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2 years ago

14

A Cessna aircraft has a lift off speed of 147 km/h. What minimum constant acceleration does this require if the aircraft is to b

e airborne after a take off run of 208 m?

1 answer:

netineya [11]2 years ago

6 0

Answer:

The acceleration is $a =51945 \ km/h^2$

Explanation:

From the question we are told that

The lift up speed is $v = 147 \ km/h$

The distance covered for the take off run is $s = 208 m = 0.208 \ km$

Generally from kinematic equation we have that

$v^2 = u^2 + 2as$

Here u is the initial speed of the aircraft with value 0 m/ s give that the aircraft started from rest

So

$147^2 = 0^2 + 2* a* 0.208$

=> $a =51945 \ km/h^2$

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Resistivity = 231.481 K Ohm

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Explanation:

Solution:

At 300K:

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So, plugging in the values, we get:

σ = C x ( $M_{h}$ + $M_{e}$ ) e

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3 years ago

A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum

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Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

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$\rho (r) \ \alpha \ \delta (r -R)$

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$\mathbf{\rho (r) = \delta* \delta (r -R) \ \ at \ \ (r=R)}$

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$\mathbf{\rho (r) =\sigma \ * \ \delta (r-R)}$

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The integrated charge Q

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$Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr$

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$Q = 4 \pi \sigma *R^2$ since $( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )$

$\mathbf{Q = 4 \pi R^2 \sigma }$

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3 years ago