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3 years ago

An object is dropped from a platform 400 ft high. Ignoring wind resistance, how long will it take to reach the ground?

1 answer:

8 0

Let's convert the initial height from feet to meters first:
$h=400 ft=121.9 m$

The object is in free fall, so it is moving by uniformly accelerated motion with constant acceleration $g=9.81 m/s^2$ (gravitational acceleration) from initial height h=121.9 m. Its vertical position at time t is given by
$y(t)=h- \frac{1}{2}gt^2$

We want to know how long it will take to reach the ground, therefore we should calculate the time t at which the vertical position y(t) becomes zero:
$0=h- \frac{1}{2}gt^2$
rearranging,
$t= \sqrt{ \frac{2h}{g} }= \sqrt{ \frac{2(121.9 m)}{9.81 m/s^2} }=4.98 s$

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Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu

vlada-n [284]

Answer:

$\frac{1}{10}$ M

Explanation:

To apply the concept of angular momentum conservation, there should be no external torque before and after

As the asteroid is travelling directly towards the center of the Earth, after impact ,it does not impose any torque on earth's rotation, So angular momentum of earth is conserved

⇒ $I_{1} \times W_{1} =I_{2} \times W_{2}$

$I_{1}$ is the moment of interia of earth before impact
$W_{1}$ is the angular velocity of earth about an axis passing through the center of earth before impact
$I_{2}$ is moment of interia of earth and asteroid system
$W_{2}$ is the angular velocity of earth and asteroid system about the same axis

let $W_{1}=W$

since $\text{Time period of rotation}∝$ $\frac{1}{\text{Angular velocity}}$

⇒ if time period is to increase by 25%, which is $\frac{5}{4}$ times, the angular velocity decreases 25% which is $\frac{4}{5}$ times

therefore $W_{1}$ $=$ $\frac{4}{5} \times W_{1}$

$I_{1}=\frac{2}{5} \times M\times R^{2}$ (moment of inertia of solid sphere)

where M is mass of earth

R is radius of earth

$I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}$

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where $M_{1}$ is mass of asteroid

⇒ $\frac{2}{5} \times M\times R^{2} \times W_{1}=}(\frac{2}{5} \times M\times R^{2}$ $+$ $M_{1}\times R^{2})\times(\frac{4}{5} \times W_{1})$

$\frac{1}{2} \times M\times R^{2}$ = $(\frac{2}{5} \times M\times R^{2}$ + $M_{1}\times R^{2})$

$M_{1}\times R^{2}=$ $\frac{1}{10} \times M\times R^{2}$

⇒ $M_{1}=}$ $\frac{1}{10} \times M$

3 0

3 years ago

A piece of wire 29 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral tria

skelet666 [1.2K]

Answer:

Explanation:

Total length of the wire is 29 m.

Let the length of one piece is d and of another piece is 29 - d.

Let d is used to make a square.

And 29 - d is used to make an equilateral triangle.

(a)

Area of square = d²

Area of equilateral triangle = √3(29 - d)²/4

Total area,

$A = d^{2}+\frac{\sqrt3}{4}\left ( 29-d \right )^{2}$

Differentiate both sides with respect to d.

$\frac{dA}{dt}=2d- \frac{\sqrt3}{4}\times 2(29-d)$

For maxima and minima, dA/dt = 0

d = 8.76 m

Differentiate again we get the

$\frac{d^{2}A}{dt^{2}}= + ve$

(a) So, the area is maximum when the side of square is 29 m

(b) so, the area is minimum when the side of square is 8.76 m

8 0

3 years ago

noname [10]

The wind will blow from higher pressure over the water to lower pressure over the land causing sea breeze. The sea breeze strength will vary depending on the temperature difference between the land and the ocean

5 0

4 years ago

Look at the graph pic and answer the question correctly!

SCORPION-xisa [38]

Answer:

Explanation:

That the time period of which they stop.

4 0

3 years ago

Read 2 more answers

........................ is The force affecting on an object during certain period of time *

Marta_Voda [28]

Answer:

I think c is the answer but I have a little concern on d too

4 0

3 years ago