Vas happenin!
Independent variable : amount of water each day
Dependent variable: water on the windsill
Hypotheses: Ben wants to try by adding water each day to two different places. Will that work? Will that effect the water?
Hope this helps you out
*smiles*
-Zayn Malik
a)
• P = F/A
P = pressure = 630 N/m^2
F = force
A = area
F = mg = 0.50 kg x 9.8 m/s^2 = 4.9 N
m= mass
g= gravity
P = F/A
A = F/P
A = 4.9 N / 630 N/m^2 = 7.778 x 10^-3 m^2
b)
• Area of a circle = pi* radius ^2
7.778 x 10^-3 m^2 = pi* radius ^2
√(7.778 x 10^-3 m^2 / pi ) = radius
radius = 0.04976 m
Answers:
a ) 7.778 x 10^-3 m^2
b) 0.04976 m
Find the intensity of the electromagnetic wave described in each case.
(a) an electromagnetic wave with a wavelength of 645 nm and a peak electric field magnitude of 8.5 V/m.
(b) an electromagnetic wave with an angular frequency of 6.3 ✕ 1018 rad/s and a peak magnetic field magnitude of 10−10 T.
Answer:
Explanation:
Velocity of a wave is describe as
velocity =Frequency × Wavelength
Mathematically
v = fλ
Hence, Frequency, F = v / λ
Wavelength λ = v/f
So, if the frequency is kept constant, wavelength of the wave becomes directly proportional to velocity of the wave.
And this implies that, as the speed double, the wavelength is double.
Answer:
Explanation:
a ) Between r = 0 and r = r₁
Electric field will be zero . It is so because no charge lies in between r = 0 and r = r₁ .
b ) From r = r₁ to r = r₂
At distance r , charge contained in the sphere of radius r
volume charge density x 4/3 π r³
q = Q x r³ / R³
Applying Gauss's law
4πr² E = q / ε₀
4πr² E = Q x r³ / ε₀R³
E= Q x r / (4πε₀R³)
E ∝ r .
c )
Outside of r = r₂
charge contained in the sphere of radius r = Q
Applying Gauss's law
4πr² E = q / ε₀
4πr² E = Q / ε₀
E = Q / 4πε₀r²
E ∝ 1 / r² .
