PLEASE HELP ME ANSWER DUE SOON BRAINLIEST FOR WHOEVER

A portable music player, operating with four 1.5 V cells connected in series, provides a resistance of 15 000 Ω. What amount of

nata0808 [166]

Answer:

Explanation:

Given that,

A portable music player is operating with 4 cell batteries connected in series, and each cell has a P.D of 1.5V.

Then,

Total potential difference is

P.D_total = V1 + V2 + V3 + V4

P.D_total = 1.5 + 1.5 + 1.5 + 1.4

P.D_total = 6V.

The music player provides a resistance of 15,000Ω

R = 15,000Ω

We want to find the current (I) flowing through the music player?

Using ohms law

V = IR

Where

V is the potential difference

I is the current

R is the resistance

Therefore,

I = V/R

I = 6 / 15,000

I = 4 × 10^-4 A

I = 0.4 × 10^-3 A

I = 0.4 mA.

So, 0.4mA is passing through the music player

7 0

3 years ago

Two thin concentric spherical shells of radii r1 and r2 (r1 < r2) contain uniform surface charge densities V1 and V2, respect

Lyrx [107]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

So,

a) 0 < r < r1 :

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

Hence, E = 0 for r < r1

b) r1 < r < r2:

Electric field =?

Let, us consider the Gaussian Surface,

E x 4 $\pi$ $r^{2}$ = $\frac{Q1}{E_{0} }$

So,

Rearranging the above equation to get Electric field, we will get:

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$

Multiply and divide by $r1^{2}$

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ x $\frac{r1^{2} }{r1^{2} }$

Rearranging the above equation, we will get Electric Field for r1 < r < r2:

E= (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ )

c) r > r2 :

Electric Field = ?

E x 4 $\pi$ $r^{2}$ = $\frac{Q1 + Q2}{E_{0} }$

Rearranging the above equation for E:

E = $\frac{Q1+Q2}{E_{0} . 4 \pi. r^{2} }$

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ + $\frac{Q2}{E_{0} . 4 \pi. r^{2} }$

As we know from above, that:

$\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ = (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ )

Then, Similarly,

$\frac{Q2}{E_{0} . 4 \pi. r^{2} }$ = (σ2 x $r2^{2}$ ) /( $E_{0}$ x $r^{2}$ )

So,

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ + $\frac{Q2}{E_{0} . 4 \pi. r^{2} }$

Replacing the above equations to get E:

E = (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ ) + (σ2 x $r2^{2}$ ) /( $E_{0}$ x $r^{2}$ )

Now, for

d) Under what conditions, E = 0, for r > r2?

For r > r2, E =0 if

σ1 x $r1^{2}$ = - σ2 x $r2^{2}$

4 0

3 years ago

If a conductor wire gets heated. What happens with its resistance?

yawa3891 [41]

If a conductor wire is heated, its resistance decreases and conductivity increases.

4 0

3 years ago

ANSWER ASAPPPPPPPP BEING TIMED

Dmitrij [34]

C) above, the Earth's core is about the melting point of iron.

8 0

3 years ago

Read 2 more answers

The pirate ship tie at the amusement park is a giant pendulum that riders sit in. It swings back and forth, with a maximum veloc

kkurt [141]

Here as we know that there is no loss of energy

so we can say that maximum kinetic energy will become gravitational potential energy at its maximum height

So here we have

$\frac{1}{2}mv^2 = mgh$

here we have

v = 20 m/s

m = 8000 kg

now from above equation we have

$\frac{1}{2}(8000)(20^2) = (8000)(9.8)h$

$h = \frac{200}{9.8)$

$h = 20.4 m$

so maximum height is 20.4 m

4 0

3 years ago

Read 2 more answers