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2 years ago

8

For a moving object, the force acting on the object varies directly with the object's acceleration. When a force of 21 N acts on

a certain object, the acceleration
of the object is 7 mis. f the acceleration of the object becomes 2 m/s?, what is the force?

1 answer:

boyakko [2]2 years ago

6 0

Answer: i think its 65n

Step-by-step explanation:

You might be interested in

Simplify and solve each equation for the x

Olegator [25]

Answer:

19y - 7y = 1 + 4(x+2x)

y = 1/12 + x

x = y - 1/12

I hope this helps!

8 0

2 years ago

The Amaco Middle School bookstore sells pencils costing a whole number of cents. Some seventh graders each bought a pencil, payi

vodka [1.7K]

Answer:

12

Step-by-step explanation:

4 0

2 years ago

The area of a rectangular garden is given by the trinomial x2 + x – 42. What are the possible dimensions of the rectangle? Use f

Murrr4er [49]

Answer:

Step-by-step explanation:

Hello, please consider the following.

$x^2+x-42\\\\\text{The sum of the zeroes is -1=-7+6 and the product is -42=-7*6.}\\\\\text{So, we can factorise.}\\\\x^2+x-42\\\\=x^2+7x-6x-42\\\\=x(x+7)-6(x+7)\\\\=(x+7)(x-6)$

So, the possible dimensions of the rectangle are (x+7) and (x-6).

Thank you.

3 0

3 years ago

Read 2 more answers

Fine length of BC on the following photo.

MrMuchimi

Answer:

$BC=4\sqrt{5}\ units$

Step-by-step explanation:

see the attached figure with letters to better understand the problem

step 1

In the right triangle ACD

Find the length side AC

Applying the Pythagorean Theorem

$AC^2=AD^2+DC^2$

substitute the given values

$AC^2=16^2+8^2$

$AC^2=320$

$AC=\sqrt{320}\ units$

simplify

$AC=8\sqrt{5}\ units$

step 2

In the right triangle ACD

Find the cosine of angle CAD

$cos(\angle CAD)=\frac{AD}{AC}$

substitute the given values

$cos(\angle CAD)=\frac{16}{8\sqrt{5}}$

$cos(\angle CAD)=\frac{2}{\sqrt{5}}$ ----> equation A

step 3

In the right triangle ABC

Find the cosine of angle BAC

$cos(\angle BAC)=\frac{AC}{AB}$

substitute the given values

$cos(\angle BAC)=\frac{8\sqrt{5}}{16+x}$ ----> equation B

step 4

Find the value of x

In this problem

$\angle CAD=\angle BAC$ ----> is the same angle

so

equate equation A and equation B

$\frac{8\sqrt{5}}{16+x}=\frac{2}{\sqrt{5}}$

solve for x

Multiply in cross

$(8\sqrt{5})(\sqrt{5})=(16+x)(2)\\\\40=32+2x\\\\2x=40-32\\\\2x=8\\\\x=4\ units$

$DB=4\ units$

step 5

Find the length of BC

In the right triangle BCD

Applying the Pythagorean Theorem

$BC^2=DC^2+DB^2$

substitute the given values

$BC^2=8^2+4^2$

$BC^2=80$

$BC=\sqrt{80}\ units$

simplify

$BC=4\sqrt{5}\ units$

7 0

2 years ago

Convert a markup of 15% on selling price to its equivalent markup on cost.

spayn [35]

Answer:

option: D is correct. ( (D) a markup of 15% on selling price is equivalent to a 85% markup on cost ).

Step-by-step explanation:

" Markup is the percentage difference between a product's cost and its selling price "

Since we are asked to Convert a markup of 15% on selling price to its equivalent markup on cost.

As there is a markup of 15% on selling price. Hence the cost of the product would be (100-15)%=85% of the selling price.

<em>Hence, a markup of 15% on selling price is equivalent to a 85% markup on cost.</em>

<em>Hence, option D is true.</em>

5 0

3 years ago