The length of a spring varies directly with the mass of an object that is attached to it. when a 30-gram object is attached, the spring stretches 9 centimeters. which equation relates the mass of the object, m, and the length of the spring, s?

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The length of a spring varies directly with the mass of an object that is attached to it. when a 30-gram object is attached, the spring stretches 9 centimeters. which equation relates the mass of the object, m, and the length of the spring, s?

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Step-by-step explanation:

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[tex]s = \frac{3}{10}m[/tex]

I hope this helps!

s=3/10m

Step-by-step explanation:

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Step-by-step explanation:

The third is the answer. Hope this helps.

[tex]s = \frac{3}{10}m[/tex]

Step-by-step explanation:

Given

Mass = 30g

Spring Stretch = 9cm

Variation: Direct Variation

Required

Equation that relates the mass of the object, m, and the length of the spring, s

Let m represents mass and s represents length of the spring

Given that s varies directly to m;

This implies that

[tex]s\ \alpha\ m[/tex]

Convert the above to an equation

[tex]s = km[/tex]; where k is the constant of variation;

The next step is to solve for k

Divide both sides by m

[tex]\frac{s}{m} = \frac{km}{m}[/tex]

[tex]\frac{s}{m} = k[/tex]

[tex]k = \frac{s}{m}[/tex]

From the given parameters;

when m = 30; s = 9.

[tex]k = \frac{9}{30}[/tex]

Divide the numerator and denominator by 3

[tex]k = \frac{3}{10}[/tex]

To get the equation that relates the mass of the object to the length of the spring

Substitute [tex]k = \frac{3}{10}[/tex] in [tex]s = km[/tex]

This becomes

[tex]s = \frac{3}{10}m[/tex]

Hence, the equation that relates the mass and length is [tex]s = \frac{3}{10}m[/tex]

s = (3 cm)/(10 g)*m

Step-by-step explanation:

The length of a spring varies directly with the mass means that the equation has the next form: s = k*m, where m refers to the mass of the object and s refers to the length of the spring. If a 30-gram object is attached, the spring stretches 9 centimeters, then:

s = k*m

9 cm = k*30 g

(9 cm)/(30 g) = k

(3 cm)/(10 g) = k

Therefore, the equation is s = (3 cm)/(10 g)*m

M/s=30/9 =10/3

so, 3m=10s

Hi.

your answer is s=3/10m

hope this helps!!!

The required equation is [tex]s=\frac{3}{10}m[/tex]

Step-by-step explanation:

Given: The length of a spring varies directly with the mass of an object that is attached to it. When a 30-gram object is attached, the spring stretches 9 centimeters.

To find : Which equation relates the mass of the object, m, and the length of the spring, s?

Solution :

Let the mass of the object m,

Length of the spring s,

When a 30-gram object is attached, the spring stretches 9 centimeters.

i.e, the times length of the object is attached to spring is [tex]\frac{9}{30}=\frac{3}{10}[/tex]

The length of a spring varies directly with the mass of an object that is attached to it.

According to question,

[tex]s=\frac{3}{10}m[/tex]

i.e, length of the spring is 3 by 10 times the mass of the object.

Therefore, The required equation is [tex]s=\frac{3}{10}m[/tex]

We have been given that the length (s) of a spring varies directly with me mass (m) of an object. We are asked to find the equation that represents the relation, when a 30-gram object is attached the spring stretches 9 centimeters.

We know that two directly proportional quantities are in form [tex]y=kx[/tex], where y is directly proportional to x and k is constant of proportionality.

Let us find constant of proportionality for our relation.

[tex]s=k\cdot m[/tex]

[tex]9=k\cdot 30[/tex]

[tex]\frac{9}{30}=\frac{k\cdot 30}{30}[/tex]

[tex]\frac{3}{10}=k[/tex]

[tex]0.3=k[/tex]

Therefore, our required equation would be [tex]s=0.30m[/tex].