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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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Answer: The correct option is that all of the sugar will come out of solution, and pure water will float to the top

Explanation:

Solution in the field of Chemistry is usually made up of two or more substances which contains a solute that dissolves in a solvent.

A solution can either be:

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A saturated solution is a solution with solutes that dissolves until it is unable to dissolve anymore leaving the undissolved solute beneath.

When there is mixture of a solute and a solvent in a solution the reactions that occurs are called crystallization and dissolution. Crystallization causes solid solutes to remain undissolved while dissolution is simply the dissolving process of the solute.

When Ryan added more sugar after reaching the saturation point of the mixture, the process of crystallization set in which surpassed the process of dissolution of the sugar solute leading to precipitation of the solute of out the solution.

6 0
3 years ago
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!
jeyben [28]

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For a relative frequency distribution, relative frequency is computed is computed as ____________.
Yuliya22 [10]

Answer:

For a relative frequency distribution, relative frequency is computed as the class frequency divided by the number of observations.

6 0
2 years ago
A jet accelerates from rest down a runway at 1.75m/s² for a distance of 1500 m before takeoff.
babunello [35]
A. Using the third equation of motion:
v2 = u2 + 2as
from the question;
the jet was initially at rest
hence u = 0
a = 1.75m/s2
s = 1500m
v2 = 02 + 2(1.75)(1500)
v2 = 5250
v = √5250
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b. s = ut + 1/2at2
1500 = 0(t) + 1/2(1.75)t2
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Read more on Brainly.com - brainly.com/question/18743384#readmore
8 0
3 years ago
A mason is shot with a constant speed of 7.5 x 10²m/sinto a region, when an electric field produces acceleration on the mason of
nadya68 [22]

Answer:

Distance, d = 0.1 m

It is given that,

Initial velocity of meson,

Finally, the meson is coming to rest v = 0

Acceleration of the meson,  (opposite to initial velocity)

Using third equation of motion as :

s is the distance the meson travelled before coming to rest.

So,

 

s = 0.1 m

The meson will cover the distance of 0.1 m before coming to rest. Hence, this is the required solution.  

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