All categories

3 years ago

Please help, is an electric lamp a luminous object?

Physics

2 answers:

bezimeni [28]3 years ago

8 0

Electric lamp why yes it is

VMariaS [17]3 years ago

6 0

Electric bulbs are definitely luminous objects so yes, i presume so!

You might be interested in

A 22.0 nF capacitor is connected across an AC generator that produces a peak voltage of 5.80 V. part a

mariarad [96]

Answer:

Explanation:

Impedence of the circuit = peak voltage / peak current

= 5.8 / 51 x 10⁻³

= 113.725 ohm.

1 / wC =113.725

w = 1 / (113.725 x 22 x 10⁻⁹ )

= 10⁹ / 2.5 x 10³

=10⁶ / 2.5

40 x 10⁴

frequency n = 40 x 10⁴ / 2 x 3.14

6.37 x 10⁴ Hz.

b ) charge on the capacitor = 1 C

V = Q / C

= Charge / capacitor

= 1 / 22 x 10⁻⁹

4.54 x 10⁷ V.

4 0

3 years ago

The position of a ball as a function of time is given by

Andre45 [30]

Answer:

x = -6.5 meters

Explanation:

The position of a ball as a function of time t is given by :

$x=3\ m+(-5\ m/s)t$ ..................(1)

Where t is time in seconds

We need to find the position of the ball at 1.9 s. It can be simply calculated putting t = 1.9 s in equation (1) as :

$x=3+(-5)(1.9)$

x = -6.5 meters

So, the position of the ball at 1.9 seconds is -6.5 meters. Hence, this is the required solution.

6 0

4 years ago

Question 5 of 5: Someone texting or talking spans an average of 27 seconds after they put the phone down are still thinking abou

AVprozaik [17]

The answer is true. Distraction “latency” lasts for about 27 seconds.

This means that even after driver put down the phone or stop fooling with the navigation system; he or she isn’t fully committed with the driving task. Talking on a cell phone and texting are frequent what people associate with distracted driving, but there are so much more activities behind distracted driving.

7 0

4 years ago

In a generator, mechanical energy is converted into which of the following?

Serggg [28]

In a generator, mechanical energy is converted electrical energy. The answer is letter D. the mechanical energy in the generator produces energy by making the machines work and then transmitting this into the power lines of commercial industries, communities and others.

8 0

4 years ago

Read 2 more answers

A particle of charge +2q is placed at the origin and particle of charge -q is placed on the x-axis at x = 2a. Where on the x-axi

zzz [600]

Answer:

$r_{31}=\frac{2a\sqrt{2}}{1+\sqrt{2}}$

Explanation:

We know that the electric force equation is:

$F=k\frac{q_{1}*q_{2}}{r^{2}}$

k is the electric constant $9*10^{9} Nm^{2}/C^{2}$
r is the distance between the particles
q1 and q2 are the particle

Now, we have three particles, the first one at x=0, the second one at x=2a and the third in some place between these two particle.

1. Let's find the electric force between the first particle and the third particle.

$F_{31}=k\frac{q_{3}*q_{1}}{r_{31}^{2}}$

$F_{31}=k\frac{q*2q}{r_{31}^{2}}$

$F_{31}=k\frac{2q^{2}}{r_{31}^{2}}$

r(31) is the distance between 3 and 1

2. Now, let's find the electric force between the third particle and the second particle.

$F_{32}=k\frac{q_{3}*q_{2}}{x_{32}^{2}}$

$F_{32}=k\frac{q*(-q)}{r_{32}^{2}}$

$F_{32}=-k\frac{q^{2}}{r_{32}^{2}}$

r(32) is the distance between 3 and 2.

Now, $r_{31}+r_{32}=2a$ or $r_{32}=2a-r_{31}$

The net force must be zero so:

$F_{31}+F_{32}=0[\tex][tex]k\frac{2q^{2}}{r_{31}^{2}}-k\frac{q^{2}}{r_{32}^{2}}=0[\tex] [tex]kq^{2}(\frac{2}{r_{31}^{2}}-\frac{1}{r_{32}^{2}})=0[\tex] [tex]kq^{2}(\frac{2}{r_{31}^{2}}-\frac{1}{(2a-r_{31})^{2}})=0[\tex] It means that:[tex]\frac{2}{r_{31}^{2}}-\frac{1}{(2a-r_{31})^{2}}$

We just need to solve it for r(31)

$r_{31}^{2}=2(2a-r_{31})^{2}$

$r_{31}=\frac{2a\sqrt{2}}{1+\sqrt{2}}$

Therefore the distance from the origin will be:

$r_{31}=\frac{2a\sqrt{2}}{1+\sqrt{2}}$

I hope it helps you!

4 0

4 years ago

Please help, is an electric lamp a luminous object?​

Please help, is an electric lamp a luminous object?