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4vir4ik [10]
2 years ago
5

Newton’s Second Law of Motion states that the acceleration of an object is directly proportional to the force exerted on the obj

ect. If an object accelerates at a rate of 3 m/s2, it will exert a force of 15 N. How much force will the same object exert if it accelerates at a rate of 7 m/s^2 ?
Mathematics
1 answer:
aev [14]2 years ago
4 0

The force the same object will exert if it accelerates at a rate of 7 m/s² is 35N

From the question,

We are to determine the force the object will exert.

From the information,

Newton’s Second Law of Motion states that the acceleration of an object is directly proportional to the force exerted on the object

Let acceleration be a

and

Force be F

Then we can write that,

F ∝ a

Then,

Introducing the constant, mass, m

We get

F = ma

Now, we will determine the mass of the object

F = 15 N

a = 3 m/s²

Putting the parameters into the formula, we get

15 = m × 3

∴ m = 15 ÷ 3

m = 5 kg

The mass of the object is 5 kg.

Now, to determine the force the same object will exert if it accelerates at a rate of 7 m/s²

That is,

a = 7 m/s²

and

m = 5 kg

Putting the parameters into the formula, we get

F = 5 × 7

F = 35 N

Hence, the force the same object will exert if it accelerates at a rate of 7 m/s² is 35N

Learn more here: brainly.com/question/13590154

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Given:

P(S\cap F)=0.44\\P(S\cap F^{c})=0.13\\P(S^{c}\cap F^{c}) = 0.32\\P(S^{c}\cap F) = 0.11

The rule of total probability states that:

P(A) = P(A\cap B) + P(A\cap B^{c})

Compute the individual probabilities as follows:

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Conditional probability of an event A given B is:

P(A|B)=\frac{P(A\cap B)}{P(B)}

  • Compute the value of P(S|F):

         P(S|F)=\frac{P(S\cap F)}{P(F)}\\=\frac{0.44}{0.55}\\=0.80

  • Compute the value of P(S|F^{c})

        P(S|F^{c})=\frac{P(S\cap F^{c})}{P(F^{c})}\\=\frac{0.13}{0.45}\\=0.289

  • Compute the value of P(S^{c}|F)

        P(S^{c}|F)=\frac{P(S^{c}\cap F)}{P(F}\\=\frac{0.11}{0.55}\\=0.20

  • Compute the value ofP(S^{c}|F^{c})

       P(S^{c}|F^{c})=\frac{P(S^{c}\cap F^{c})}{P(F^{c})}\\=\frac{0.32}{0.45}\\=0.711

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