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2 years ago

Small, icy bodies that have highly eccentric orbits and can be found in the Oort cloud or the Kuiper belt are called ____.

Physics

1 answer:

irina1246 [14]2 years ago

4 0

Answer:

Small, icy bodies that have highly eccentric orbits and can be found in the Oort cloud or the Kuiper belt are called COMETS.

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Oh no! The Hulk just fell off the Empire State Building! Calculate how long it took him to fall straight down from the top of th

nekit [7.7K]

Answer:it takes approximately 148.8 seconds to achieve. The average person in a free-fall will hit the ground going at 9.66 m/s from the top of the Empire State Building.

Explanation:

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2 years ago

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qwelly [4]

Answer:

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1 year ago

Which transition metals have more in common than other

jeka94

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2 years ago

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A sky diver with a mass of 70kg jumps from an aircraft. The aerodynamic drag force acting on the sky diver is known to be Fd=kV^

xeze [42]

Answer:

$v_{max}=52.38\frac{m}{s}$

$v_{100}=33.81$

Explanation:

the maximum speed is reached when the drag force and the weight are at equilibrium, therefore:

$\sum{F}=0=F_d-W$

$F_d=W$

$kv_{max}^2=m*g$

$v_{max}=\sqrt{\frac{m*g}{k}} =\sqrt{\frac{70*9.8}{0.25}}=52.38\frac{m}{s}$

To calculate the velocity after 100 meters, we can no longer assume equilibrium, therefore:

$\sum{F}=ma=W-F_d$

$ma=W-F_d$

$ma=mg-kv_{100}^2$

$a=g-\frac{kv_{100}^2}{m}$ (1)

consider the next equation of motion:

$a = \frac{(v_{x}-v_0)^2}{2x}$

If assuming initial velocity=0:

$a = \frac{v_{100}^2}{2x}$ (2)

joining (1) and (2):

$\frac{v_{100}^2}{2x}=g-\frac{kv_{100}^2}{m}$

$\frac{v_{100}^2}{2x}+\frac{kv_{100}^2}{m}=g$

$v_{100}^2(\frac{1}{2x}+\frac{k}{m})=g$

$v_{100}^2=\frac{g}{(\frac{1}{2x}+\frac{k}{m})}$

$v_{100}=\sqrt{\frac{g}{(\frac{1}{2x}+\frac{k}{m})}}$ (3)

$v_{100}=\sqrt{\frac{9.8}{(\frac{1}{2*100}+\frac{0.25}{70})}}$

$v_{100}=\sqrt{\frac{9.8}{(\frac{1}{200}+\frac{1}{280})}}$

$v_{100}=\sqrt{\frac{9.8}{(\frac{3}{350})}}$

$v_{100}=\sqrt{1,143.3}$

$v_{100}=33.81$

To plot velocity as a function of distance, just plot equation (3).

To plot velocity as a function of time, you have to consider the next equation of motion:

$v = v_0 +at$

as stated before, the initial velocity is 0:

$v =at$ (4)

joining (1) and (4) and reducing you will get:

$\frac{kt}{m}v^2+v-gt=0$

solving for v:

$v=\frac{ \sqrt{1+\frac{4gk}{m}t^2}-1}{\frac{2kt}{m} }$

Plots:

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2 years ago

Bill and Ted are standing on a bridge 40 ft above a river. Bill drops a stone, while Ted decides to throw a stone downward at 10

USPshnik [31]

Answer:

Explanation:

In order to know how long after Bill released his rock should Ted throw his if they want the stones to hit the water simultanously, we need to calculate the time needed to hit the water to both rocks independent each other, and just take the difference.

For the rock dropped by Bill, as the only influence on it is gravity (accelerating it downwards with an acceleration equal to g), and v₀ =0, we can use the following kinematic equation:

$y = \frac{1}{2} * g * t^{2}$

where y = height = 40 ft.

As all the parameters are given in SI units, it is advisable to convert this value to m, as follows:

$y = 40 ft*\frac{0.3048m}{1 ft} = 12.2 m$

Now, we can solve for t, as follows:

$t = \sqrt{\frac{2*y}{g}} = \sqrt{\frac{2*12.2m}{9.8m/s2}} = 1.58 s$

For the rock thrown down at 10 m/s, the kinematic equation we just have used becomes:

$y = v0*t +\frac{1}{2} * g * t^{2}$

This leaves us a quadratic equation on t, as follows:

$t = \frac{-10m/s}{9.8m/s2} +/- \sqrt{10m/s^{2} -4*4.9m/s2*(-12.2m)} = -1.02 s +/- 1.88s$

Taking the positive root, we have:

t = -1.02 s + 1.88 s = 0.86 s

So, in order to get that both rocks hit the water at the same time, Ted will need to wait the difference between both times:

Δt = 0.86 s - 1.58s = -0.72 s

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2 years ago