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4 years ago

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Physics

1 answer:

Aleonysh [2.5K]4 years ago

3 0

Answer:

8.0 mm

Explanation:

Force of a spring is:

F = kx

where k is the spring constant and x is the displacement.

When the force is equal to Ben's weight:

mg = kx

(180 kg) (9.8 m/s²) = (220,000 N/m) x

x = 0.0080 m

The spring compresses 0.0080 m, or 8.0 mm.

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A cylindrical rod with length (1.7 m) and radius (2 cm) is fixed from one end and a mass of (20 kg) is attached to the other fre

belka [17]

Answer:

0.44

Explanation:

Poisson's ratio is:

ν = (3K − E) / 6K

where K is the bulk modulus and E is Young's modulus.

Young's modulus is:

E = FL / (AΔL)

where F is the force, L is the initial length, A is the cross sectional area, and ΔL is the change in length.

E = (20 kg × 9.8 m/s²) (1.7 m) / (π (0.02 m)² × 0.0005 m)

E = 0.530×10⁹ Pa

Bulk modulus is:

K = -ΔP / (ΔV/V)

where ΔP is the change in pressure, ΔV is the change in volume, and V is the initial volume.

K = -(180 atm × 101325 Pa/atm) / (-0.012)

K = 1.52×10⁹ Pa

Therefore, the Poisson's ratio is:

ν = (3(1.52×10⁹ Pa) − 0.530×10⁹ Pa) / 6(1.52×10⁹ Pa)

ν = (3(1.52) − 0.530) / 6(1.52)

ν = 0.442

Rounded to 2 significant figures, the Poisson's ratio is 0.44.

3 0

3 years ago

A thin, uniformly charged insulating rod has a linear charge density λ = 3 nC/m and lies along the x axis from x = 1m to x = 3m.

Contact [7]

Answer:

A) $V_A = 11.93~V$

B) The vector definition of E-field is

$\vec{E} = -1.13\^x + 2.41\^y$

where magnitude is E = 2.66 N/m.

Explanation:

The potential of a uniformly charged rod can be found by the method of integration. We will first choose an infinitesimal part on the rod. We will compute the potential of this part at point A. Then we will integrate this potential over the entire rod.

We will use the following formula for electric potential:

$V = \frac{1}{4\pi \epsilon_0}\frac{Q}{r}$

Let us choose the infinitesimal part a distance 'x' from the origin. Then the distance between this point and point A is

$r = \sqrt{x^2+4^2}$

The infinitesimal length is 'dx', and the potential of this length is dV. Let's apply the formula:

$dV = \frac{1}{4\pi\epsilon_0}\frac{\lambda dx}{\sqrt{x^2 + 4^2}}$

Here, the charge Q is equal to the charge density multiplied by the length. Q = λdx

Now we have to integrate this infinitesimal potential over the rod:

$V = \int\limits^3_1 {dV} \, dx = \frac{1}{4\pi \epsilon_0}\int\limits^3_1 {\frac{\lambda}{\sqrt{x^2 + 16}} \, dx$

By using an integral table, this can be calculated:

$V = \frac{3\times 10^{-9}}{4\pi\epsilon_0}\ln(|\sqrt{x^2+16}+x|)\left \{ {{x=3} \atop {x=1}} \right. \\V = 11.93~V$

B) The electric field can be found by a similar approach, but a different formula:

$\vec{E} = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}\^r$

Let's apply this formula to the infinitesimal part we have chosen.

$dE_x = \frac{1}{4\pi\epsilon_0}\frac{\lambda dx}{x^2 + 4^2}\cos(\theta)\\dE_y = \frac{1}{4\pi\epsilon_0}\frac{\lambda dx}{x^2 + 4^2}\sin(\theta)$

By the geometry sine and cosine terms can be found:

$\sin(\theta) = \frac{4}{\sqrt{x^2+16}}\\\cos(\theta) = \frac{x}{\sqrt{x^2 + 16}}$

The x- and y-components of the E-field can be found separately by integrating the infinitesimal parts over the entire rod.

$E_x = \int\limits^3_1 {dE_x} \, dx = \frac{\lambda}{4\pi\epsilon_0}\int\limits^3_1 {\frac{x}{(x^2+16)^{3/2}}} \, dx = 1.13(-\^x)\\E_y = \int\limits^3_1 {dE_y} \, dx = \frac{4\lambda}{4\pi\epsilon_0}\int\limits^3_1 {\frac{1}{(x^2+16)^{3/2}}} \, dx = 2.41(\^y)$