All categories

3 years ago

If a single-load Series circuit has one 20 Ohm bulb, and a

Physics

1 answer:

Alex Ar [27]3 years ago

5 0

At the instant the second bulb is connected ...

... the brightness of the first bulb doesn't change

... the brightness of the second bulb changes from dark (no brightness) to equal to the first bulb

You might be interested in

What is the maximum power that can be delivered by a 1.4-cm-diameter laser beam propagating through air

Aleksandr-060686 [28]

Complete Question

The maximum electric field strength in air is 4.0 MV/m. Stronger electric fields ionize the air and create a spark. What is the maximum power that can be delivered by a 1.4-cm-diameter laser beam propagating through air

Answer:

The value is $P = 3.270960 *10^{6} \ W$

Explanation:

From the question we are told that

The electric field strength is $E = 4.0 \ M \ V/m = 4.0 *10^6 \ V/m$

The diameter is $d = 1.4 \ cm = 0.014 \ m$

Generally the radius is mathematically represented as

$r = \frac{d}{2}$

=> $r = \frac{0.014}{2}$

=> $r = 0.007 \ m$

Generally the cross-sectional area is mathematically represented as

$A = \pi r^2$

$A = 3.142 * (0.007)^2$

$A = 0.000154 \ m^2$

Generally the maximum power that can be delivered is mathematically represented as

$P = \frac{c * \epsilon_o * E^2 * A }{2}$

Here c is the speed of light with value $c = 3.0*10^{8} \ m/s$

$\epsilon_o$ is the permittivity of free space with value $\epsilon_o = 8.85 *10^{-12} \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2$

$P = \frac{3,0*10^8 * 8.85*10^{-12} * (4 *10^6)^2 * 0.00154}{ 2}$

$P = 3.270960 *10^{6} \ W$

5 0

2 years ago

What best defines work?

MrRa [10]

Moving the body to do something

4 0

2 years ago

a car accelerates from rest at 3 m / s^2 along a straight road. how far has the car travelled after 4 s

evablogger [386]

Answer:

24 m

Explanation:

The motion of the car is a uniformly accelerated motion (=at constant acceleration), therefore we can find the distance covered by using the following suvat equation:

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