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ddd [48]
2 years ago
10

A fisherman uses a spring scale to weigh a tilapia fish. He records the fish weight as a kilograms and notices that the spring s

tretches b centimeters. Which expression represents the spring constant (1 =9.8 )? A). 980ab B). 9.8ab C). 9.8ab D). 980ab
Mathematics
1 answer:
belka [17]2 years ago
4 0

Answer:

k = \frac{980a}{b}

Step-by-step explanation:

Fisherman noticed a stretch in the spring = 'b' centimetres

Weight of the fish = a kilograms

If force applied on a spring scale makes a stretch in the spring then Hook's law for the force applied is,

F = kΔx

Where k = spring constant

Δx = stretch in the spring

F = weight applied

F = mg

Here 'm' = mass of the fish

g = gravitational constant

F = a(9.8)

  = 9.8a

Δx = b centimetres = 0.01b meters

Therefore, 9.8a = k(0.01b)

k = \frac{9.8a}{0.01b}

k = \frac{980a}{b}

Therefore, spring constant of the spring will be determined by the expression, k = \frac{980a}{b}

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3

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1

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8

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18

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15

⎤

⎥

⎦

The easiest way to obtain a 1 in row 2 of column 1 is to interchange \displaystyle {R}_{2}R

​2

​​  and \displaystyle {R}_{3}R

​3

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Interchange

R

2

and

R

3

→

⎡

⎢

⎣

1

−

1

1

8

0

1

−

12

−

15

0

5

−

3

−

18

⎤

⎥

⎦

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−

5

R

2

+

R

3

=

R

3

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⎡

⎢

⎣

1

−

1

1

0

1

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12

0

0

57

|

8

−

15

57

⎤

⎥

⎦

−

1

57

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3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

1

|

8

−

15

1

⎤

⎥

⎦

The last matrix represents the equivalent system.

x

−

y

+

z

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8

y

−

12

z

=

−

15

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1

Using back-substitution, we obtain the solution as \displaystyle \left(4,-3,1\right)(4,−3,1).First, we write the augmented matrix.

⎡

⎢

⎣

1

−

1

1

2

3

−

1

3

−

2

−

9

|

8

−

2

9

⎤

⎥

⎦

Next, we perform row operations to obtain row-echelon form.

−

2

R

1

+

R

2

=

R

2

→

⎡

⎢

⎣

1

−

1

1

0

5

−

3

3

−

2

−

9

|

8

−

18

9

⎤

⎥

⎦

−

3

R

1

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

5

−

3

0

1

−

12

|

8

−

18

−

15

⎤

⎥

⎦

The easiest way to obtain a 1 in row 2 of column 1 is to interchange \displaystyle {R}_{2}R

​2

​​  and \displaystyle {R}_{3}R

​3

​​ .

Interchange

R

2

and

R

3

→

⎡

⎢

⎣

1

−

1

1

8

0

1

−

12

−

15

0

5

−

3

−

18

⎤

⎥

⎦

Then

−

5

R

2

+

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

57

|

8

−

15

57

⎤

⎥

⎦

−

1

57

R

3

=

R

3

→

⎡

⎢

⎣

1

−

1

1

0

1

−

12

0

0

1

|

8

−

15

1

⎤

⎥

⎦

The last matrix represents the equivalent system.

x

−

y

+

z

=

8

y

−

12

z

=

−

15

z=1

Using back-substitution, we obtain the solution as \displaystyle \left(4,-3,1\right)(4,−3,1).

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