Answer:
x = 5[km]
Explanation:
We must convert the time from minutes to hours.
![t=30[min]*\frac{1h}{60min}= 0.5[h]\\](https://tex.z-dn.net/?f=t%3D30%5Bmin%5D%2A%5Cfrac%7B1h%7D%7B60min%7D%3D%200.5%5Bh%5D%5C%5C)
We know that speed is defined as the relationship between space and time.

where:
x = space [m]
t = time = 0.5 [h]
v = velocity [m/s]
Now replacing:
![x = 10[\frac{km}{h} ]*0.5[h]\\x=5[km]](https://tex.z-dn.net/?f=x%20%3D%2010%5B%5Cfrac%7Bkm%7D%7Bh%7D%20%5D%2A0.5%5Bh%5D%5C%5Cx%3D5%5Bkm%5D)
The energy transfer in terms of work has the equation:
W = mΔ(PV)
To be consistent with units, let's convert them first as follows:
P₁ = 80 lbf/in² * (1 ft/12 in)² = 5/9 lbf/ft²
P₂ = 20 lbf/in² * (1 ft/12 in)² = 5/36 lbf/ft²
V₁ = 4 ft³/lbm
V₂ = 11 ft³/lbm
W = m(P₂V₂ - P₁V₁)
W = (14.5 lbm)[(5/36 lbf/ft²)(4 ft³/lbm) - (5/9 lbf/ft²)(11 lbm/ft³)]
W = -80.556 ft·lbf
In 1 Btu, there is 779 ft·lbf. Thus, work in Btu is:
W = -80.556 ft·lbf(1 Btu/779 ft·lbf)
<em>W = -0.1034 BTU</em>
It is known as silicon dioxide or silica!
Hope this helps!
Answer:
Eo = 9.796 x 10^2 N/C
Bo = 3.266 x 10^-6 T
Explanation:
Given
Wavelength λ = 633 nm
Diameter of the beam D = 1.0 mm
Power P = 1.0 mW
Solution
Radius of the beam r = D/2 = 0.5 mm = 0.0005 m
Area of cross section

Intensity

Amplitude of Electric Field

Amplitude of Magnetic Field
