An antiparticle has the same _______ as its elementary particle.

Describe wha scientists can learn about the past by studying fossils

inna [77]

What humans looked like and tools they used and the animals that have been alive through the centuries.

3 0

3 years ago

a student hypothesizes that the bubbles on this water plant are oxygen gas (o2). however, identifying a gas as o2 with direct ev

serious [3.7K]

A)<span>the production of bubbles by the plant when it is in darkness because in the dark the plant respires and when it does so it produces CO2 gas not O2</span>

6 0

3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Ivanshal [37]

If the wavelength increases (gets longer), then the frequency <em>decreases</em>.

(A wave occurs less often.)

7 0

3 years ago

A clothes dryer in a home draws a current of 10 amps when connected on a special 220-volts household circuit.what is the resista

aliya0001 [1]

Answer:

22Ω

Explanation:

if V ⇒ voltage

I ⇒ current

R ⇒ resistance

V = IR

220 = 10 x R

220 / 10 = R

22 = R

8 0

2 years ago

A solid conducting sphere with radius RR that carries positive charge QQ is concentric with a very thin insulating shell of radi

svetlana [45]

Answer:

$E=0$ at r < R;

$E=\frac{1}{4\pi\epsilon}\frac{Q}{r^{2}}$ at 2R > r > R;

$E=\frac{1}{4\pi\epsilon} \frac{2Q}{r^{2}}$ at r >= 2R

Explanation:

Since we have a spherically symmetric system of charged bodies, the best approach is to use Guass' Theorem which is given by,

$\int {E} \, dA=\frac{Q_{enclosed}}{\epsilon}$ (integral over a closed surface)

where,

$E$ = Electric field

$Q_{enclosed}$ = charged enclosed within the closed surface

$\epsilon$ = permittivity of free space

Now, looking at the system we can say that a sphere(concentric with the conducting and non-conducting spheres) would be the best choice of a Gaussian surface. Let the radius of the sphere be r .

at r < R,

$Q_{enclosed}$ = 0 and hence $E$ = 0 (since the sphere is conducting, all the charges get repelled towards the surface)

at 2R > r > R,

$Q_{enclosed}$ = Q,

therefore,

$E\times4\pi r^{2}=\frac{Q_{enclosed}}{\epsilon}$

(Since the system is spherically symmetric, E is constant at any given r and so we have taken it out of the integral. Also, the surface integral of a sphere gives us the area of a sphere which is equal to $4\pi r^{2}$ )

or, $E=\frac{1}{4\pi\epsilon}\frac{Q}{r^{2}}$

at r >= 2R

$Q_{enclosed}$ = 2Q

Hence, by similar calculations, we get,

$E=\frac{1}{4\pi\epsilon} \frac{2Q}{r^{2}}$

4 0

4 years ago