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
Explanation:
Given that,
Mass of a freight car, 
Speed of a freight car, 
Mass of a scrap metal, 
(a) Let us assume that the final velocity of the loaded freight car is V. The momentum of the system will remain conserved as follows :
So, the final velocity of the loaded freight car is 0.182 m/s.
(b) Lost on kinetic energy = final kinetic energy - initial kinetic energy
![\Delta K=\dfrac{1}{2}[(m_1+m_2)V^2-m_1u_1^2)]\\\\=\dfrac{1}{2}\times [(30,000+110,000 )0.182^2-30000(0.85)^2]\\\\=-8518.82\ J](https://tex.z-dn.net/?f=%5CDelta%20K%3D%5Cdfrac%7B1%7D%7B2%7D%5B%28m_1%2Bm_2%29V%5E2-m_1u_1%5E2%29%5D%5C%5C%5C%5C%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%20%5B%2830%2C000%2B110%2C000%20%290.182%5E2-30000%280.85%29%5E2%5D%5C%5C%5C%5C%3D-8518.82%5C%20J)
Lost in kinetic energy is 8518.82. Negative sign shows loss.
Answer:
They are...if I'm correct Chemically combined, sorry if I'm wrong.
Answer : The final volume of the balloon at this temperature and pressure is, 17582.4 L
Solution :
Using combined gas equation is,
where,
= initial pressure of gas = 1 atm
= final pressure of gas = 0.3 atm
= initial volume of gas = 6000 L
= final volume of gas = ?
= initial temperature of gas = 273 K
= final temperature of gas = 240 K
Now put all the given values in the above equation, we get the final pressure of gas.
Therefore, the final volume of the balloon at this temperature and pressure is, 17582.4 L
Answer:
0.8712 m/s²
Explanation:
We are given;
Velocity of first car; v1 = 33 m/s
Distance; d = 2.5 km = 2500 m
Acceleration of first car; a1 = 0 m/s² (constant acceleration)
Velocity of second car; v2 = 0 m/s (since the second car starts from rest)
From Newton's equation of motion, we know that;
d = ut + ½at²
Thus,for first car, we have;
d = v1•t + ½(a1)t²
Plugging in the relevant values, we have;
d = 33t + 0
d = 33t
For second car, we have;
d = v2•t + ½(a2)•t²
Plugging in the relevant values, we have;
d = 0 + ½(a2)t²
d = ½(a2)t²
Since they meet at the next exit, then;
33t = ½(a2)t²
simplifying to get;
33 = ½(a2)t
Now, we also know that;
t = distance/speed = d/v1 = 2500/33
Thus;
33 = ½ × (a2) × (2500/33)
Rearranging, we have;
a2 = (33 × 33 × 2)/2500
a2 = 0.8712 m/s²
