Moving force of air flows through areas of high pressure to areas of low pressure

Nataly [62]

Answer:

1.for the first one 100centimeters make 1 meter therefore

100cm-1m

340cm-x

340/100

=3.4

the answer is supposed to be 3.4, maybe there's a mistake with the question or the answer

2.the weight of a body is given by the formula

mass×gravity,in this case the mass is 75kg and the gravity is 9.8

weight=75×9.8

=735N

3.for this one the mass of a body is given by the formula

mass=weight/gravity

=420/9.8

=42.8kg

I hope this helps

4 0

3 years ago

What to do if carbon monoxide detector is going off

Ksivusya [100]

Get To Safety And Call A Fire Department,

Be Alert Of The Signals Of The Presence Carbon Monoxide

Get Routine Checks Of Security Tools

5 0

3 years ago

1. A 180 kg motorcycle travels in a straight line on a horizontal road. The relationship between

Sergio039 [100]

Eh not really sure bout this one

5 0

2 years ago

Positive charge Q is distributed uniformly along the x-axis fromx=0 to x=a. A positive point charge q is located on the positive

dybincka [34]

Answer:

a. b- x= y

dx = -dy

b. F = $\frac{-kQqi}{r (a+r)}$

c. F = $\frac{-kQqi}{r^{2} }$

Explanation:

a. x components:

$dE = \frac{kdq}{(a+r-x^{2}) } \\$

= $\frac{kQdx}{(a(a+r-x)^2}$

Integrating and solving gives:

b- x= y

dx = -dy

b. the force is given by the equation derived from (a.):

F = $\frac{-kQqi}{r (a+r)}$

c. Given that r>>a, the expression becomes:

F = $\frac{-kQqi}{r^{2} }$

Explanation:

When the size of the charge distribution is less than the distance to the deviation point of the charge then the charge distribution would produce the same effect such as a linear charge.

6 0

3 years ago

Monochromatic light from a distant source is incident on a slit 0.750 mm wide. On a screen 2.00 m away, the distance from the ce

jasenka [17]

Answer:

$\lambda= 506.25 nm$

Explanation:

Diffraction is observed when a wave is distorted by an obstacle whose dimensions are comparable to the wavelength. The simplest case corresponds to the Fraunhofer diffraction, in which the obstacle is a long, narrow slit, so we can ignore the effects of extremes.

This is a simple case, in which we can use the Fraunhofer single slit diffraction equation:

$y=\frac{m \lambda D}{a}$

Where:

$y=Displacement\hspace{3}from\hspace{3} the\hspace{3} centerline \hspace{3}for \hspace{3}minimum\hspace{3} intensity =1.35mm\\\lambda=Light\hspace{3} wavelength \\D=Distance\hspace{3}between\hspace{3}the\hspace{3}screen\hspace{3}and\hspace{3}the\hspace{3}slit=2m\\a=width\hspace{3}of\hspace{3}the\hspace{3}slit=0.750mm\\m=Order\hspace{3}number=1$

Solving for λ:

$\lambda=\frac{y*a}{mD}$

Replacing the data provided by the problem:

$\lambda=\frac{(1.35\times 10^{-3})*(0.750\times 10^{-3})}{1*2} =5.0625\times 10^{-7}m =506.25nm$

7 0

2 years ago