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horrorfan [7]
3 years ago
14

3000 J of work are done by an elevator on a man lifting him to a height of 2.5 m in 0.4 seconds​? how much energy has the man ga

ined in being lifted?
Physics
1 answer:
ANEK [815]3 years ago
3 0

The energy that the man has gained in being lifted is 3000J which is in the form of potential energy.

<u>Explanation:</u>

Given:

Work done, W = 3000J

Height, h = 2.5m

Time, t = 0.4 seconds

Energy gained = ?

As the man gets lifted, it gains potential energy.

Potential energy will be equal to the work done by the elevator.

We know,

potential energy = mgh

Thus, the energy that the man has gained in being lifted is 3000J which is in the form of potential energy.

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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)

So, the final E-field is

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

The magnitude of the E-field is

E = 2.66 N/m

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A meter stick is balanced at the 50 cm mark. You tie a 20 N weight at the 20 cm mark. Where should a 30 N weight be placed so th
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Answer:

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Explanation:

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Balance point on 50 cm

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As the force is conserved we have

\dfrac{20}{30}\times 30=20\ cm

The distance will be 20 cm to the right of the center or 20+50 = 70 cm from the left side.

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