Answer: A proton is a positively charged particle located in the nucleus of an atom.
Answer:
Explanation:
Suppose v is the initial velocity and
is the angle of inclination
distance traveled in vertical direction in t=1 s
When gravity is present
![y=vt+\frac{1}{2}at^2](https://tex.z-dn.net/?f=y%3Dvt%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2)
where ![y=vertical\ distance](https://tex.z-dn.net/?f=y%3Dvertical%5C%20distance)
![a=acceleration](https://tex.z-dn.net/?f=a%3Dacceleration)
![t=time](https://tex.z-dn.net/?f=t%3Dtime)
![v=initial\ velocity](https://tex.z-dn.net/?f=v%3Dinitial%5C%20velocity)
here initial velocity is v\sin \theta [/tex] so
![y=v\sin \theta \times 1-\frac{1}{2}gt^2](https://tex.z-dn.net/?f=y%3Dv%5Csin%20%5Ctheta%20%5Ctimes%201-%5Cfrac%7B1%7D%7B2%7Dgt%5E2)
![y=v\sin \theta -0.5g](https://tex.z-dn.net/?f=y%3Dv%5Csin%20%5Ctheta%20-0.5g)
In absence of gravity
![y_2=v\sin \theta \times t](https://tex.z-dn.net/?f=y_2%3Dv%5Csin%20%5Ctheta%20%5Ctimes%20t)
![y_2=v\sin \theta \times 1](https://tex.z-dn.net/?f=y_2%3Dv%5Csin%20%5Ctheta%20%5Ctimes%201)
![\Delta y=y_2-y=v\sin \theta -v\sin \theta +0.5 g=4.9\ m](https://tex.z-dn.net/?f=%5CDelta%20y%3Dy_2-y%3Dv%5Csin%20%5Ctheta%20-v%5Csin%20%5Ctheta%20%2B0.5%20g%3D4.9%5C%20m)
Answer:
only the weight of the ball will act on the ball
Explanation: There is no contact force on the ball. Also there is no air resistance on the ball so the friction force on the ball due to air is not shown
When boat is sunk into the liquid the net buoyancy on the boat is counterbalanced by weight of the boat
So here weight of the boat = Buoyancy force
let say boat is sunk by distance "h"
now we can say
![F_b = \rho * V * g](https://tex.z-dn.net/?f=F_b%20%3D%20%5Crho%20%2A%20V%20%2A%20g)
![F_b = 1000*0.10 * 0.08 * h * 9.8](https://tex.z-dn.net/?f=F_b%20%3D%201000%2A0.10%20%2A%200.08%20%2A%20h%20%2A%209.8)
now by above force balance equation we can write
![m*g = F_b](https://tex.z-dn.net/?f=m%2Ag%20%3D%20F_b)
![0.040 * 9.8 = 1000 * 0.10 * 0.08 * h * 9.8](https://tex.z-dn.net/?f=0.040%20%2A%209.8%20%3D%201000%20%2A%200.10%20%2A%200.08%20%2A%20h%20%2A%209.8)
![0.040 = 8h](https://tex.z-dn.net/?f=0.040%20%3D%208h)
![h = 5 * 10^{-3} m](https://tex.z-dn.net/?f=h%20%3D%205%20%2A%2010%5E%7B-3%7D%20m)
so boat will sunk by total 5 mm distance