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Marianna [84]
3 years ago
6

A spring has a natural length of 14 cm. If a 23-N force is required to keep it stretched to a length of 20 cm, how much work W i

s required to stretch it from 14 cm to 17 cm? (Round your answer to two decimal places.)
Physics
1 answer:
lubasha [3.4K]3 years ago
4 0

Answer:

1.78 J

Explanation:

Find the spring coefficient using Hooke's law:

F = k Δx

23 N = k (0.20 m − 0.14 m)

k = 383.33 N/m

The work is the change in energy:

W = PE₂ − PE₁

W = ½ kx₂² − ½ kx₁²

W = ½ k (x₂² − x₁²)

W = ½ (383.33 N/m) ((0.17 m)² − (0.14 m)²)

W = 1.78 J

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Explanation:

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Density=\frac {Mass}{Volume}

The Si unit for density is kg/m³

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Specific volume is defined as the ration of the volume to the mass. It is the reciprocal of density.

The BG unit of specific volume is ft³ / lb.

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Which of the following statements is true? A. Both warming up and cooling down or important. B. It is more important warm up the
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A car drives 215km east and then 45km north. What is the magnitude of the cars displacement? Round you answer to nearest whole n
kifflom [539]

Answer: 219.65\ \text{km}

Explanation:

Given

Car drives 215 km east and then 45 km North

Displacement is East direction is

\Rightarrow \vec{r_1}=215\hat{i}

Now, the displacement from that to 45 km North is given by

\Rightarrow \vec{r_{21}}=45\hat{j}

Net displacement is \vec{r_2}

\Rightarrow \vec{r_2}=\vec{r_1}+\vec{r_{21}}\\\Rightarrow \vec{r_2}=215\hat{i}+45\hat{j}

Magnitude of the displacement is

 \Rightarrow \left | r_{2} \right |=\sqrt{\left ( 215 \right )^2+\left ( 45 \right )^2}\\\Rightarrow  \left | r_{2} \right |=\sqrt{48250}\\\Rightarrow  \left | r_{2} \right |=219.65\ \text{km}

5 0
3 years ago
An 88 kg person steps into a car of mass 2002 kg, causing it to sink 5.36 cm on itssprings. Assuming no damping, with what fre-q
Marina CMI [18]

Answer:

The required frequency = 0.442 Hz

Explanation:

Frequency f  = ( \dfrac{1}{2 \pi}) \omega

where;

\omega = \sqrt{\dfrac{k}{m} }

Then;

f = \Bigg ( \dfrac{1}{2 \pi}  \Bigg )   \Bigg( \sqrt{\dfrac{k}{m} }  \Bigg )

However;

k  = \dfrac{F}{x} and;

mass m = m_{car } + m_{person}

f = \Bigg ( \dfrac{1}{2 \pi}  \Bigg )   \Bigg( \sqrt{\dfrac{\dfrac{F}{x}}{m_{car}+m_{person}} }  \Bigg )

f = \Bigg ( \dfrac{1}{2 \pi}  \Bigg )   \Bigg( \sqrt{\dfrac{{F}}{x(m_{car}+m_{person})} }  \Bigg )

where;

F = m_{person}g

Then;

f = \Bigg ( \dfrac{1}{2 \pi}  \Bigg )   \Bigg( \sqrt{\dfrac{ {m_{person}g }}{x(m_{car}+m_{person})} }  \Bigg )

replacing the values;

f = \Bigg ( \dfrac{1}{2 \pi}  \Bigg )   \Bigg( \sqrt{\dfrac{ {(88 \ kg)* (9.81 \ m/s^2) }}{(5.36 \times 10^{-2} \ m) (2002 \ kg +88 \ kg)} }  \Bigg )

\mathbf{f = 0.442 \ Hz}

8 0
3 years ago
The force in Newtons on a particle directed along the x-axis is given by F(x)=exp(−(x/2)+6) for x≥0 where x is in meters. The pa
HACTEHA [7]

To find the work done on the particle, the following is the solution:

Dw = F dx

W = integral over the path ( F(x) dx)

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W = 135 J

The work done is 135 J.

3 0
3 years ago
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