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3 years ago

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Arm abcd is pinned at b and undergoes reciprocating motion such that θ=(0.3 sin 4t) rad, where t is measured in seconds and the

argument for the sine is in radiaus. determine the largest speed of point a during the motion and the magnitude of the acceleration of point d at this instant.

1 answer:

storchak [24]3 years ago

8 0

<span>θ=0.3sin(4t)
w=0.3cost(4t)(4)=1.2cost(4t)
a=-4.8sin(4t)

cos4t max will always be 1 (refer to cos graph), for same reason, sin4t will always be 0

therefore, wmax=1.2rad/s

vAmax=r*w=250*1.2=300mm/s
(may be different if your picture/radius is from a different picture)

adt=a*r=200*-4.8sin(4t)=0 (sin(4t)=0)

adn=r*w^2=200*1.2^2=288

ad= square root of adt^2+adn^2 = 288mm/s^2</span>

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Two stationary point charges of 3.00 nC and 2.00 nC are separated by a distance of 50.0 cm. An electron is released from rest at

andrew-mc [135]

Answer:

1. the electric potential energy of the electron when it is at the midpoint is - 2.9 x $10^{-17}$ J

2. the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge is - 5.04 x $10^{-17}$ J

Explanation:

given information:

$q_{1}$ = 3 nC = 3 x $10^{-9}$ C

$q_{2}$ = 2 nC = 2 x $10^{-9}$ C

r = 50 cm = 0.5 m

the electric potential energy of the electron when it is at the midpoint

potential energy of the charge, F

F = k $\frac{q_{e}q}{r}$

where

k = constant (8.99 x $10^{9} Nm^{2} /C^{2}$ )

electron charge, $q_{e}$ = - 1.6 x $10^{-19}$ C

since it is measured at the midpoint,

r = $\frac{0.5}{2}$

= 0.25 m

thus,

F = $F_{1}+ F_{2}$

= k $\frac{q_{e} q_{1} }{r}$ + k $\frac{q_{e} q_{2} }{r}$

= $\frac{kq_{e} }{r}$ ( $q_{1} +q_{2}$ )

= (8.99 x $10^{9}$ )( - 1.6 x $10^{-19}$ )(3 x $10^{-9}$ +2 x $10^{-9}$ )/0.25

= - 2.9 x $10^{-17}$ J

the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge

$r_{1}$ = 10 cm = 0.1 m

$r_{2}$ = 0.5 - 0.1 = 0.4 m

F = k $\frac{q_{e} q_{1} }{r}$ + k $\frac{q_{e} q_{2} }{r}$

= $kq_{e}$ ( $\frac{q_{1} }{r_{1} }$ + $\frac{q_{2} }{r_{2} }$ )

= (8.99 x $10^{9}$ )( - 1.6 x $10^{-19}$ )(3 x $10^{-9}$ /0.1+2 x $10^{-9}$ /0.4)

= - 5.04 x $10^{-17}$ J

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As a laudably skeptical physics student, you want to test Coulomb's law. For this purpose, you set up a measurement in which a p

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Explanation:

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In this case we have an electron (-e) and a proton (e), so:

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4) This is true because electrons move around the nucleus in diffuse areas known as orbitals.

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