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8_murik_8 [283]
2 years ago
7

At the beginning of its route, a garbage truck accelerates at 10ms2 as it is driven down the road. By the end of the route, the

truck's mass has quadrupled. If the truck's force remains constant, what is the truck's acceleration at the end of the route?
select one
- 2.5 m/s^2
- 10 m/s^2
- 20 m/s^2
- 5 m/s^2
Mathematics
1 answer:
Dmitry [639]2 years ago
8 0

Answer:

2.5\ m/s^2

Step-by-step explanation:

Newton's Second Law: Force on a body is equal to the product of mass and acceleration of the centre of the mass of the body.

Force(F)=Mass(m)\times Acceleration(a)

Initially:

Let\ mass=m\ kg\\\\acceleration=10\ m/s^2\\\\Force(F)=m\times 10=10m

At the end of the road:

Let\ mass=4\times m\ kg\\\\acceleration=a\ m/s^2\\\\Force(F)=4m\times a=4ma

Both\ forces\ are\ equal\\\\4ma=10m\\\\divide\ both\ the\ sides\ by\ m\\\\4a=10\\\\a=\frac{10}{4}\\\\a=2.5\ m/s^2

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Butoxors [25]
Hello!

Change each mixed number to an improper fraction to make it easier

2 3/8 = 19/8

5 3/4 = 23/4

Now, multiply

19/8  × 23/4 = 437/32 = 13 21/32

Answer ⇒ 13 21/32

4 0
3 years ago
Read 2 more answers
Write a quadratic function, in standard form, that fits the set of points. Solve it as a system of three equations. (-4, 9), (0,
Tresset [83]

Answer:

\boxed{y=2x^2+4x-7}

Step-by-step explanation:

Let the quadratic function be

y=ax^2+bx+c


We substitute (-4,9) into the equation to obtain;


9=a(-4)^2+b(-4)+c


\Rightarrow 9=16a-4b+c---(1)


We substitute (0,-7) to obtain;


-7=a(0)^2+b(0)^2+c


\Rightarrow c=-7---(2)


We finally substitute (1,-1) to obtain;


-1=a(1)^2+b(1)^2+c


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We put equation (2) into equation (1) to get;


9=16a-4b-7


16a-4b=16


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\Rightarrow -1=a+b-7


\Rightarrow a+b=6---(5)


We add equation (4) and (5) to get;

4a+a=6+4



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\Rightarrow a=2


We put a=2 into equation (5) to get;


2+b=6


\Rightarrow b=6-2


\Rightarrow b=4


The reqiured quadratic function is

y=2x^2+4x-7

7 0
2 years ago
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<span>width = x
</span><span>length = 6x + 8 

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Answer:

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Step-by-step explanation:

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