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3 years ago

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which of the following locations has the largest electric field? A)0.4 cm from a +3.0 uC point charge B)0.2 cm frm a +1.5 uC poi

nt charge C)0.4 cm from a +6.0 uC point charge D)0.2 cm from a +3.0 uC point charge

1 answer:

JulsSmile [24]3 years ago

7 0

<span>Electric field is proportional to q/d^2, where q is the magnitude of the charge and d is the distance. Since all the given units are identical, we can just compare their relative magnitudes without calculating for the exact values.
A) 3/(0.4)^2 = 18.75
B) 1.5/(0.2)^2 = 37.5
C) 6/(0.4)^2 = 37.5
D) 3/(0.2)^2 = 75
Therefore, choice D has the largest electric field of all.
</span>

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6. A 2-kg ball B is traveling around in a circle of radius r1 = 1 m with a speed (vB)1 = 2 m/s. If the attached cord is pulled d

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Answer:

Explanation:

Given that,

Mass of ball m = 2kg

Ball traveling a radius of r1= 1m.

Speed of ball is Vb = 2m/s

Attached cord pulled down at a speed of Vr = 0.5m/s

Final speed V = 4m/s

Let find the transverse component of the final speed using

V² = Vr²+ Vθ²

4² = 0.5² + Vθ²

Vθ² = 4²—0.5²

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Vθ =√15.75

Vθ = 3.97 m/s.

Using the conservation of angular momentum,

(HA)1 = (HA)2

Mb • Vb • r1 = Mb • Vθ • r2

Mb cancels out

Vb • r1 = Vθ • r2

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The distance r2 to the hole for the ball to reach a maximum speed of 4m/s is 0.504m

The required time,

Using equation of motion

V = ∆r/t

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3 years ago

Compare the circular velocity of a particle orbiting in the Encke Division, whose distance from Saturn 133,370 km, to a particle

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Answer:

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

Explanation:

The circular velocity is define as:

$v = \frac{2 \pi r}{T}$

Where r is the radius of the trajectory and T is the orbital period

To determine the circular velocity of both particles it is necessary to know the orbital period of each one. That can be done by means of the Kepler’s third law:

$T^{2} = r^{3}$

Where T is orbital period and r is the radius of the trajectory.

Case for the particle in the Encke Division:

$T^{2} = r^{3}$

$T = \sqrt{(133370 Km)^{3}}$

$T = \sqrt{(2.372x10^{15} Km)}$

$T = 4.870x10^{7} Km$

It is necessary to pass from kilometers to astronomical unit (AU), where 1 AU is equivalent to 150.000.000 Km ( $1.50x10^{8} Km$ )

1 AU is defined as the distance between the earth and the sun.

$\frac{4.870x10^{7} Km}{1.50x10^{8}Km} . 1AU$

$T = 0.324 AU$

But 1 year is equivalent to 1 AU according with Kepler’s third law, since 1 year is the orbital period of the earth.

$T = \frac{0.324 AU}{1 AU} . 1 year$

$T = 0.324 year$

That can be expressed in units of days

$T = \frac{0.324 year}{1 year} . 365.25 days$

$T = 118.60 days$

<em>Circular velocity for the particle in the </em><em>Encke Division</em><em>:</em>

$v = \frac{2 \pi r}{T}$

$v = \frac{2 \pi (133370 Km)}{(118.60 days)}$

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

$118.60 days .\frac{86400 s}{1 day}$ ⇒ 10247040 s

$133370 Km .\frac{1000 m}{1 Km}$ ⇒ 133370000 m

$v = \frac{2 \pi (133370000 m)}{(10247040 s)}$

$v = 81.778 m/s$

Case for the particle in the D Ring:

For the case of the particle in the D Ring, the same approach used above can be followed

$T^{2} = r^{3}$

$T = \sqrt{(69000 Km)^{3}}$

$T = \sqrt{(3.285x10^{14} Km)}$

$T = 1.812x10^{7} Km$

$\frac{1.812x10^{7} Km}{1.50x10^{8}Km} . 1 AU$

$T = 0.120 AU$

$T = \frac{0.120 AU}{1 AU} . 1 year$

$T = 0.120 year$

$T = \frac{0.120 year}{1 year} . 365.25 days$

$T = 43.83 days$

<em>Circular velocity for the particle in </em><em>D Ring</em><em>:</em>

$v = \frac{2 \pi r}{T}$

$v = \frac{2 \pi (69000 Km)}{(43.83 days)}$

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

$43.83 days . \frac{86400 s}{1 day}$ ⇒ 3786912 s

$69000 Km . \frac{1000 m}{ 1 Km}$ ⇒ 69000000 m

$v = \frac{2 \pi (69000000 m)}{(3786912 s)}$

$v = 114.483 m/s$

$\frac{114.483 m/s}{81.778 m/s} = 1.399$

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

7 0

3 years ago

The platform height for Olympic divers is 10 m. A 60 kg diver steps off the platform to begin his dive.

azamat

Answer:

a) Ep = 5886[J]; b) v = 14[m/s]; c) W = 5886[J]; d) F = 1763.4[N]

Explanation:

a)

The potential energy can be found using the following expression, we will take the ground level as the reference point where the potential energy is equal to zero.

$E_{p} =m*g*h\\where:\\m = mass = 60[kg]\\g = gravity = 9.81[m/s^2]\\h = elevation = 10 [m]\\E_{p}=60*9.81*10\\E_{p}=5886[J]$

b)

Since energy is conserved, that is, potential energy is transformed into kinetic energy, the moment the harpsichord touches water, all potential energy is transformed into kinetic energy.

$E_{p} = E_{k} \\5886 =0.5*m*v^{2} \\v = \sqrt{\frac{5886}{0.5*60} }\\v = 14[m/s]$

c)

The work is equal to

W = 5886 [J]

d)

We need to use the following equation and find the deceleration of the diver at the moment when he stops his velocity is zero.

$v_{f} ^{2}= v_{o} ^{2}-2*a*d\\where:\\d = 2.5[m]\\v_{f}=0\\v_{o} =14[m/s]\\Therefore\\a = \frac{14^{2} }{2*2.5} \\a = 39.2[m/s^2]$

By performing a sum of forces equal to the product of mass by acceleration (newton's second law), we can find the force that acts to reduce the speed of the diver to zero.

m*g - F = m*a

F = m*a - m*g

F = (60*39.2) - (60*9.81)

F = 1763.4 [N]

3 0

3 years ago