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3 years ago

Help can’t find the answer no where

Physics

1 answer:

katen-ka-za [31]3 years ago

5 0

i'm stuck on that question also

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Hurryyyyy plzzzzzz..........the one in the middle...

zheka24 [161]

Answer:

Friction force is independent of the direction of the contacting surfaces

Explanation:

It can go any way depending on how much force is being out on it.

5 0

3 years ago

(please help i gotta turn this in a few minutes 10 points!)

pogonyaev

Answer:

3a, 2b,4c,1d

Explanation:

what do I need to explain just something you know

7 0

3 years ago

Four copper wires of equal length are connected in series. Their cross-sectional areas are 0.7 cm2 , 2.5 cm2 , 2.2 cm2 , and 3 c

Travka [436]

Answer:

22.1 V

Explanation:

We are given that

$A_1=0.7 cm^2=0.7\times 10^{-4} m^2$

$A_2=2.5 cm^2=2.5\times 10^{-4} m^2$

$A_3=2.2 cm^2=2.2\times 10^{-4} m^2$

$A_4=3 cm^2=3\times 10^{-4} m^2$

Using $1cm^2=10^{-4} m^2$

We know that

$R=\frac{\rho l}{A}$

In series

$R=R_1+R_2+R_3+R_4$

$R=\frac{\rho l}{A_1}+\frac{\rho l}{A_2}+\frac{\rho l}{A_3}+\frac{\rho l}{A_4}$

$R=\frac{\rho l}{\frac{1}{A_1}+\frac{1}{A_2}+\frac{1}{A_3}+\frac{1}{A_4}}$

Substitute the values

$R=\rho A(\frac{1}{0.7\times 10^{-4}}+\frac{1}{2.5\times 10^{-4}}+\frac{1}{2.2\times 10^{-4}}+\frac{1}{3\times 10^{-4}})$

$R=\rho l(2.62\times 10^4)$

$V=145 V$

$I=\frac{V}{R}=\frac{145}{\rho l(2.62\times 10^4)}$

Voltage across the 2.5 square cm wire= $IR=I\times \frac{\rho l}{A_2}$

Voltage across the 2.5 square cm wire= $\frac{145}{\rho l(2.62\times 10^4)}\times \frac{\rho l}{2.5\times 10^{-4}}=22.1 V$

Voltage across the 2.5 square cm wire=22.1 V

6 0

3 years ago

The answer to question 1

Oxana [17]

I have to guess it is force

6 0

3 years ago

INT Raindrops acquire an electric charge as they fall. Suppose a 2.0-mm-diameter drop has a charge of 12 pC; these are both very

horsena [70]

Answer:

W = 2.3 10² $F_{e}$

Explanation:

The force of the weight is

W = m g

let's use the concept of density

ρ= m / v

the volume of a sphere is

V = $\frac{4}{3}$ π r³

V = $\frac{4}{3}$ π (1.0 10⁻³)³

V = 4.1887 10⁻⁹ m³

the density of water ρ = 1000 kg / m³

m = ρ V

m = 1000 4.1887 10⁻⁹

m = 4.1887 10⁻⁶ kg

therefore the out of gravity is

W = 4.1887 10⁻⁶ 9.8

W = 41.05 10⁻⁶ N

now let's look for the electric force

F_e = q E

F_e = 12 10⁻¹² 15000

F_e = 1.8 10⁻⁷ N

the relationship between these two quantities is

$\frac{W}{F_e}$ = 41.05 10⁻⁶ / 1.8 10⁻⁷

\frac{W}{F_e} = 2,281 10²

W = 2.3 10² $F_{e}$

therefore the weight of the drop is much greater than the electric force

3 0

2 years ago