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3 years ago

How much force was applied to a 5 kg object if it accelerated by 15 m/s/s

Mathematics

2 answers:

REY [17]3 years ago

8 0

Answer:

We use Newton's second law of motion, which states that,

where:

is the net force applied in newtons

is the mass of the object in kilograms

is the acceleration of the object in meters per second squared

So, we get:

⋅

m/s

(

∵

≡

kg m/s

)

Step-by-step explanation:

raketka [301]3 years ago

8 0

Answer:

75 Newtons

Step-by-step explanation:

By Newtons Second Law

Force F = mass * acceleration

So F = 5 * 15 = 75 N.

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How to find the vertex calculus 2What is the vertex, focus and directrix of x^2 = 6y

son4ous [18]

Solution:

Given:

$x^2=6y$

Part A:

The vertex of an up-down facing parabola of the form;

$\begin{gathered} y=ax^2+bx+c \\ is \\ x_v=-\frac{b}{2a} \end{gathered}$

Rewriting the equation given;

$\begin{gathered} 6y=x^2 \\ y=\frac{1}{6}x^2 \\ \\ \text{Hence,} \\ a=\frac{1}{6} \\ b=0 \\ c=0 \\ \\ \text{Hence,} \\ x_v=-\frac{b}{2a} \\ x_v=-\frac{0}{2(\frac{1}{6})} \\ x_v=0 \\ \\ _{} \\ \text{Substituting the value of x into y,} \\ y=\frac{1}{6}x^2 \\ y_v=\frac{1}{6}(0^2) \\ y_v=0 \\ \\ \text{Hence, the vertex is;} \\ (x_v,y_v)=(h,k)=(0,0) \end{gathered}$

Therefore, the vertex is (0,0)

Part B:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

$\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}$

Rewriting the equation in standard form,

$\begin{gathered} x^2=6y \\ 6y=x^2 \\ 4(\frac{3}{2})(y-k)=(x-h)^2 \\ \text{putting (h,k)=(0,0)} \\ 4(\frac{3}{2})(y-0)=(x-0)^2 \\ Comparing\text{to the standard form;} \\ p=\frac{3}{2} \end{gathered}$

Since the parabola is symmetric around the y-axis, the focus is a distance p from the center (0,0)

Hence,

$\begin{gathered} Focus\text{ is;} \\ (0,0+p) \\ =(0,0+\frac{3}{2}) \\ =(0,\frac{3}{2}) \end{gathered}$

Therefore, the focus is;

$(0,\frac{3}{2})$

Part C:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

$\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}$

Rewriting the equation in standard form,

Since the parabola is symmetric around the y-axis, the directrix is a line parallel to the x-axis at a distance p from the center (0,0).

Hence,

$\begin{gathered} Directrix\text{ is;} \\ y=0-p \\ y=0-\frac{3}{2} \\ y=-\frac{3}{2} \end{gathered}$

Therefore, the directrix is;

$y=-\frac{3}{2}$

3 0

1 year ago

What is equivalent to (x+1) (x-2)(x+4)(3x+7)

marta [7]

Answer: 6x*9

Step-by-step explanation:

open parathesises to

x+1*x-3*x+4*3x+7

combine like terms

6x*9

you can’t simplify further so bam

hope i did this right and it helps :)

8 0

3 years ago

Factor 15z^2+17z-18 please show work

siniylev [52]

There is no common factor between 15, 17 and 18 because 17 is prime.

15z² +17z - 18 =

d = 17² - 4.15.-18

d = 289 + 1080

d = 1369

z = (-17 +/- \/1369) : 2 * 15

z = (-17 + 37) : 30

z' = 20 : 30 = 2/3

z" = (-17 - 37) : 30

z" = -54 : 30 = -1.8

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3 years ago

Please help, What is 1+1?

Veseljchak [2.6K]

1+1 is 2
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