Why the specific heat capacity of the sun remain constant<br><br>

Nitrogen is used to proctect steel from water because nitrigen has?

blondinia [14]

<span>Nitrogen is used to protect steel from water because nitrogen has stable physical properties that enables the nitrogen to be resistant to water molecules. Nitrogen also serves as a coat for the steel to not be corroded easily</span>

8 0

3 years ago

Which statement about the ocean is true?

svet-max [94.6K]

Answer:

most evaporation and precipitation in the water cycle occus over the ocean

7 0

3 years ago

Read 2 more answers

A feather and a minivan are dropped vertically downward from a height of twenty meters and both are subject to air drag as they

Luda [366]

Answer:

maybe C i'm not 100% sure

I'm sorry if this didn't help

Explanation:

6 0

2 years ago

The helium-neon lasers most commonly used in student physics laboratories have average power outputs of 0.250 mW.

Nookie1986 [14]

(a) $72.3 W/m^2$

First of all, we need to find the area of the circular spot, which is given by:

$A=\pi r^2$

where r is the radius of the spot, which is half the diameter, therefore

$r=\frac{d}{2}=\frac{2.10 mm}{2}=1.05 mm=1.05\cdot 10^{-3} m$

So, the area of the spot is

$A=\pi (1.05\cdot 10^{-3}m)^2=3.46\cdot 10^{-6} m^2$

We know that the power output of the laser is

$P=0.250 mW=2.5\cdot 10^{-4} W$

So the intensity of the laser beam is

$I=\frac{P}{A}=\frac{2.5\cdot 10^{-4} W}{3.46\cdot 10^{-6} m^2}=72.3 W/m^2$

(b) $7.8\cdot 10^{-7}T$

The average intensity of the laser is related to the peak magnetic field strength by

$I=\frac{cB_0^2}{2\mu_0}$

where

c is the speed of light

$B_0$ is the peak magnetic field strength

$\mu_0=1.257\cdot 10^{-6} H/m$ is the vacuum magnetic permeability

Solving the formula for $B_0$ , we find

$B_0 = \sqrt{\frac{2I\mu_0}{c}}=\sqrt{\frac{2(72.3 W/m^2)(1.257\cdot 10^{-6} H/m)}{3\cdot 10^8 m/s}}=7.8\cdot 10^{-7}T$

(c) 234 V/m

The relationship between magnetic field and electric field in an electromagnetic wave is

$E_0=cB_0$

where

$E_0$ is the peak electric field strength

c is the speed of light

$B_0$ is the peak magnetic field strength

Substituting numbers into the formula, we find

$E_0=(3\cdot 10^8 m/s)(7.8\cdot 10^{-7} T)=234 V/m$

7 0

3 years ago

Consider the following distribution of objects: a 5.00-kg object with its center of gravity at (0, 0) m, a 3.00-kg object at (0,

Minchanka [31]

Answer:

(-1.5,-1.5)m

Explanation:

we know that:

$X_{cm} = \frac{m_1x_1+m_2x_2....m_nX_n}{m_1+m_2...m_n}$

where $X_{cm}$ is the location of the center of gravity in the axis x, $m_i$ is the mass of the object i and $x_i$ the first coordinate of center of gravity of object i.

so:

$0 = \frac{(5kg)(0)+(3kg)(0)+(4kg)(3)+(8kg)x_4}{5kg+3kg+4kg+8kg}$

Where $x_4$ is the first coordinate of the center of gravity for the fourth object.

Therefore, solving for $x_4$ , we get:

$x_4 = -1.5m$

At the same way:

$Y_{cm} = \frac{m_1y_1+m_2y_2....m_ny_n}{m_1+m_2...m_n}$

where $Y_{cm}$ is the location of the center of gravity in the axis y, $m_i$ is the mass of the object i and $y_i$ the second coordinate of center of gravity of object i. replacing values we get: