Which statement could be defended using data from the graph ?

A weight of 200 n is hung from a spring with a spring constant of 2500 n/m and lowered slowly. How much will the spring stretch?

melisa1 [442]

The length by which the spring got stretched will be 0.08 m. The force is directly propotional to the distance by which the spring stretched.

<h3>What is spring force?</h3>

The force required to extend or compress a spring by some distance scales linearly with respect to that distance is known as the spring force. Its formula is

F = kx

The given data in the problem is;

F is the spring force =200

K is the spring constant= 2500 N/m

x is the length by which spring got stretched =?

The stretch of the spring is found as;

$\rm F=kx \\\\ x = \frac{F}{k} \\\\\ x= \frac{200}{2500} \\\\ x=0.08 \ m$

Hence the length by which the spring got stretched will be 0.08 m.

To learn more about the spring force refer to the link;

brainly.com/question/4291098

#SPJ4

8 0

2 years ago

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How much force is needed to accelerate a 3kg book in 15m/s

kotegsom [21]

F=M×A
F=3 ×15
F=45N
so the force is 45 N

7 0

3 years ago

Struggling on this, really need help

Bad White [126]

Answer:

Guessing you just need help with the definition but if it's the question I can still help you.

7 0

2 years ago

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A fisherman notices that his boat is moving up and down periodically, owing to waves on the surface of the water. It takes 2.5 s

blondinia [14]

Answer:

1.2 m/s

0.31 m

0.15 m

Explanation:

Time period is

$T=2.5\times 2\\\Rightarrow T=5\ s$

Frequency is

$f=\dfrac{1}{T}\\\Rightarrow f=\dfrac{1}{5}\\\Rightarrow f=0.2\ Hz$

Velocity is given by

$v=f\lambda\\\Rightarrow v=0.2\times 6\\\Rightarrow v=1.2\ m/s$

The waves are traveling at 1.2 m/s

Amplitude is given by

$A=\dfrac{d}{2}\\\Rightarrow A=\dfrac{0.62}{2}\\\Rightarrow A=0.31\ m$

Amplitude is 0.31 m

If d = 0.3 m

$A=\dfrac{0.3}{2}=0.15\ m$

The amplitude would be 0.15 m. The speed would remain the same.

8 0

3 years ago

the length of iron rod at 100 C is 300.36 cm and at 159 C is 300.54 cm.Calculate its length at 0 c and coefficient of linear exp

Ugo [173]

Answer:

The length at 0 °C is 300.05 cm

Coefficient of linear expansion of iron is 1.02×10¯⁵ C¯¹

Explanation:

From the question given above, the following data were obtained:

Length (L₁) at 100 °C = 300.36 cm

Temperature 1 (θ₁) = 100 °C

Length (L₂) at 159 °C = 300.54 cm

Temperature 2 (θ₂) = 159 °C

Length (L₀) at 0 °C =?

Coefficient of linear expansion (α) =?

L₁ = L₀ (1 + θ₁α)

300.36 = L₀ (1 + 100α) ....(1)

L₂ = L₀ (1 + θ₂α)

300.54 = L₀ (1 + 159α) ..... (2)

Divide equation (2) by (1)

300.54 / 300.36 = L₀ (1 + 159α) / L₀ (1 + 100α)

1.0006 = (1 + 159α) / (1 + 100α)

Cross multiply

1.0006 (1 + 100α) = (1 + 159α)

1.0006 + 100.06α = 1 + 159α

Collect like terms

1.0006 – 1 = 159α – 100.06α

0.0006 = 58.94α

Divide both side by 58.94

α = 0.0006 / 58.94

α = 1.02×10¯⁵ C¯¹

Substitute the value of α into anything of the equation to obtain L₀. Here we shall use equation (2).

300.54 = L₀ (1 + 159α)

α = 1.02×10¯⁵ C¯¹

300.54 = L₀ (1 + 159 ×1.02×10¯⁵)

300.54 = L₀ (1 + 0.0016218)

300.54 = L₀ (1.0016218)

Divide both side by 1.0016

L₀ = 300.54 / 1.0016

L₀ = 300.05 cm

Summary:

The length at 0 °C is 300.05 cm

Coefficient of linear expansion of iron is 1.02×10¯⁵ C¯¹

6 0

3 years ago