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kipiarov [429]
3 years ago
7

A Boeing 777 aircraft has a mass of 300,000 kg. At a certain instant during its landing, its speed is 27.0 m/s. If the braking f

orce is 445,000 N, what is the speed of the airplane 10.0 s later?
Physics
1 answer:
GarryVolchara [31]3 years ago
6 0

Answer:

Speed of the airplane 10.0 s later = 12.2 m/s

Explanation:

Mass of Boeing 777 aircraft = 300,000 kg

Braking force = 445,000 N

Deceleration

            a=\frac{445000}{300000}=1.48m/s^2

Initial velocity, u = 27 m/s

Time , t = 10 s

We have equation of motion, v =u +at

            v = 27 + (-1.48) x 10 = 27 - 14.8 = 12.2 m/s

Speed of the airplane 10.0 s later = 12.2 m/s

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Answer:

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Explanation:

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d = Vertical height from the ground

F = Force = Weight = mg

Net work done would be

W_n=W_{up}+W_{down}\\\Rightarrow W_n=Fdcos180+Fdcos0\\\Rightarrow W_n=-mgd+mgd\\\Rightarrow W_n=0

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Explanation:

Im not sure rn but ill get back to you.

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2 years ago
You are riding a bicycle. If you apply a forward force of 150 N, and you and the bicycle have a combined mass of 90 kg, what wil
bekas [8.4K]
<h3>Hello There!!</h3>

<h3><u>Given</u>,</h3>

Force(F) = 150N

Mass(m) = 90kg

<h3><u>To </u><u>Find,</u></h3>

Acceleration(a) = ?

<h3><u>We know,</u></h3>

F= m×a

150 = 90 \times  \text{a} \\  \\  \text{a} =  \frac{150}{90}  \\  \fbox{cancelling by 3} \\  \\   \text{ a}  = \cancel \frac{150}{90} \\  \\ \text{ a}  =  \frac{5}{3}  = 1.67 \text{m/s} {}^{2}

\therefore  \text{Option A= 1.67 m/s² is the correct answer}

<h3>Hope this helps</h3>
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2 years ago
A 0.125 kg hockey puck moving at 20.0 m/s is caught and held by an 85.0 kg goalkeeper at rest. After catching the puck, with wha
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3 years ago
A pulsar is a rapidly rotating neutron star that emits a radio beam the way a lighthouse emits a light beam. We receive a radio
DanielleElmas [232]

Answer:

\alpha =-2.2669642\times^{-10}rad/s^2

Explanation:

Angular acceleration is defined by \alpha =\frac{\Delta \omega}{\Delta t}=\frac{\omega_f-\omega_i}{\Delta t}

Angular velocity is related to the period by \omega=\frac{2\pi}{T}

Putting all together:

\alpha =\frac{\frac{2\pi}{T_f}-\frac{2\pi}{T_i}}{\Delta t}=\frac{2\pi}{\Delta t}(\frac{1}{T_f}-\frac{1}{T_i})

Taking our initial (i) point now and our final (f) point one year later, we would have:

\Delta t=1\ year=(365)(24)(60)(60)s=31536000&#10;s

T_i=0.0786s

T_f=0.0786s+7.03\times10^{-6}s

So for our values we have:

\alpha =\frac{2\pi}{\Delta t}(\frac{1}{T_f}-\frac{1}{T_i})=\frac{2\pi}{31536000s}(\frac{1}{0.0786s+7.03\times10^{-6}s}-\frac{1}{0.0786s})=-2.2669642\times^{-10}rad/s^2

Where the minus sign indicates it is decelerating.

8 0
3 years ago
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