Collisions tend to happen when one driver is going faster or slower than other cars on the road.

A 1500 kg weather rocket accelerates upward at 10m/s2. It explodes 2.0 s after liftoff and breaks into two fragments, one twice

ladessa [460]

Answer:

Approximately $\rm 19.8\; m\cdot s^{-1}$ (downwards.)

Assumptions:

the rocket started from rest;
the gravitational acceleration is constantly $\rm -9.8\; m \cdot s^{-2}$ ;
there's no air resistance on the rocket and the two fragments.
Both fragments traveled without horizontal velocity.

Explanation:

The upward speed of the rocket increases by $\rm 10\; m \cdot s^{-1}$ . If the rocket started from rest, the vertical speed of the rocket should be equal to $\rm 20\; m \cdot s^{-1}$ .

The mass of the rocket (before it exploded) is 1500 kilograms. At 20 m/s, its momentum will be equal to $\rm 20 \times 1500 = 30,000\; kg \cdot m\cdot s^{-1}$ .

What's the initial upward velocity, $u$ , of the lighter fragment?

The upward velocity of the lighter fragment is equal to $v = 0$ once it reached its maximum height of $x = \rm 530\; m$ .

$v^2 - u^2 = 2g \cdot x$ .

$\begin{aligned}u &= \sqrt{v^2 - 2g\cdot x} \\ &= \sqrt{-2 (-9.8) \times 530}\\ &\approx \rm 101.922\; m \cdot s^{-1}\end{aligned}$ .

Mass of the two fragments:

Lighter fragments: $\displaystyle \frac{1}{1 + 2} \times 1500 =\rm 500\; kg$ .
Heavier fragment: $\displaystyle \frac{2}{1 + 2} \times 1500 =\rm 1000\; kg$ .

Initial momentum of the lighter fragment:

$m \cdot v = \rm 10192.2\; kg \cdot m \cdot s^{-1}$ .

If there's no air resistance, momentum shall conserve. The momentum of the lighter fragment, plus that of the heavier fragment, should be equal to that of the rocket before it exploded.

The initial momentum of the heavier fragment should thus be equal to the momentum of the two pieces, combined, minus the initial momentum of the lighter fragment.

$\rm 30000 - 10192.2 = 19807.8\;kg \cdot m \cdot s^{-1}$ .

Velocity of the heavier fragment:

$\displaystyle \rm \frac{19807.8\;kg \cdot m \cdot s^{-1}}{1000\; kg} \approx 19.8\; m \cdot s^{-1}$ .

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4 years ago

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You are throwing a ball straight up in the air. At the highest point, the ball’s

Afina-wow [57]

The answer is C. Acceleration is nonzero, but it’s velocity is zero

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4 years ago

What is the magnification of a virtual image if the image is 60.0 cm from a

Hatshy [7]

Answer:

4cm

Explanation:

Magnification of the virtual image

= image distance / object distance

Given

Image distance = 60.0cm

Object distance = 15.0cm

Therefore,

Magnification = 60.0/15.0

= 4 cm

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3 years ago

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Why echo sound cannot be heard in small room

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They can't hear an echo in small room because in it the sound can't be reflected back. For an echo of a sound to be heard,the minimum distance between the source of sound and the walls of the roomshould be 17.2 m. Obviously,in asmall room echoes cannot beheard.

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Mangrove forests play an important role in _______.

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D. All of the above

At high tide fish will feed among the mangrove roots - rich fishing ground

The trees trap sediment and soil in the river that would flow out to sea which also helps stop erosion

Wildlife utilise almost every part of the tree, with insects and birds, monkeys and lizards in the branches, shrimps and fish in the roots, and snails and clams in the soil

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2 years ago