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4 years ago

What happens to gravity when someone jumps up?

Physics

1 answer:

Alla [95]4 years ago

5 0

Answer:

The direct answer to the question as written is as follows: nothing happens to gravity when someone jumps up - gravity continues exerting a force on the body of that particular someone proportional to (mass of someone) x (mass of Earth) / (distance squared). What you might be asking, however, is what is the net force acting on the body of someone jumping up. At the moment of someone jumping up there is an upward acceleration, i.e., an upward-directed force which counteracts the gravitational force - this is the net force ( a result of the jump force minus gravity). From that moment on, only gravity acts on the body. The someone moves upward gradually decelerating to the downward gravitational acceleration until they reaches the peak of the jump (zero velocity). Then, back to Earth.

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igor_vitrenko [27]

Answer: (a) α = $\frac{3I}{2.\pi.R^{3}}$

(b) For r≤R: B(r) = μ_0. $(\frac{I.r^{2}}{2.\pi.R^{3}})$

For r≥R: B(r) = μ_0. $(\frac{I}{2.\pi.r})$

Explanation:

(a) The current I enclosed in a straight wire with current density not constant is calculated by:

$I_{c} = \int {J} \, dA$

where:

dA is the cross section.

In this case, a circular cross section of radius R, so it translates as:

$I_{c} = \int\limits^R_0 {\alpha.r.2.\pi.r } \, dr$

$I_{c} = 2.\pi.\alpha \int\limits^R_0 {r^{2}} \, dr$

$I_{c} = 2.\pi.\alpha.\frac{r^{3}}{3}$

$\alpha = \frac{3I}{2.\pi.R^{3}}$

For these circunstances, α = $\frac{3I}{2.\pi.R^{3}}$

(b) Ampere's Law to calculate magnetic field B is given by:

$\int\ {B} \, dl =$ μ_0. $I_{c}$

(i) First, first find $I_{c}$ for r ≤ R:

$I_{c} = \int\limits^r_0 {\alpha.r.2\pi.r} \, dr$

$I_{c} = 2.\pi.\frac{3I}{2.\pi.R^{3}} \int\limits^r_0 {r^{2}} \, dr$

$I_{c} = \frac{I}{R^{3}}\int\limits^r_0 {r^{2}} \, dr$

$I_{c} = \frac{3I}{R^{3}}\frac{r^{3}}{3}$

$I_{c} = \frac{I.r^{3}}{R^{3}}$

Calculating B(r), using Ampere's Law:

$\int\ {B} \, dl =$ μ_0. $I_{c}$

$B.2.\pi.r = (\frac{Ir^{3}}{R^{3}} )$ .μ_0

B(r) = $(\frac{Ir^{3}}{R^{3}2.\pi.r})$ .μ_0

B(r) = $(\frac{Ir^{2}}{2.\pi.R^{3}} )$ .μ_0

For r ≤ R, magnetic field is B(r) = $(\frac{Ir^{2}}{2.\pi.R^{3}} )$ .μ_0

(ii) For r ≥ R:

$I_{c} = \int\limits^R_0 {\alpha.2,\pi.r.r} \, dr$

So, as calculated before:

$I_{c} = \frac{3I}{R^{3}}\frac{R^{3}}{3}$

$I_{c} =$ I

Using Ampere:

B.2.π.r = μ_0.I

B(r) = $(\frac{I}{2.\pi.r} )$ .μ_0

For r ≥ R, magnetic field is; B(r) = $(\frac{I}{2.\pi.r} )$ .μ_0.

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