Ask question

Ask question

All categories

2 years ago

12

A piece of plastic is uniformly charged with surface charge density eta1. The plastic is then broken into a large piece with sur

face charge density eta2 and a small piece with surface charge density eta3. Rank in order, from largest to smallest, the surface charge densities eta1 to eta3.?
a. eta1>eta2>eta3
b. eta1>eta2=eta3
c. eta1=eta2=eta3
d. eta2=eta3>eta1

1 answer:

Amiraneli [1.4K]2 years ago

3 0

C is the answer for this problem

You might be interested in

Will give correct answer brainliest<br><br>5 kg m/s<br>8kg m/s<br>80 kg m/s<br>200 kg m/s

o-na [289]

Answer: Here this will help you..

Explanation:

1 kg-m/s to kilogram-force meter/second = 1 kilogram-force meter/second

5 kg-m/s to kilogram-force meter/second = 5 kilogram-force meter/second

10 kg-m/s to kilogram-force meter/second = 10 kilogram-force meter/second

20 kg-m/s to kilogram-force meter/second = 20 kilogram-force meter/second

30 kg-m/s to kilogram-force meter/second = 30 kilogram-force meter/second

40 kg-m/s to kilogram-force meter/second = 40 kilogram-force meter/second

50 kg-m/s to kilogram-force meter/second = 50 kilogram-force meter/second

75 kg-m/s to kilogram-force meter/second = 75 kilogram-force meter/second

100 kg-m/s to kilogram-force meter/second = 100 kilogram-force meter/second

8 0

1 year ago

A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 3.9 m/s. The car is a distanc

sammy [17]

If it takes $t$ seconds to reach the car, then the distance $d$ is $3.9t$ .

The bear's distance from the tourist's starting point is
$6t-23$

For maximum $d$ , we set the equations equal to each other:
$3.9t=6t-23$
$\Rightarrow -2.1t=-23$
$\Rightarrow t=\frac{23}{2.1}$
so the distance is
$d=3.9(\frac{23}{2.1})\approx42.741\ m$

6 0

2 years ago

Mohammad has been saving 8 each week. Today he spent 142 of the savings, and he now has 50 left. For how many weeks has he been

poizon [28]

Answer:

24 weeks

Explanation:

142+50=192

192/8=24

3 0

2 years ago

A child\'s top is held in place, upright on a frictionless surface. The axle has a radius of r = 2.96 mm. Two strings are wrappe

sladkih [1.3K]

Answer:

Explanation:

Given

radius r=2.96 mm

Tension T=2.4 N

time taken=0.74 s

Let $\alpha$ be the angular acceleration

$2 T\times r=I\times \alpha$

$2\times 2.4\times 2.96\times 10^{-3}=0.5\times m\times (2.96\times 10^{-3})^2\times \alpha$

$\alpha =\frac{4\times 2.4}{m\times 2.96\times 10^{-3}}$

$\alpha =\frac{3.24\times 10^3}{m} rad/s^2$

$\omega =\omega _0+\alpha \cdot t$

$\omega =0+\frac{3.24\times 10^3}{m}\times 0.74$

$\omega =\frac{2.4\times 10^3}{m} rad/s$

Angular momentum

$L=I\omega$

$L=0.5\times mr^2\times \omega$

$L=0.5\times m\times (2.96\times 10^{-3})^2\times \frac{2.4\times 10^3}{m}$

$L=0.01051 kg-m^2/s$

4 0

2 years ago

A 5000-lb wrecking ball hangs from a 50-ft cable of density 10 lb/ft attached to a crane. Calculate the work done if the crane l

aniked [119]

Answer:

total work is = 52450 J

Explanation:

given data

mass = 5000-lb

density = 10 lb/ft

height = 50 ft

solution

as we will treat here cable and ball are separate

and

here work need to lift cable is

w = (10Δy )(9.8 y ) j

and

now summing all segment of cable

so passing limit Δy to 0

so total work need

= $\int\limits^{10}_0 {98y} \, dy$

= $[49 y^2]^{50}_0$

= 2450J

so lifting 5000 lb wrcking 50 m required additional 5000 + 2450

so total work is = 52450 J

3 0

2 years ago