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Sia ? 6 years, 6 months ago
Let the usual speed of the plane b x km/hr.
Distance to the destination = 1500 km
Case (i):
{tex}\text{we know that,} \ Speed = {Distance\over Time} \\ \Rightarrow Time = {Distance\over speed}{/tex}
So, in case(i) Time = {tex}1500 \over x{/tex}Hrs
Case (iI)
Distance to the destination = 1500 km
Increased speed = 100 km/hr
So, speed = x+100
So, in case(ii) Time = {tex}1500 \over {x+100}{/tex}Hrs
So, according to the question
{tex}\therefore{/tex} {tex}\frac{1500}{x}{/tex} - {tex}\frac{1500}{x + 100}{/tex} = {tex}\frac{30}{60}{/tex}
{tex}\Rightarrow{/tex} x2 + 100x - 300000 = 0
{tex}\Rightarrow{/tex} x2 + 600x - 500x - 300000 = 0
{tex}\Rightarrow{/tex} (x + 600)(x - 500) = 0
{tex}\Rightarrow{/tex}x = 500 or x = -600
Since, speed can not be negative, x = 500
Therefore, Speed of plane = 500 km/hr.
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Sia ? 6 years, 5 months ago
{tex}F _ { b } = \rho \mathrm { gV } = \rho \mathrm { gh } \mathrm { A }{/tex}
<i>Fb</i> = buoyant force of a liquid acting on an object (N)
<i>ρ</i> = density of the liquid(kg/m3)
<i>g</i> = gravitational acceleration(9.80 m/s2)
<i>V</i> = volume of liquid displaced (m3 or liters, where 1 m3 = 1000 L)
<i>h</i> = height of water displaced by a floating object(m)
<i>A</i> = surface area of a floating object(m2)
Posted by Abhishek Kushwaha 5 years, 8 months ago
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Jasmeet Kaur 7 years, 9 months ago
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