1) 15 / 12 = 1.25 ratio
2) to increase acceleration 1.25 times (with same F, or same engine) you have to lower mass 1.25 times
3) 1515/1.25 = 1212 kg
choose A
Answer:
I don't really know
Explanation:
I really wanted to help you, but then I realized i didnt know how to
Answer:
We know that Force = mass × acceleration
By substituting the values we get,
30 N = 15 kg × a (where a is acceleration)
Or we can write it as
15 kg × a = 30 N
Transposing 15 to RHS,
a = 30 ÷ 15 m/s²
Therefore, acceleration = 2 m/s²
pls give brainliest for the answer
The acceleration of the car would be 0.33 first and then it would be 0.17.
<u>Explanation:</u>
An applied force is a force that is applied to an object by an individual or another item. On the off chance that an individual is pushing a work area over the room, at that point there is an applied power following up on the article. The applied power is the power applied on the work area by the individual.
The net force applied to the object rises to the mass of the article increased by the measure of its acceleration. The net power following up on the soccer ball is equivalent to the mass of the soccer ball duplicated by its adjustment in speed each second (its acceleration).
Answer:
Explanation:
Let the first height be h . second height .75h
third height .75h . fourth height .75²h
fifth height .75²h , sixthth height .75³ and so on
Total distance consists of two geometric series as follows
1 ) first series
h + .75h + .75²h + .75³h......
2 ) second series
.75h +.75²h +.75³h + .75⁴h .......
Sum of first series :
first term a = h , commom ratio r = .75
sum = a / (1 - r )
= h / 1 - .75
= h / .25
4h
sum of second series :--
first term a = .75 h , commom ratio r = .75
sum = a / (1 - r )
= .75h / 1 - .75
= .75h / .25
3h
Total of both the series
= 4h + 3h
= 7h .
h = 1 m
Total distance = 7 m
