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3 years ago

A passenger jet flies from one airport to another 1,273 miles away in 2.2 h. find its average speed.

Physics

2 answers:

oee [108]3 years ago

3 0

Speed : distance divide by time . 1273/2.2 = ...

Use calculator .

Shtirlitz [24]3 years ago

3 0

Answer:

V_avg = 578.6364 miles/hr

Explanation:

Given:

- The distance between airports d = 1,273 miles

- The time taken for the flight t = 2.2 hr

Find:

The average speed of the jet

Solution:

- The average speed of the jet is given by V_avg:

V_avg = Total distance traveled / Total time

V_avg = 1273 / 2.2

V_avg = 578.6364 miles/hr

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Answer:

$h=25.52\times10^6 m$

Explanation:

Initial speed, v = 10 x 10^3 m/s

Mass of the earth, M = 6 x 10^24 kg

Radius of the earth, R = 6.4 x 10^6 m

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= $-2.50625\times10^7 J$

$=- 8\times10^{14} / ( R + h )=-2.50625\times 10^7$

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3 years ago

A 0.5 m diameter wagon wheel consists of a thin rim having a mass of 7 kg and six spokes, each with a mass of 1.2 kg. 1.2 kg 7 k

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Explanation:

It is given that,

Mass of the rim of wheel, m₁ = 7 kg

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Let I is the the moment of inertia of the wagon wheel for rotation about its axis.

We know that the moment of inertia of the ring is given by :

$I_1=m_1r^2$

$I_1=7\times (0.25)^2=0.437\ kgm^2$

The moment of inertia of the rod about one end is given by :

$I_2=\dfrac{m_2l^2}{3}$

l = r

$I_2=\dfrac{m_2r^2}{3}$

$I_2=\dfrac{1.2\times (0.25)^2}{3}=0.025\ kgm^2$

For 6 spokes, $I_2=0.025\times 6=0.15\ kgm^2$

So, the net moment of inertia of the wagon is :

$I=I_1+I_2$

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4 years ago

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How large a force is necessary to stretch a 4.0-mm-diameter steel wire from its original length by 1.0%?

jekas [21]

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The given parameters include;

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final length of the wire, L₂ = 1.01 x L₁ (increase of 1% = 101%)

extension of the wire e = L₂ - L₁ = 1.01L₁ - L₁ = 0.01L₁

the Youngs modulus of steel, E = 200 Gpa

The area of the steel wire is calculated as follows;

$A = \pi r^2\\\\ A= 3.142 \times (0.002)^2\\\\ A= 1.257 \times 10^{-5} \ m^2$

The force needed to stretch the wire is calculated from Youngs modulus of elasticity given as;

$E = \frac{stress}{strain} = \frac{F/A}{e/L} = \frac{FL}{Ae} \\\\F = \frac{EAe}{L}$

$F = \frac{200 \times 10^9\ \times\ 1.257\times 10^{-5}\ \times \ 0.01l_1}{l_1} \\\\F = 25,140\ N$

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Learn more here: brainly.com/question/21413915

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