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IceJOKER [234]
3 years ago
12

For a moving object, the force acting on the object varies directly with the object's acceleration. When a force of 10 N acts on

a certain object, the acceleration of the object is 5 /ms2 . If the acceleration of the object becomes 6 /ms2 , what is the force
Mathematics
1 answer:
Step2247 [10]3 years ago
5 0

Answer:

The force is 12 N

Step-by-step explanation:

* Lets explain how to solve the problem

- Direct variation is a relationship between two variables that can

 be expressed by an equation in which one variable is equal to a

 constant times the other

- If y ∝ x , then y = kx , where k is the constant of variation

* Lets solve the problem

- The force acting on the object varies directly with the object's

  acceleration

∵ The force is F in newtons and a is the acceleration is m/s²

∴ F ∝ a

∴ F = ka

- To find k substitute F and a by the initial values of them

∵ A force of 10 N acts on a certain object, the acceleration of the

  object is 5 m/s²

∵ F = 10 N when a = 5 m/s²

∵ F = ka

∴ 10 = k(5)

- Divide both sides by 5

∴ K = 2

- Substitute the value of k in the equation

∴ F = 2a

- Lets find the force when the acceleration is 6 m/s²

∵ a = 6 m/s²

∴ F = 2(6) = 12

* The force is 12 N

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Gelneren [198K]

Answer:

A. y = \frac{1}{2}x + 10

Step-by-step explanation:

Given;

  • The equation y = -7x - 5
  • and another equation have a solution of (-2, 9)

If we equate the value of y in the first equation and the value of y in the second equation, then the solution of x should be -2 . Let's test this with equation A:

-7x - 5 = \frac{1}{2}x + 10

-7x - \frac{1}{2}x = 10 + 5

-\frac{15}{2}x = 15

x = (15 × -2) ÷ 15 = -2

*<em>This proves that equation A has the same value of x as our first equation!</em>

<em />

Let's test if equation A has a solution; y = 9:

y = \frac{1}{2}(-2) + 10 = 9

*<em>This proves that equation A has the same value of y as our first equation!</em>

<em />

So the second equation is,

A. y = \frac{1}{2}x + 10

<em />

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