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3 years ago

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An average human weighs about 600 N. If two such generic humans each carried 1.5 coulomb of excess charge, one positive and one

negative, how far apart would they have to be for the electric attraction between them to equal their 600 N weight?

1 answer:

MakcuM [25]3 years ago

6 0

Answer:If two such generichumans each carried 1.0 coulomb of excess charge, one posit... ... Charge, One Positive Andone Negative, How Far Apart Would They Have To Be For The Electricattraction Between Them To Equal Their 650-N Weight? ... 21.5 An average human weighs about 650 N. If two such generichumans ...

Explanation:

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A 1640 kg merry-go-round with a radius of 7.50 m accelerates from rest to a rate of 1.00 revolution per 8.00 s. Estimate the mer

son4ous [18]

Solution :

Given data :

Mass of the merry-go-round, m= 1640 kg

Radius of the merry-go-round, r = 7.50 m

Angular speed, $\omega = \frac{1}{8}$ rev/sec

$=\frac{2 \pi \times 7.5}{8}$ rad/sec

= 5.89 rad/sec

Therefore, force required,

$F=m.\omega^2.r$

$$$=1640 \times (5.89)^2 \times 7.5$

= 427126.9 N

Thus, the net work done for the acceleration is given by :

W = F x r

= 427126.9 x 7.5

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3 years ago

A light bulb dissipates 100 Watts of power when it is supplied a voltage of 220 volts.

ycow [4]

Given Information:

Power = P = 100 Watts

Voltage = V = 220 Volts

Required Information:

a) Current = I = ?

b) Resistance = R = ?

Answer:

a) Current = I = 0.4545 A

b) Resistance = R = 484 Ω

Explanation:

According to the Ohm’s law, the power dissipated in the light bulb is given by

$P = VI$

Where V is the voltage across the light bulb, I is the current flowing through the light bulb and P is the power dissipated in the light bulb.

Re-arranging the above equation for current I yields,

$I = \frac{P}{V} \\\\I = \frac{100}{220} \\\\I = 0.4545 \: A \\\\$

Therefore, 0.4545 A current is flowing through the light bulb.

According to the Ohm’s law, the voltage across the light bulb is given by

$V = IR$

Where V is the voltage across the light bulb, I is the current flowing through the light bulb and R is the resistance of the light bulb.

Re-arranging the above equation for resistance R yields,

$R = \frac{V}{I} \\\\R = \frac{220}{0.4545} \\\\R = 484 \: \Omega$

Therefore, the resistance of the bulb is 484 Ω

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3 years ago

Read 2 more answers

Me podrían dar la respuesta y el proceso?

Nat2105 [25]

Answer:

I don’t understand Espanol

Explanation:

sorry

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3 years ago

A disk of radius R = 11 cm is pulled along a frictionless surface with a force of F = 16 N by a string wrapped around the edge.A

kenny6666 [7]

Answer: 1.76 Nm

Explanation:

If the force pulls horizontally, this means that the force is tangent to the disk at any point of the string unwinding process, so the distance d is irrelevant.

In this case, the torque is directly given by the product of the force times the distance perpendicular to the center of the disk, which is just the radius, as follows:

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4 years ago

Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface.

SSSSS [86.1K]

Answer:

a = 9.94 m/s²

Explanation:

given,

density at center= 1.6 x 10⁴ kg/m³

density at the surface = 2100 Kg/m³

volume mass density as function of distance

$\rho(r) = ar^2 - br^3$

r is the radius of the spherical shell

dr is the thickness

volume of shell

$dV = 4 \pi r^2 dr$

mass of shell

$dM = \rho(r)dV$

$\rho = \rho_0 - br$

now,

$dM = (\rho_0 - br)(4 \pi r^2)dr$

integrating both side

$M = \int_0^{R} (\rho_0 - br)(4 \pi r^2)dr$

$M = \dfrac{4\pi}{3}R^3\rho_0 - \pi R^4(\dfrac{\rho_0-\rho}{R})$

$M = \pi R^3(\dfrac{\rho_0}{3}+\rho)$

we know,

$a = \dfrac{GM}{R^2}$

$a = \dfrac{G( \pi R^3(\dfrac{\rho_0}{3}+\rho))}{R^2}$

$a =\pi RG(\dfrac{\rho_0}{3}+\rho)$

$a =\pi (6.674\times 10^{-11}\times 6.38 \times 10^6)(\dfrac{1.60\times 10^4}{3}+2.1\times 10^3)$

a = 9.94 m/s²

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3 years ago