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2 years ago

How can resistance be useful and where can they be useful?

1 answer:

3 0

-- Resistance can be useful among the population of a repressive government.
Although it can be dangerous for those who resist, it can also exert pressure
against the regime to alter its repressive practices.

-- Resistance can also be useful in electronic circuits. "Lumped" components with
known numerical values of resistance are used to divide voltage, limit current, and
dissipate controlled amounts of electrical energy.

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A new planet is discovered in an approximately circular orbit beyond Pluto. It moves at a rate of approximately 1° per year. It'

Mazyrski [523]

A new planet is discovered in an approximately circular orbit beyond Pluto. It moves at a rate of approximately 1° per year. It's semi-major axis is approximately:
answer:::::50 AU

6 0

3 years ago

A 0.0400-g positive charged ball with charge q = 6.40 μC is resting on a flat, frictionless horizontal surface. For a time of t

Leto [7]

Answer:

The height is 0.1014 m

Explanation:

Given that,

Mass = 0.0400 g

Charge $q= 6.40\ \mu C$

Time t = 0.0420 s

Electric field $E=7.80\times10^{2}\ N/C$

We need to calculate the electric force on the particle

Using formula of electric force

$F=qE$

Put the value into the formula

$F_{e}=6.40\times10^{-6}\times7.80\times10^{2}$

$F_{e}=0.004992= 0.499\times10^{-2}\ N$

We need to calculate the gravitational force

Using formula of force

$F=mg$

Put the value into the formula

$F_{g}=0.0400\times10^{-3}\times9.8$

$F_{g}=0.000392 = 0.392\times10^{-3}\ N$

We need to calculate the net force

$F_{net}=F_{e}-F_{g}$

$F_{net}=0.499\times10^{-2}-0.392\times10^{-3}$

$F_{net}=0.004598= 0.4598\times10^{-2}\ N$

We need to calculate the acceleration

Using newton's law

$F = ma$

$a = \dfrac{F}{m}$

$a=\dfrac{0.4598\times10^{-2}}{0.0400\times10^{-3}}$

$a =114.95\ m/s^2$

We need to calculate the height

Using equation of motion

$s = ut+\dfrac{1}{2}at^2$

$s=0+\dfrac{1}{2}\times114.95\times(0.0420)^2$

$s=0.1014\ m$

Hence, The height is 0.1014 m

3 0

3 years ago

Does a ball ever bounce back to it's drop height?

Valentin [98]

Answer: No

Explanation: Once you bounce the ball it will go higher and then it will slowly come back down

5 0

3 years ago

Read 2 more answers

A spherical, non-conducting shell of inner radius = 10 cm and outer radius = 15 cm carries a total charge Q = 13 μC distributed

Vaselesa [24]

Answer:

E = 1580594.95 N/C

Explanation:

To find the electric field inside the the non-conducting shell for r=11.2cm you use the Gauss' law:

$\int EdS=\frac{Q_{in}}{\epsilon_o}$ (1)

dS: differential of the Gaussian surface

Qin: charge inside the Gaussian surface

εo: dielectric permittivity of vacuum = 8.85 × 10-12 C2/N ∙ m2

The electric field is parallel to the dS vector. In this case you have the surface of a sphere, thus you have:

$\int EdS=ES=E(4\pi r^2)$ (2)

Qin is calculate by using the charge density:

$Q_{in}=V_{in}\rho=\frac{4}{3}(r^3-a^3)\rho$ (3)

Vin is the volume of the spherical shell enclosed by the surface. a is the inner radius.

The charge density is given by:

$\rho=\frac{Q}{V}=\frac{13*10^{-6}C}{\frac{4}{3}\pi((0.15m)^3-(0.10m)^3)}\\\\\rho=1.30*10^{-3}\frac{C}{m^3}$

Next, you use the results of (3), (2) and (1):

$E(4\pi r^2)=\frac{4}{3\epsilon_o}(r^3-a^3)\rho\\\\E=\frac{\rho}{3\epsilo_o}(r-\frac{a^3}{r^2})$

Finally, you replace the values of all parameters, and for r = 11.2cm = 0.112m you obtain:

$E=\frac{1.30*10^{-3}C/m^3}{3(8.85*10^{-12}C^2/Nm^2)}((0.112m)-\frac{(0.10)^3}{(0.112m)^2})\\\\E=1,580,594.95\frac{N}{C}$

hence, the electric field is 1580594.95 N/C

7 0

3 years ago

Assume that the surface temperature of the lightbulb filament when it is plugged into an outlet of 120 V is about 3000 K and the

myrzilka [38]

Answer:

A= $8.7\times 10^{-7}m^{2}$

Explanation:

We have given power =40 W but only 10% of this power is used so actual power $P=\frac{40\times 10}{100}=4 W$

We know that from Stefan's law $p=\sigma AT^4$ where $\sigma$ is Boltzmann constant which value is $5.6\times 10^{-8}Wm^{-2}K^{-4}$

So $4=5.67\times 10^{-8}\times A\times 3000^4$

A= $8.7\times 10^{-7}m^{2}$

4 0

3 years ago

Read 2 more answers