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?
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?
30
Step-by-step explanation:
huhvujhuug gu hhu gbc y6
[tex]s = \frac{3}{10}m[/tex]
I hope this helps!
s=3/10m
Step-by-step explanation:
option a) on edge
om is
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].