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lisabon 2012 [21]
3 years ago
11

A student builds a rocket-propelled cart for a science project. Its acceleration is not quite high enough to win a prize, so he

uses a larger rocket engine that provides 39% more thrust, although doing so increases the mass of the cart by 13%. By what percentage does the cart's acceleration increase?
Physics
1 answer:
irina [24]3 years ago
5 0

Answer:

Explanation:

a = F / m

where a is acceleration , F is thrust and m is mass

taking log and differentiating

da / a = dF / F - dm / m

(da / a)x 100 = (dF / F)x100 - (dm / m) x100

percentage increase in a = percentage increase in F - percentage increase in m

= percentage increase in acceleration a   = 39 - 13 = 26 %

required increase = 26 %.

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3 years ago
a projectile whose masss is 10g is fired directly upward from ground level with an initial velocity of 1000m/s neglecting the ef
sammy [17]

Answer:

<em>The speed of the projectile when it impacts the ground is 1000 m/s</em>

Explanation:

<u>Vertical Launch</u>

When an object is launched vertically and upwards it starts to move at an initial speed vo, then the acceleration of gravity makes that speed to reduce until it reaches 0. The object has reached its maximum height. Then, it starts to move downwards in free fall, with initial speed zero and gradually increasing it until it reaches the ground level. We will demonstrate that the speed it has when impacts the ground is the same (and opposite) as the initial speed vo.

The speed when the object is moving upwards is given by

v_f=v_o-g.t

The time it takes to reach the maximum height is when vf=0, i.e.

0=v_o-g.t

solving for t

\displaystyle t=\frac{v_o}{g}

The maximum height reached is

\displaystyle y=\frac{gt^2}{2}=\frac{v_o^2}{2g}

Then, the object starts to fall. The object's height is given by

\displaystyle y=\frac{v_o^2}{2g}-\frac{gt'^2}{2}

where t' is the time the object has traveled downwards. The height will be 0 again when

\displaystyle \frac{v_o^2}{2g}=\frac{gt'^2}{2}

Solving for t'

\displaystyle t'=\frac{v_o}{g}

We can see the time it takes to reach the maximum height is the same it takes to return to ground level. Of course, the speed when it happens is

v_f=g.t'=v_o

Thus, the speed of the projectile when it impacts the ground is 1000 m/s

4 0
4 years ago
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