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3 years ago

What will happen when a light bulb is removed from the circuit below, leaving a gap?

Physics

2 answers:

Charra [1.4K]3 years ago

7 0

Answer:

The remaining light bulbs will go out.

Explanation:

The light bulb that was taken out routed power to the other light bulbs, there for, not giving power to the next light bulbs will make them turn off or, "go out". This may be incorrect, as you did not provide a picture of the circuit.

Furkat [3]3 years ago

3 0

Answer:

The light bulbs will go out

Explanation:

I took the test

You might be interested in

What, roughly, is the percent uncertainty in the volume of a spherical beach ball whose radius is 5.66 0.09 m?

iren2701 [21]

Answer:

4.77 %

Explanation:

We know that the volume V for a sphere of radius r is

$V(r) = \frac{4}{3} \ \pi \ r^3$

If we got an uncertainty $\Delta r$ the formula for the uncertainty of V is:

$\Delta V(r) = \sqrt{ (\frac{dV}{dr} \Delta r)^2 }$

We can calculate this uncertainty, first we obtain the derivative:

$\frac{dV}{dr} = 3 * \frac{4}{3} \ \pi \ r^2$

$\frac{dV}{dr} = 4 \ \pi \ r^2$

And using it in the formula:

$\Delta V(r) = \sqrt{ (4 \ \pi \ r^2\Delta r)^2 }$

$\Delta V(r) = \sqrt{ 4^2 \ \pi^2 \ r^4 \Delta r^2 }$

$\Delta V(r) = 4 \ \pi \ r^2 \Delta r$

The relative uncertainty is:

$\frac{\Delta V(r)}{V(r)}$

$\frac{ 4 \ \pi \ r^2 \Delta r }{ \frac{4}{3} \ \pi \ r^3}$

$\frac{ 3 \Delta r }{ r}$

Using the values for the problem:

$\frac{ 3 * 0.09 m }{ 5.66 m} = 0.0477$

This is, a percent uncertainty of 4.77 %

4 0

2 years ago

Which of the following biomes might be found at a latitude of 33° South?

cupoosta [38]

The answer is going be desert.

5 0

3 years ago

1) La longitud del brazo de potencia de una palanca es de 0,0035 hectómetros y la del brazo de resistencia es de 55 centímetros.

Arlecino [84]

Answer:

Entonces seria 127 para vencer.

Explanation:

espero averte ayudado:-)

6 0

3 years ago

how much force would be required to produce 88 j of work when pushing a box 1.1meters at an angle of 10 degrees?

ycow [4]

Answer:81.235N

Explanation:

Work=88J

theta=10°

distance=1.1 meters

work=force x cos(theta) x distance

88=force x cos10 x 1.1 cos10=0.9848

88=force x 0.9848 x 1.1

88=force x 1.08328

Divide both sides by 1.08328

88/1.08328=(force x 1.08328)/1.08328

81.235=force

Force=81.235

5 0

3 years ago

Death Star has a diameter of 160,000m and a mass of 5.1e17kg. Millennium Falcon has a mass of 1.36e6kg (data from Wookieepedia)

lions [1.4K]

Answer:

7229 N

Explanation:

The gravitational force between the Death Star and the Millenium Falcon is given by:

$F=G\frac{mM}{R^2}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2}$ is the gravitational constant

$M=5.1\cdot 10^{17} kg$ is the mass of the Death Star

$m=1.36\cdot 10^6 kg$ is the mass of the Millennium Falcon

$R=\frac{160,000 m}{2}=80,000 m$ is the radius of the Death Star

Substituting numbers into the equation, we find the force

$F=(6.67\cdot 10^{-11})\frac{(1.36\cdot 10^6 kg)(5.1\cdot 10^{17} kg)}{(80,000 m)^2}=7229 N$

5 0

3 years ago