A At one constant temp and another at a constant pressure
To solve this problem we will apply the concepts related to the intensity included as the power transferred per unit area, where the area is the perpendicular plane in the direction of energy propagation.
Since the propagation occurs in an area of spherical figure we will have to


Replacing with the given power of the Bulb of 100W and the radius of 2.5m we have that


The relation between intensity I and 

Here,
= Permeability constant
c = Speed of light
Rearranging for the Maximum Energy and substituting we have then,




Finally the maximum magnetic field is given as the change in the Energy per light speed, that is,



Therefore the maximum value of the magnetic field is 
False. They are arranged in a structure called a crystal lattice
Happy Holidays!
Recall that:
Impulse = Change in Momentum = mass × change in velocity
Since both cars are identical and have the same initial velocity of 60 mph, them breaking to a stop means that they both experience the same change in velocity.
Thus, both of the cars' impulses are equal.
Answer: 29.50 m
Explanation: In order to calculate the higher accelation to stop a train without moving the crates inside the wagon which is traveling at constat speed we have to use the second Newton law so that:
f=μ*N the friction force is equal to coefficient of static friction multiply the normal force (m*g).
f=m.a=μ*N= m*a= μ*m*g= m*a
then
a=μ*g=0.32*9.8m/s^2= 3.14 m/s^2
With this value we can determine the short distance to stop the train
as follows:
x= vo*t- (a/2)* t^2
Vf=0= vo-a*t then t=vo/a
Finally; x=vo*vo/a-a/2*(vo/a)^2=vo^2/2a= (49*1000/3600)^2/(2*3.14)=29.50 m