we have to calculate the moments of inertia. Looking at tables, can be found that the moment of inertia of a rod that rotates around an end is

I = 1/3 ML²

Therefore since the mass M is the same before and after the expansion

I₁ = 1/3 ML₁² , I₂ = 1/3 ML₂²  → I₂/I₁ = (1/3 ML₂²)/(1/3 ML₁²)= L₂²/L₁²= (L₂/L₁)²

since

L₂= L₁ (1+αΔT) , L₂/L₁=1+αΔT  , where ΔT is the change in temperature

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T₂/T₁=I₂/I₁=(L₂/L₁)² = (1+αΔT) ²

finally

%change in period =(T₂-T₁)/T₁ = T₂/T₁ - 1 = (1+αΔT) ² -1

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Answer:

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