The octopus’s tentacle keeps _ right after it is bitten off ? a. Moving b. Breathing c. Growing

When does a lightbulb carry more current, (a) immediately after it is turned on and the glow of the metal filament is increasing

emmasim [6.3K]

A lightbulb carries more current immediately after it is turned on and the glow of the metal filament is increasing; option A.

<h3>What is current?</h3>

Current refers to the flow of electric charges typically electrons.

Current flowing through a metallic material decreases with increase in temperature of the material.

This is because the resistance of the metal increases with increase in temperature.

Therefore, for a light bulb, the current flow through it will be maximum when it is just turned on because the temperature, and hence the resistance of the filament is at its lowest.

In conclusion, current flow decreases with increase in resistance.

Learn more about current and resistance at: brainly.com/question/24858512

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8 0

2 years ago

One string of a certain musical instrument is 70.0 cm long and has a mass of 8.79 g . It is being played in a room where the spe

Svetach [21]

To solve this problem we will apply the concepts of linear mass density, and the expression of the wavelength with which we can find the frequency of the string. With these values it will be possible to find the voltage value. Later we will apply concepts related to harmonic waves in order to find the fundamental frequency.

The linear mass density is given as,

$\mu = \frac{m}{l}$

$\mu = \frac{8.79*10^{-3}}{70*10^{-2}}$

$\mu = 0.01255kg/m$

The expression for the wavelength of the standing wave for the second overtone is

$\lambda = \frac{2}{3} l$

Replacing we have

$\lambda = \frac{2}{3} (70*10^{-2})$

$\lambda = 0.466m$

The frequency of the sound wave is

$f_s = \frac{v}{\lambda_s}$

$f_s = \frac{344}{0.768}$

$f_s = 448Hz$

Now the velocity of the wave would be

$v = f_s \lambda$

$v = (448)(0.466)$

$v = 208.768m/s$

The expression that relates the velocity of the wave, tension on the string and linear mass density is

$v = \sqrt{\frac{T}{\mu}}$

$v^2 = \frac{T}{\mu}$

$T= \mu v^2$

$T = (0.01255kg/m)(208.768m/s)^2$

$T = 547N$

The tension in the string is 547N

PART B) The relation between the fundamental frequency and the $n^{th}$ harmonic frequency is

$f_n = nf_1$

Overtone is the resonant frequency above the fundamental frequency. The second overtone is the second resonant frequency after the fundamental frequency. Therefore

$n=3$

Then,

$f_3 = 3f_1$

Rearranging to find the fundamental frequency

$f_1 = \frac{f_3}{3}$

$f_1 = \frac{448Hz}{3}$

$f_1 = 149.9Hz$

7 0

3 years ago

During the Water Wheel simulation you were able to change the flow of water

Law Incorporation [45]

Answer:

Explanation Home Page, THE FLOW OF ENERGY OUT OF THE SUN, Please sign our mailing list ... install itself when you run it on your computer ... To show how spectral lines are formed by random processes of absorption and re-emission in its ... One set of simulations deals with the scattering of photons in the interior of a star.n:

4 0

3 years ago

How is position time and motion related

rusak2 [61]

When you make a motion you change position over time.

5 0

3 years ago

One end of a horizontal spring with force constant 130.0 N/m is attached to a vertical wall. A 3.00 kg block sitting on the floo

Nuetrik [128]

Answer:

a) v = 0

b) The aceleration is 1.41 $m/s^{2}$

c) The block is accelerating away from the wall.

Explanation:

First, you need to think about the effect this constant force is causing in the spring: it causes a displacement in the equilibrium point of the system, therefore we need to know where it sits now:

At equilibrium no movement is present reducing friction to 0:

$\sum{F} = 0 = F_{spring} - F_{external}$

$F_{spring} = F_{external}$

$Kx = F_{external}$

$x = \frac{F_{external}}{K}=\frac{88}{130}=0.68m=68cm$

This means that the spring can be compressed with the single force up to 68 cm, Any further compression will cause an unbalanced system and the occilation of the mass.

The spring can't be compressed by the given force to 80 cm, therefore it must have been compressed by another force and then released.

In this case, the instantanous speed is 0, since the block has just been released.

In the same instant we can stimate the free body diagram of forces by the next two equations:

$\sum_y{F}={F_N-W}=0\\\sum_x{F}={F_{spring}-F_{external}-F_{friction}}=ma$

For the y axis:

$F_N = W = mg = 3*9.8 = 29.4N$

To calculate the force of friction:

$F_{friction} = \mu_k F_N=0.4*29.4 = 11.76N$

Therefore for x axis:

${Kx-F_{external}-F_{friction}}=ma$

$a = \frac{130*0.8-88-11.76}{3} = \frac{104-88-11.76}{3}=\frac{4.24}{3}=1.41\frac{m}{s^2}$

7 0

3 years ago