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3 years ago

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A rocket is launched upward with a constant acceleration of 165 m/s^2. After 8.00 seconds of ascension a passenger on the rocket

drops a rock out of it. How long does it take for this rock to hit the ground?

1 answer:

Novay_Z [31]3 years ago

5 0

To solve this problem we will apply the concepts related to the linear kinematic movement. We will start by finding the speed of the body from time and the acceleration given.

Through the position equations we will calculate the distance traveled.

Finally, using this same position relationship and considering the previously found speed, we can determine the time to reach your goal.

For time (t) and acceleration (a) we have to,

$t = 8s, a = 165m/s^2$

The velocity would be,

$u = a*t \\u = 165*8\\u = 1320m/s$

Now the position is,

$h= \frac{1}{2} at^2$

$h = \frac{1}{2} 165*8^2$

$h = 5280m$

Now with the initial speed and position found we will have the time is,

$h=ut +\frac{1}{2} at^2$

$-5280=1320t - \frac{1}{2} 9.8t^2$

$4.9t^2-1320t-5280=0$

Solving the polynomian we have,

$t = 273.33s = 4.56minutes$

Therefore the rocket will take to hit the ground around to 4.56min

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Answer:

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The free body diagram of this setup is on the first uploaded image

From the question

The diameter of closely packed cones in the fovea of the eye is = $2 \mu m$

The distance of separation by one cone(not excited ) is $d = 4\mu m = 4*10^{-4}cm$

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The distance of the two point-like object from the near point of the eye is A = 28 cm

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3 years ago

Read 2 more answers

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Answer:

The water level will drop by about 1.24 cm in 1 day.

Explanation:

Here Mass flux of water vapour is given as

$j_{H_2O}=\frac{D}{l} \bigtriangleup c$

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$j_{H_2O}$ is the mass flux of the water which is to be calculated.

D is diffusion coefficient which is given as $0.25 cm^2/s$

l is the thickness of the film which is 0.15 cm thick.
$\bigtriangleup c$ is given as

$\bigtriangleup c= \frac{P_{sat}-P_a}{RT}$

In this

$P_{sat}$ is the saturated water pressure, which is look up from the saturated water property at 20°C and 0.5 saturation given as 2.34 Pa

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T is the temperature in Kelvin scale which is $20+273= 293K$

By substituting values in the equation

$\bigtriangleup c= \frac{P_{sat}-P_a}{RT} \\ \bigtriangleup c= \frac{P_{sat}-0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5 \times 2.34}{8.314 \times 293} \\\bigtriangleup c= 0.48 mol/m^3$

Converting $\bigtriangleup c$ into $cm^3/cm^3$

As 1 mole of water 18 $cm^3$ so

$\bigtriangleup c= 0.48 mol/m^3 \\ \bigtriangleup c= 0.48 \times 18 \times 10^{-6} cm^3/cm^3 \\ \bigtriangleup c= 8.64 \times 10^{-6} cm^3/cm^3$

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For calculation of water level drop in a day, converting mass flux as

$j_{H_2O}=14.4 \times 10^{-6} \times 24 \times 3600 cm/day\\ j_{H_2O}=1.24 cm/day$

So the water level will drop by about 1.24 cm in 1 day.

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4 years ago

A capacitor with initial charge q0 is discharged through a resistor. What multiple of the time constant τ gives the time the cap

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4 years ago