Answer:
The work done on the wagon is 37 joules.
Explanation:
Given that,
The force applied by Charlie to the right, F = 37.2 N
The force applied by Sara to the left, F' = 22.4 N
We need to find the work done on the wagon after it has moved 2.50 meters to the right. The net force acting on the wagon is :
![F_n=F-F'](https://tex.z-dn.net/?f=F_n%3DF-F%27)
![F_n=37.2-22.4](https://tex.z-dn.net/?f=F_n%3D37.2-22.4)
![F_n=14.8\ N](https://tex.z-dn.net/?f=F_n%3D14.8%5C%20N)
Work done on the wagon is given by the product of net force and displacement. It is given by :
![W=F_n\times d](https://tex.z-dn.net/?f=W%3DF_n%5Ctimes%20d)
![W=14.8\ N\times 2.5\ m](https://tex.z-dn.net/?f=W%3D14.8%5C%20N%5Ctimes%202.5%5C%20m)
W = 37 Joules
So, the work done on the wagon is 37 joules. Hence, this is the required solution.
Answer:
a) 30 degree
Explanation:
As we know that time of flight of the projectile depends on the the vertical component of the velocity always
It is given as
![T = \frac{2v_y}{g}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B2v_y%7D%7Bg%7D)
now we know that
![v_y = vsin\theta](https://tex.z-dn.net/?f=v_y%20%3D%20vsin%5Ctheta)
so we will have
![T = \frac{2vsin\theta}{g}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B2vsin%5Ctheta%7D%7Bg%7D)
since we know that two projectiles are projected at same speed but different angles
so smaller the angle will take smaller time
so it would be 30 degree projectile which will take smaller time
Without the velocity of the wave you can't get the frequency.
Answer:
An indirect measurement of the speed of molecules is temperature
Answer:
![Q(t) = 0.18\times 10^{-6} cos(5\times 10^3 t)](https://tex.z-dn.net/?f=Q%28t%29%20%3D%200.18%5Ctimes%2010%5E%7B-6%7D%20cos%285%5Ctimes%2010%5E3%20t%29)
Explanation:
We knwo that Kirchoff law
![L\frac{DI(t)}{dt} + \frac{1}{C}Q = 0](https://tex.z-dn.net/?f=L%5Cfrac%7BDI%28t%29%7D%7Bdt%7D%20%2B%20%5Cfrac%7B1%7D%7BC%7DQ%20%3D%200)
where
![I(t) = \frac{dQ}{dt}](https://tex.z-dn.net/?f=I%28t%29%20%3D%20%5Cfrac%7BdQ%7D%7Bdt%7D)
hence
![LQ" + \frac{Q}{C} = 0](https://tex.z-dn.net/?f=LQ%22%20%2B%20%5Cfrac%7BQ%7D%7BC%7D%20%3D%200)
C is given as 0.04\times 10^{6} F
L= 1 H , so we have
![Q" + 25\times 10^6Q = 0](https://tex.z-dn.net/?f=Q%22%20%2B%20%2025%5Ctimes%2010%5E6Q%20%3D%200)
the characteristic equation of this differential equation is
![r^2 + 25\times 10^6 = 0](https://tex.z-dn.net/?f=r%5E2%20%2B%2025%5Ctimes%2010%5E6%20%3D%200)
![r = \pm 5\times 10^3 i](https://tex.z-dn.net/?f=r%20%3D%20%5Cpm%205%5Ctimes%2010%5E3%20i)
Therefore differential equation is
![Q(t) = c_1 cos(5000t) + c_2sin(5000t)](https://tex.z-dn.net/?f=Q%28t%29%20%3D%20%20c_1%20cos%285000t%29%20%2B%20c_2sin%285000t%29)
we know initial value if capacitor is given as ![0.18\times 10^{-6} C](https://tex.z-dn.net/?f=0.18%5Ctimes%2010%5E%7B-6%7D%20C)
Therefore
![0.18\times 10^{-6} = Q(0) = c_1 cos(0) + c_2sin(0)](https://tex.z-dn.net/?f=0.18%5Ctimes%2010%5E%7B-6%7D%20%3D%20Q%280%29%20%3D%20%20c_1%20cos%280%29%20%2B%20c_2sin%280%29)
![c_1 = 0.18\times 10^{-6}](https://tex.z-dn.net/?f=c_1%20%3D%200.18%5Ctimes%2010%5E%7B-6%7D)
if no inital current is present then we hvae I(0) = Q'(0) = 0
![Q'(T) = -5000 C_1 sin(5000t) + 5000 c_2cos(5000t)](https://tex.z-dn.net/?f=Q%27%28T%29%20%3D%20-5000%20C_1%20sin%285000t%29%20%2B%20%205000%20c_2cos%285000t%29)
therefre
![c_2 =0](https://tex.z-dn.net/?f=c_2%20%3D0%20)
hence charge is
![Q(t) = 0.18\times 10^{-6} cos(5\times 10^3 t)](https://tex.z-dn.net/?f=Q%28t%29%20%3D%200.18%5Ctimes%2010%5E%7B-6%7D%20cos%285%5Ctimes%2010%5E3%20t%29)