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1 year ago

eight points are in/on the circle of radius 10cm. show that distance between some two points is less than 1cm.

Physics

1 answer:

noname [10]1 year ago

6 0

Here we have shown that the distance between some two points is less than 1cm, in the circle of radius of 10 cm.

considering the points be p₁,p₂,…,p₈, now placing the point p8 at the center of the circle and the other seven points on the circle's periphery, at equal distances. we can see we have seven points in the periphery and one single point in the center. after joining the seven points, we get a heptagon having its side equal in length. The length of the side of the heptagon is less than 2π/7, which is almost less than 1 cm. if we replace the position of any points come up with a pair of points that are apart from each from a length that is less than 1 cm

To know more about circle refer to the link brainly.com/question/11833983?referrer=searchResults.

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An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el

Tresset [83]

Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is $Q =2.094 C$

Explanation:

From the question we are told that

The diameter of the wire is $d = 0.205cm = 0.00205 \ m$

The radius of the wire is $r = \frac{0.00205}{2} = 0.001025 \ m$

The resistivity of aluminum is $2.75*10^{-8} \ ohm-meters.$

The electric field change is mathematically defied as

$E (t) = 0.0004t^2 - 0.0001 +0.0004$

Generally the charge is mathematically represented as

$Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt$

Where A is the area which is mathematically represented as

$A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2$

$\frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega$

Therefore

$Q = 120 \int\limits^{t}_{0} { E(t) } \, dt$

substituting values

$Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt$

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.$

From the question we are told that t = 5 sec

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.$

$Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }$

$Q =2.094 C$

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4 years ago

After you pick up a spare, your bowling ball rolls without slipping back toward the ball rack with a linear speed of 2.85 m/s. T

motikmotik

Answer:

V' = 0.84 m/s

Explanation:

given,

Linear speed of the ball, v = 2.85 m/s

rise of the ball, h = 0.53 m

Linear speed of the ball, v' = ?

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using conservation of energy

$KE_r = \dfrac{1}{2}( \dfrac{2}{5}MR^2)(\dfrac{V}{R})2$

$KE_r = \dfrac{1}{5}MV^2$

Applying conservation of energy

Initial Linear KE + Initial roational KE = Final Linear KE + Final roational KE + Potential energy

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V'² = 0.7025

V' = 0.84 m/s

the linear speed of the ball at the top of ramp is equal to 0.84 m/s

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3 years ago

A bicyclist of mass 90 kg drives around a circle with a centripetal acceleration

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given,

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3 years ago

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4 years ago

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Answer:

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Let the height at the lowest point of the be h=0, the energy of the mass will be:

2) $E=\frac{1}{2}mv^2$

The energy when the mass comes to a stop will be:

3) $E=mgh$

Setting equations 2 and 3 equal and solving for height h will give:

4) $h=\frac{v^2}{2g}$

The angle ∅ of the string with the vertical with the mass at the highest point will be given by:

5) $cos\phi=\frac{l-h}{l}$

where l is the lenght of the string.

Combining equations 4 and 5 and solving for ∅:

6) $\phi={cos}^{-1}(\frac{l-h}{l})={cos}^{-1}(1-\frac{h}{l})={cos}^{-1}(1-\frac{v^2}{2gl})$

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4 years ago