Answer:
![d = 2.26 km](https://tex.z-dn.net/?f=d%20%3D%202.26%20km)
Explanation:
Let the child is moving with speed same as the speed of water flow
So here the position of child with respect to flow must be zero
And if the boat start at an angle with the vertical
so its relative speed with flow of water is given as
![v_x = 24.8 sin\theta](https://tex.z-dn.net/?f=v_x%20%3D%2024.8%20sin%5Ctheta)
![v_y = 24.8 cos\theta](https://tex.z-dn.net/?f=v_y%20%3D%2024.8%20cos%5Ctheta)
now the time to reach the child is given as
![\frac{0.6}{24.8 cos\theta} = \frac{2.5}{24.8 sin\theta }](https://tex.z-dn.net/?f=%5Cfrac%7B0.6%7D%7B24.8%20cos%5Ctheta%7D%20%3D%20%5Cfrac%7B2.5%7D%7B24.8%20sin%5Ctheta%20%7D)
so now we have
![\theta = 76.5 degree](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2076.5%20degree)
So the time to catch the child is given as
![t = \frac{0.6}{24.8 cos78.2}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B0.6%7D%7B24.8%20cos78.2%7D)
![t = 0.104 h](https://tex.z-dn.net/?f=t%20%3D%200.104%20h)
So distance moved by it in 0.104 h
distance moved by the boat in upstream direction given as
![x = (24.8 sin 74.8 - 3.1)(0.104)](https://tex.z-dn.net/?f=x%20%3D%20%2824.8%20sin%2074.8%20-%203.1%29%280.104%29)
![x = 2.18 km](https://tex.z-dn.net/?f=x%20%3D%202.18%20km)
In y direction the displacement of boat is
![y = 0.6 km](https://tex.z-dn.net/?f=y%20%3D%200.6%20km)
net displacement of the ball is given as
![d = \sqrt{x^2 + y^2}](https://tex.z-dn.net/?f=d%20%3D%20%5Csqrt%7Bx%5E2%20%2B%20y%5E2%7D)
![d = \sqrt{2.18^2 + 0.6^2}](https://tex.z-dn.net/?f=d%20%3D%20%5Csqrt%7B2.18%5E2%20%2B%200.6%5E2%7D)
![d = 2.26 km](https://tex.z-dn.net/?f=d%20%3D%202.26%20km)
Answer:
i know the questin but i got to try and find it
Explanation:
Answer:
a. one line down one line to the right one live to the northwest from the object
b. t1=190 t2=310
Explanation:
Answer:
Q at the center of the distribution.
Explanation:
- The Gauss's law is the law that relates to the distribution of electrical charges to the resulting electrical field. It states that a flux of electricity outside the arabatory closed surface is proportional to the electricitical harg enclosed by the surface.
Answer:
![F = 1.5 \times 10^{-16} N](https://tex.z-dn.net/?f=F%20%3D%201.5%20%5Ctimes%2010%5E%7B-16%7D%20N)
this force is
times more than the gravitational force
Explanation:
Kinetic Energy of the electron is given as
![KE = 1 keV](https://tex.z-dn.net/?f=KE%20%3D%201%20keV)
![KE = 1.6 \times 10^{-16} J](https://tex.z-dn.net/?f=KE%20%3D%201.6%20%5Ctimes%2010%5E%7B-16%7D%20J)
now the speed of electron is given as
![KE = \frac{1}{2}mv^2](https://tex.z-dn.net/?f=KE%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
now we have
![v = \sqrt{\frac{2 KE}{m}}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B%5Cfrac%7B2%20KE%7D%7Bm%7D%7D)
![v = 1.87 \times 10^7 m/s](https://tex.z-dn.net/?f=v%20%3D%201.87%20%5Ctimes%2010%5E7%20m%2Fs)
now the maximum force due to magnetic field is given as
![F = qvB](https://tex.z-dn.net/?f=F%20%3D%20qvB)
![F = (1.6\times 10^{-19})(1.87 \times 10^7)(0.5 \times 10^{-4})](https://tex.z-dn.net/?f=F%20%3D%20%281.6%5Ctimes%2010%5E%7B-19%7D%29%281.87%20%5Ctimes%2010%5E7%29%280.5%20%5Ctimes%2010%5E%7B-4%7D%29)
![F = 1.5 \times 10^{-16} N](https://tex.z-dn.net/?f=F%20%3D%201.5%20%5Ctimes%2010%5E%7B-16%7D%20N)
Now if this force is compared by the gravitational force on the electron then it is
![\frac{F}{F_g} = \frac{1.5 \times 10^{-16}}{9.1 \times 10^{-31} (9.8)}](https://tex.z-dn.net/?f=%5Cfrac%7BF%7D%7BF_g%7D%20%3D%20%5Cfrac%7B1.5%20%5Ctimes%2010%5E%7B-16%7D%7D%7B9.1%20%5Ctimes%2010%5E%7B-31%7D%20%289.8%29%7D)
![\frac{F}{F_g} = 1.68 \times 10^{13}](https://tex.z-dn.net/?f=%5Cfrac%7BF%7D%7BF_g%7D%20%3D%201.68%20%5Ctimes%2010%5E%7B13%7D)
so this force is
times more than the gravitational force