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3 years ago

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Two circular rods, one steel and the other copper, are both 0.780 m long and 1.50 cm in diameter. Each is subjected to a force w

ith magnitude 4350 N that compresses the rod. What is the difference in the length of the two rods when compressed?

1 answer:

astra-53 [7]3 years ago

8 0

Answer:

The difference in the length of the two rods when compressed is $5.4\times10^{-5}\ m$ .

Explanation:

Given that,

Length = 0.780 m

Diameter = 1.50 cm

Force = 4350 N

(a). For steel rod

We know ,

The young modulus for steel rod

$Y=2\times10^{11}$

Using formula of young modulus

$e_{s}=\dfrac{Fl}{AY}$

$e_{s}=\dfrac{4350\times0.780}{3.14\times(0.75\times10^{-2})^2\times2\times10^{11}}$

$e_{s}=9.6\times10^{-5}\ m$

(b). For copper rod

We know ,

The young modulus for steel rod

$Y=1.1\times10^{11}$

Using formula of young modulus

$e_{c}=\dfrac{Fl}{AY}$

$e_{c}=\dfrac{4350\times0.780}{3.14\times(0.75\times10^{-2})^2\times1.1\times10^{11}}$

$e_{c}=1.5\times10^{-4}\ m$

The difference in the length of the two rods when compressed is

$difference\ in\ length=e_{c}-e_{s}$

$difference\ in\ length=1.5\times10^{-4}-9.6\times10^{-5}$

$difference\ in\ length =5.4\times10^{-5}\ m$

Hence, The difference in the length of the two rods when compressed is $5.4\times10^{-5}\ m$ .

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cupoosta [38]

Answer:

a) The x and y components of the momentum are $8.731\,\frac{kg\cdot m}{s}$ and $-11.661\,\frac{kg\cdot m}{s}$ , respectively.

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Explanation:

a) The vectorial equation of momentum is represented by the following expression:

$\vec p = m\cdot \vec v$ (1)

Where:

$\vec p$ - Vector momentum, measured in kilogram-meters per second.

$m$ - Mass of the particle, measured in kilograms.

$\vec v$ - Vector velocity, measured in meters per second.

If we know that $m = 2.93\,kg$ and $\vec v = 2.98\,\hat{i}-3.98\,\hat{j}\,\,\,\left[\frac{m}{s} \right]$ , then the momentum is:

$\vec p = (2.93)\cdot (2.98\,\hat{i}-3.98\,\hat{j})\,\,\,\left[\frac{kg\cdot m}{s} \right]$

$\vec p = 8.731\,\hat{i}-11.661\,\hat{j}\,\,\,\left[\frac{kg\cdot m}{s} \right]$

The x and y components of the momentum are $8.731\,\frac{kg\cdot m}{s}$ and $-11.661\,\frac{kg\cdot m}{s}$ , respectively.

b) The magnitude and direction of momentum are represented by the following expressions:

$\|\vec p \| = \sqrt{p_{x}^{2}+p_{y}^{2}}$ (2)

$\theta = \tan^{-1}\left(\frac{p_{y}}{p_{x}} \right)$ (3)

Where:

$\|\vec p\|$ - Magnitude of momentum, measured in kilogram-meters per second.

$\theta$ - Direction of momentum, measured in sexagesimal degrees.

If we know that $p_{x} = 8.731\,\frac{kg\cdot m}{s}$ and $p_{y} = -11.661\,\frac{kg\cdot m}{s}$ , then the magnitude and direction of momentum are, respectively:

$\|\vec p\| = \sqrt{\left(8.731\,\frac{kg\cdot m}{s} \right)^{2}+\left(-11.661\,\frac{kg\cdot m}{s} \right)^{2}}$

$\|\vec p\| \approx 14.567\,\frac{kg\cdot m}{s}$

$\theta =\tan^{-1}\left(\frac{-11.661\,\frac{kg\cdot m}{s} }{8.731\,\frac{kg\cdot m}{s} } \right)$

$\theta \approx 306.823^{\circ}$

The magnitude and direction of its momentum are approximately 14.567 kilogram-meters per second and 306.823º.

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Answer:

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