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2 years ago

5

A jet plane lands with a velocity of +107 m/s and can accelerate at a maximum rate of -5.18 m/s2 as it comes to rest. From the i

nstant it touches the runway, what is the minimum time needed before it can come to rest?

1 answer:

Marrrta [24]2 years ago

3 0

Answer:

20.7 s

Explanation:

The equation to calculate the velocity for a uniform acceleration a, time t and initial velocity v₀:

v = a*t + v₀

Solve for t:

t = (v - v₀)/a

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when a circular plate of metal is heated in an oven, its radius increases at .03 cm/min, at what rate is the area increasing whe

Alinara [238K]

Answer:

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Explanation:

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Radius of the circular plate r = 52 cm

Area is given by $A=\pi r^2$

So $\frac{dA}{dt}=2\pi r\frac{dr}{dt}$

Puting the value of r and $\frac{dr}{dt}$

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2 years ago

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mr_godi [17]

Answer and Explanation:

The computation of the shortest wavelength in the series is shown below:-

$\frac{1}{\lambda} = R(\frac{1}{n_f^2} - \frac{1}{n_i^2} )$

Where

$\lambda$ represents wavelength

R represents Rydberg's constant

$n_f$ represents Final energy states

and $n_i$ represents initial energy states

Now Substitute is

$1.097\times 10^7\ m^{-1}\ for\ R, \infty for\ n_i,\ 3 for\ n_i,\\\\\ \frac{1}{\lambda} = R(\frac{1}{n_f^2} - \frac{1}{n_i^2} )$

now we will put the values into the above formula

$= 1.097\times 10^7 m^{-1}(\frac{1}{3^2} - \frac{1}{\infty^2} )\\\\ = 1.097\times10^7\ m^{-1} (\frac{1}{9} )$

$= 1218888.889 m^{-1}$

Now we will rewrite the answer in the term of $\lambda$

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3 years ago

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denis-greek [22]

Answer:

t = 6.17 s

Explanation:

For a 1 revolution movement, $\triangle \theta = 2\pi$

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Moment of Inertia, $I = 150 kg m^2$

If the wheel starts from rest, $w_{0} = 0 rad/s$

The angular displacement of the wheel can be given by the formula:

$\triangle \theta = \omega_0 t + 0.5 \alpha t^2$ ................(1)

Where $\alpha$ is the angular acceleration

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