What does the kinetic theory of matter state?

You plug a string of 100 lights in series into a 120 V power outlet, and each light has a resistance of 3.00 _. If each light ha

UNO [17]

V = IR

By completing the equation, i found that the total power equation is : 4.8,
Which means that it's not exceed the power rating.

So i believe the answer would be : The string will remain lit

hope this helps

6 0

3 years ago

Light of wavelength 436.1 nm falls on two slits spaced 0.31 mm apart. What is the required distance from the slits to the screen

kakasveta [241]

Answer:

The correct answer is "4.26 m".

Explanation:

Given:

Wavelength,

$\lambda = 436.1 \ nm$

or,

$=436.1\times 10^{-9} \ m$

Distance,

$d = 0.31 \ mm$

or,

$=0.31\times 10^{-3} \ m$

Distance between the 1st and 2nd dark fringes,

$(y_2-y_1) = 6\times 10^{-3} \ m$

As we know,

⇒ $\frac{d}{L} (y_2-y_1) = \lambda$

or,

⇒ $L=\frac{d(y_2-y_1)}{\lambda}$

By substituting the values, we get

$=\frac{0.31\times 6\times 10^{-6}}{436.1\times 10^{-9}}$

$=\frac{0.31\times 6\times 10^3}{436.1}$

$=\frac{1860}{436.1}$

$=4.26 \ m$

3 0

3 years ago

For both resonance curves and Fourier spectra, amplitude is plotted vs frequency, but these two types of plots are not the same.

andrey2020 [161]

Answer:

he peaks are the natural frequencies that coincide with the excitation frequencies and in the second case they are the natural frequencies that make up the wave.

Explanation:

In a resonance experiment, the amplitude of the system is plotted as a function of the frequency, finding maximums for the values where some natural frequency of the system coincides with the excitation frequency.

In a Fourier transform spectrum, the amplitude of the frequencies present is the signal, whereby each peak corresponds to a natural frequency of the system.

From this explanation we can see that in the first case the peaks are the natural frequencies that coincide with the excitation frequencies and in the second case they are the natural frequencies that make up the wave.

7 0

3 years ago

A rotating wheel requires 5.00 s to rotate 28.0 revolutions. Its angular velocity at the end of the 5.00-s interval is 96.0 rad/

torisob [31]

Answer:

The constant angular acceleration of the wheel is 12.16 rad/s²

Explanation:

Given;

initial angular distance, θ = 28

time of the motion, t = 5 s

initial angular velocity is calculated as;

$\omega _i = \frac{\theta}{t} = \frac{28}{5}.\frac{rev}{s} \ \times \ \ \frac{2 \pi \ rad}{1 \ \ rev} = 35.19 \ rad/s$

final angular velocity is given as, $\omega _f = 96.0 \ rad/s$

The constant angular acceleration is calculated as;

$\alpha = \frac{\omega _f - \omega _i}{t} \\\\\alpha = \frac{96 - 35.19}{5} \\\\\alpha = 12.16 \ rad/s^2$

Therefore, the constant angular acceleration of the wheel is 12.16 rad/s²

6 0

3 years ago

Determine the gravitational force that you exert on another person 1.50 m away. assume that you and the other person are point m

Aneli [31]

Given: Universal Gravitational constant = G = 6.67 x 10⁻¹¹ N m²/Kg²

Mass₁ = 70 Kg; Mass₂ = 70 Kg Radius r = 1.5 m; Force F = ?

Formula: F = Gm₁m₂/r²

F = (6.67 x 10⁻¹¹ N m²/Kg²)(70 Kg)(70 Kg)/(1.5 m)²

F = 1.45 X 10⁻⁷ N

4 0

3 years ago