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3 years ago

1.- Que distancia recorrió una carga de 2,5x10-6 coul, generando así un campo eléctrico de 55new/coul.

Physics

1 answer:

sergiy2304 [10]3 years ago

3 0

Answer:

r = 20.22 m

Explanation:

Given that,

Charge, $q=2.5\times 10^{-6}\ C$

Electric field, $E=55\ N/C$

We need to find the distance. We know that, the electric field a distance r is as follows :

$E=\dfrac{kq}{r^2}\\\\r=\sqrt{\dfrac{kq}{E}}\\\\r=\sqrt{\dfrac{9\times 10^9\times 2.5\times 10^{-6}}{55}}\\\\r=20.22\ m$

So, the required distance is 20.22 m.

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The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 2.0 rev/s in 10.0 s. At th

Arte-miy333 [17]

Answer:

22 revolutions

Explanation:

2 rev/s = 2*(2π rad/rev) = 12.57 rad/s

The angular acceleration when it starting

$\alpha_a = \frac{\Delta \omega}{\Delta t} = \frac{12.57}{10} = 1.257 rad/s^2$

The angular acceleration when it stopping:

$\alpha_o = \frac{\Delta \omega}{\Delta t} = \frac{-12.57}{12} = -1.05 rad/s^2$

The angular distance it covers when starting from rest:

$\omega^2 - 0^2 = 2\alpha_a\theta_a$

$\theta_a = \frac{\omega^2}{2\alpha_a} = \frac{12.57^2}{2*1.257} = 62.8 rad$

The angular distance it covers when coming to complete stop:

$0 - \omega^2 = 2\alpha_o\theta_o$

$\theta_o = \frac{-\omega^2}{2\alpha_o} = \frac{-12.57^2}{2*(-1.05)} = 75.4 rad$

So the total angular distance it covers within 22 s is 62.8 + 75.4 = 138.23 rad or 138.23 / (2π) = 22 revolutions

6 0

3 years ago

Particle 1 and particle 2 have masses of m1 = 2.2 × 10-8 kg and m2 = 4.8 × 10-8 kg, but they carry the same charge q. The two pa

Lorico [155]

Answer: $r_2=11.81 cm$

Explanation:

Given

$m_1=2.2\times 10^{-8} kg$

$m_2=4.8\times 10^{-8} kg$

same charge on both masses

potential Energy due to Magnetic Field =Kinetic Energy of Particle

$qV=\frac{mv^2}{2}$

$v=\sqrt{\frac{2qV}{m}}$

and we know

Force due to magnetic field will Provide centripetal Force

$qvB=\frac{mv^2}{r}$

$B=\frac{\sqrt{\frac{2Vm}{q}}}{r}$

and B is equal for both particles

thus $\frac{m}{r^2}=constant$

$\frac{m_1}{r_1^2}=\frac{m_2}{r_2^2}$

$r_2^2=\frac{4.8}{2.2}\times 8^2$

$r_2=11.81 cm$

4 0

3 years ago

True of false metals like copper are sometimes used to fill cavities in teeth

vovikov84 [41]

Yes. Copper, mercury, and tin are all used to fill in cavities.

4 0

3 years ago

A sample of 5.6 L of a gas is at a pressure of 1.5 atm. If the volume of the gas is compressed to 4.8 L, what will the new press

Effectus [21]

Answer:

1.75atm

Explanation:

According to Boyle's law, the pressure P of a fixed mass of gas is inversely proportional to it's volume V provided that the temperature remains constant.

$P\alpha \frac{1}{V}\\hence\\PV=constant$

This implies the following;

$P_1V_1=P_2V_2=...=P_nV_n............(1)$ Provided temperature is kept constant.

Given;

$P_1=1.5atm\\V_1=5.6L\\P_2=?\\V_2=4.8L$

From equation (1), we can write;

$P_1V_1=P_2V_2\\hence\\1.5*5.6=P_2*4.8\\\\P_2=\frac{1.5*5.6}{4.8}\\\\P_2=1.75atm$

Since all the units are consistent, there is no need for conversion.

3 0

3 years ago

Read 2 more answers

Shay reacts solid zinc and aqueous copper sulfate to form aqueous zinc sulfate and solid copper. If he reacts 10.1 grams of zinc

zhenek [66]

Answer:

Now since mass of reactant is equal to mass of the product after the reaction so we can say that mass conservation is applicable here

Explanation:

As we know that zinc reacts with copper sulfate

so the reaction is given as

$Zn + CuSO_4 --> ZnSO_4 + Cu$

so here we have

$Zn = 10.1 g$

$CuSO_4 = 18.6 g$

$ZnSO_4 = 20 g$

$Cu = 8.7 g$

Now total mass of reactant is given as

$M_1 = 10.1 + 18.6 = 28.7 g$

Mass of the product is given as

$M_2 = 20 + 8.7 = 28.7 g$

Now since mass of reactant is equal to mass of the product after the reaction so we can say that mass conservation is applicable here

7 0

3 years ago

1.- Que distancia recorrió una carga de 2,5x10-6 coul, generando así un campo eléctrico de 55new/coul.​

1.- Que distancia recorrió una carga de 2,5x10-6 coul, generando así un campo eléctrico de 55new/coul.