Answer:
The greater the amplitude the greater the energy.
(Think of a water wave - which carries greater energy a 1 ft wave or
a 10 ft wave)
Question:
A point charge of -2.14uC is located in the center of a spherical cavity of radius 6.55cm inside an insulating spherical charged solid. The charge density in the solid is 7.35×10−4 C/m^3.
a) Calculate the magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity.
Express your answer using two significant figures.
Answer:
The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity 
Explanation:
A point charge ,q =
is located in the center of a spherical cavity of radius ,
m inside an insulating spherical charged solid.
The charge density in the solid , d = 
Distance from the center of the cavity,R =
Volume of shell of charge= V =![(\frac{4\pi}{3})[ R^3 - r^3 ]](https://tex.z-dn.net/?f=%28%5Cfrac%7B4%5Cpi%7D%7B3%7D%29%5B%20R%5E3%20-%20r%5E3%20%5D)
Charge on the shell ,Q = 
![Q =(\frac{4\pi}{3})[ R^3 - r^3 ] \times d](https://tex.z-dn.net/?f=Q%20%3D%28%5Cfrac%7B4%5Cpi%7D%7B3%7D%29%5B%20R%5E3%20-%20r%5E3%20%5D%20%5Ctimes%20d)
![Q = 4.1888\times 10^{-4} [5.76364 ] \times 7.35 \times 10^{-4}](https://tex.z-dn.net/?f=Q%20%3D%204.1888%5Ctimes%2010%5E%7B-4%7D%20%5B5.76364%20%5D%20%5Ctimes%207.35%20%5Ctimes%2010%5E%7B-4%7D)


Electric field at
m due to shell
E1 = 

Electric field at
due to 'q' at center 
E2 =

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity
= E2- E1
![=[ 2.134 - 1.769 ]\times 10^6](https://tex.z-dn.net/?f=%3D%5B%20%202.134%20%20-%201.769%20%5D%5Ctimes%2010%5E6)

D. All of the above
At high tide fish will feed among the mangrove roots - rich fishing ground
The trees trap sediment and soil in the river that would flow out to sea which also helps stop erosion
Wildlife utilise almost every part of the tree, with insects and birds, monkeys and lizards in the branches, shrimps and fish in the roots, and snails and clams in the soil
Answer:
hurricanes,Typhoons and cyclones
Answer:
c.100 minutes
Explanation:
Total distance = 10 km
Runs for 1 km every 5 minutes
walks 1 km every 15 min
She alternates between walking and running so, Jessica will walk 5 km and run 5 Km
Time taken by Jessica for walking
: 5 km
Time taken to walk 1 km=5 minutes
Time taken to walk 5 km
=> 5 X 5
=>25 minutes
Time taken by Jessica for Running
: 5km
Time taken to run 1 km = 15 minutes
=> 5 X 15
=>75 minutes
Total time taken = Time taken by Jessica for walking + Time taken by Jessica for Running
=>25 minutes +75 minutes
=> 100 minutes