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2 years ago

15

If there were no air resistance, how long would it take a free-falling skydiver to fall from a plane at 3500 m to an altitude of

400 m , where she will pull her ripcord?

1 answer:

xxMikexx [17]2 years ago

7 0

By equation of motion we have S = ut + 0.5 a $t^{2}$

Where u = Initail velocity, S= dispalcement, t = time taken, a = acceleration

Here S = 3500-400 = 3100m, u = 0m/s, g = 9.8 m/ $s^{2}$

So, 3100 = 0+0.5*9.8* $t^{2}$

$t^{2}$ = 3100/4.9

t = $\sqrt{3100/4.9}$

t = 25.15 seconds

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A helicopter starting on the ground is rising directly into the air at a rate of 25 ft/s. You are running on the ground starting

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Answer:

The rate of change of the distance between the helicopter and yourself (in ft/s) after 5 s is $\sqrt{725}$ ft/ sec

Explanation:

Given:

h(t) = 25 ft/sec

x(t) = 10 ft/ sec

h(5) = 25 ft/sec . 5 = 125 ft

x(5) = 10 ft/sec . 5 = 50 ft

Now we can calculate the distance between the person and the helicopter by using the Pythagorean theorem

$D(t) = \sqrt{h^2 + x^2}$

Lets find the derivative of distance with respect to time

$\frac{dD}{dt} (t) = \frac{2h \cdot \frac{dh}{dt} +2x \cdot\frac{dx}{dt}} {2\sqrt{h^2 + x^2}}$

Substituting the values of h(t) and x(t) and simplifying we get,

$\frac{dD}{dt}(t) = \frac{50t \cdot \frac{dh}{dt} + 20 \cdot \frac{dx}dt}{2\sqrt{625\cdot t^2 + 100 \cdot t^2}}$

$\frac{dh}{dt} = 25ft/sec$

$\frac{dx}{dt} = 10 ft/sec$

$\frac{Dd}{dt} (t) = \frac{1250t +200t}{2\sqrt{725}t}$ = $\frac{725}{\sqrt{725}}$ = $\sqrt{725}$ ft / sec

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Answer:

$\rho = 12580.7 kg/m^3$

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As we know that the satellite revolves around the planet then the centripetal force for the satellite is due to gravitational attraction force of the planet

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here we have

$F =\frac {mv^2}{(r+ h)}$

$\frac{mv^2}{r + h} = \frac{GMm}{(r + h)^2}$

here we have

$v = \sqrt{\frac{GM}{(r + h)}}$

now we can find time period as

$T = \frac{2\pi (r + h)}{v}$

$T = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{GM}{(r + h)}}}$

$1.15 \times 3600 = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{(6.67 \times 10^{-11})(M)}{(2.05 \times 10^6 + 310 \times 10^3)}}}$

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Now the density is given as

$\rho = \frac{M}{\frac{4}{3}\pi r^3}$

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A gun shoots a bullet with a velocity of 500 m/s. The gun is aimed horizontally and fired from a height of 1.5 m. How far does t

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The bullet travels a horizontal distance of 276.5 m

The bullet is shot forward with a horizontal velocity $u_x$ . It takes a time <em>t</em> to fall a vertical distance <em>y</em> and at the same time travels a horizontal distance <em>x. </em>

The bullet's horizontal velocity remains constant since no force acts on the bullet in the horizontal direction.

The initial velocity of the bullet has no component in the vertical direction. As it falls through the vertical distance, it is accelerated due to the force of gravity.

Calculate the time taken for the bullet to fall through a vertical distance <em>y </em>using the equation,

$y=u_yt+\frac{1}{2} gt^2$

Substitute 0 m/s for $u_y$ , 9.81 m/s²for <em>g</em> and 1.5 m for <em>y</em>.

$y=u_yt+\frac{1}{2} gt^2\\ 1.5 m=(0m/s)t+\frac{1}{2} (9.81m/s^2)t^2\\ t=\sqrt{\frac{2(1.5m)}{9.81m/s^2} } =0.5530s$

The horizontal distance traveled by the bullet is given by,

$x=u_xt$

Substitute 500 m/s for $u_x$ and 0.5530s for t.

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