With which of the following is the weak nuclear force associated

Calculate the speed of light if the wavelength is 3cm and the period is 1*10^-10

ioda

First you find the frequency which is. 1/T(period) so u get 1*10^10
Now that you know frequency u can solve for speed: s= wavelength* frequency, but don't forget to convert 3cm into meters: 3/100= .03m so now you do (1*10^10)(.3)= 300,000,000 m/s or 3*10^8 m/s

Hope this helped :))

6 0

3 years ago

the loudness of a person's voice depends on the A. force with which air rushes across the vocal folds B. strength of the intrins

blsea [12.9K]

Answer: A. force with which air rushes across the vocal folds

Explanation:

The human voice is produced in the larynx, whose essential part is the glottis. This is how the air coming from the lungs is forced during expiration through the glottis, making its two pairs of vocal folds to vibrate.

It should be noted that this process can be consciously controlled by the person who speaks (or sings), since the variation in the intensity of the sound of the voice depends on the strength of the breath.

5 0

3 years ago

A cold beverage can be kept cold even a warm day if it is slipped into a porous ceramic container that has been soaked in water.

Arisa [49]

Answer:

The rate at which the container is losing water is 0.0006418 g/s.

Explanation:

Under the assumption that the can is a closed system, the conservation law applied to the system would be: $E_{in}-E_{out}=E_{change}$ , where $E_{in}$ is all energy entering the system, $E_{out}$ is the total energy leaving the system and, $E_{change}$ is the change of energy of the system.
As the purpose is to kept the beverage can at constant temperature, the change of energy ( $E_{change}$ ) would be 0.
The energy that goes into the system, is the heat transfer by radiation from the environment to the top and side surfaces of the can. This kind of transfer is described by: $Q=\varepsilon*\sigma*A_S*(T_{\infty}^4-T_S^4)$ where $\varepsilon$ is the emissivity of the surface, $\sigma=5.67*10^{-8}\frac{W}{m^2K}$ known as the Stefan–Boltzmann constant, $A_S$ is the total area of the exposed surface, $T_S$ is the temperature of the surface in Kelvin, $T_{\infty}$ is the environment temperature in Kelvin.
For the can the surface area would be ta sum of the top and the sides. The area of the top would be $A_{top}=\pi* r^2=\pi(0.0252m)^2=0.001995m^2$ , the area of the sides would be $A_{sides}=2*\pi*r*L=2*\pi*(0.0252m)*(0.09m)=0.01425m^2$ . Then the total area would be $A_{total}=A_{top}+A_{sides}=0.01624m^2$
Then the radiation heat transferred to the can would be $Q=\varepsilon*\sigma*A_S*(T_{\infty}^4-T_S^4)=1*5.67*10^{-8}\frac{W}{m^2K}*0.01624m^2*((32+273K)^4-(17+273K)^4)=1.456W$ .
The can would lost heat evaporating water, in this case would be $Q_{out}=\frac{dm}{dt}*h_{fg}$ , where $\frac{dm}{dt}$ is the rate of mass of water evaporated and, $h_{fg}$ is the heat of vaporization of the water ( $2257\frac{J}{g}$ ).
Then in the conservation balance: $Q_{in}-Q_{out}=Q_{change}$ , it would be $1.45W-\frac{dm}{dt}*2257\frac{j}{g}=0$ .
Recall that $1W=1\frac{J}{s}$ , then solving for $\frac{dm}{dt}$ : $\frac{dm}{dt}=\frac{1.45\frac{J}{s} }{2257\frac{J}{g} }=0.0006452\frac{g}{s}$

5 0

2 years ago

When summer changes to fall, what seasonal changes do plants experience? (IGNORE HIGHLIGHTED ANSWER)

lorasvet [3.4K]

Answer:

C

Explanation:

There is a decrease in temperature and daylight and plants produce less food.

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7 0

1 year ago

Which statement is the correct representation of these electric field lines?

irinina [24]

C . plate a is negatively charged and plate b is positively charged

6 0

2 years ago

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