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3 years ago

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On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a 6 iron. The acceleration due to gravity on t

he moon is 1/6 of its value on earth. Suppose he hits the ball with a speed of 14 m/s at an angle 20 ∘ above the horizontal. For how much more time was the ball in flight?

1 answer:

castortr0y [4]3 years ago

5 0

Answer:

4.86 seconds

Explanation:

Velocity of projection, u = 14 m/s

angle of projection, θ = 20°

Formula for the time of flight

$T=\frac{2uSin\theta }{g}$

For earth

Te = (2 x 14 x Sin 20) / 9.8

Te = 0.98 s

For moon

g' = g/6 = 1.64 m/s^2

Tm = ( 2 x 14 x Sin 20) / 1.64

Tm = 5.84 seconds

Tm - Te = 5.84 - 0.98 = 4.86 s

So, it takes 4.86 s more time of flight on moon than the earth.

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Explanation:

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Calculate the maximum capillary rise/fall of mercury in a 0.5 mm radius glass capillary. Assume that the surface tension for mer

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A series of optical telescopes produced an image that has a resolution of about 0.00350 arc second.

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Answer:

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From the question we are told that

The resolution of the telescope is $\theta = 0.00350 \ arc \ second$

The wavelength is $\lambda = 1.70 \mu m = 1.70 *10^{-6} \ m$

From the question we are told that

$1 arc \ sec = \frac{1}{3600^o}$

So $0.00350 \ arc \ second = x$

Therefore

$x = 0.00350 * \frac{1}{3600 }$

$x = ( 9.722*10^{-7} )^o$

Now $1^o = \frac{\pi}{180}$

So $(9.722*10^{-7})^o = \theta$

=> $\theta = (9.722*10^{-7}) * \frac{\pi}{180}$

$\theta = 1.69*10^{-8} rad$

The smallest diameter is mathematically represented as

$D = \frac{1.22 \lambda }{\theta }$

substituting values

$D = \frac{1.22 * 1.7 *10^{-6}} {1.69 *10^{-8} }$

$D =122 \ m$

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