Answer:
(a) 1.73 s
(b) 14.75 m
(c) 3.36 s
(d) double
(e) 63.32 m
Explanation:
Vertical component of initial velocity, uy = 17 m/s
Horizontal component of initial velocity, ux = 18.3 m/s
(A) At highest point of trajectory, the vertical component of velocity is zero. Let the time taken is t.
Use first equation of motion in vertical direction
vy = uy - gt
0 = 17 - 9.8 t
t = 1.73 seconds
(B) Let the highest point is at height h.
Use III equation of motion in vertical direction
![v^{2}=u^{2}-2gh](https://tex.z-dn.net/?f=v%5E%7B2%7D%3Du%5E%7B2%7D-2gh)
0 = 17 x 17 - 2 x 9.8 x h
h = 14.75 m
(C) The time taken by the ball to return to original level is T.
Use second equation of motion i vertical direction.
![h = ut + 0.5at^2](https://tex.z-dn.net/?f=h%20%3D%20ut%20%2B%200.5at%5E2)
h = 0 , u = 17 m/s
0 = 17 t - 0.5 x 9.8 t^2
t = 3.46 second
(D) It is the double of time calculated in part A
(E) Horizontal distance = horizontal velocity x total time
d = 18.3 x 3.46 = 63.32 m
Answer:
Explanation:
We know that at pointed corner of a conductor , charge accumulate ie electric field is maximum . Ar pointed corner , radius of curvature will be least . That is why charge leakage is greatest at pointed corner of a charged conductor.
a ) Electric field will be 26.0 kN/C at the point where radius of curvature is greatest .
b ) So electric field will be 66 kN /C at the point where radius of curvature is smallest .
Answer:
0.68 s
Explanation:
We are given that
Initial velocity of box=![u=13m/s](https://tex.z-dn.net/?f=u%3D13m%2Fs)
Final velocity of box=v=11.5 m/s
Distance=d=8.5 m
We have to find the time taken by box to slow by this amount.
We know that
![v^2-u^2=2as](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2as)
Substitute the values
![(11.5)^2-(13)^2=2a(8.5)](https://tex.z-dn.net/?f=%2811.5%29%5E2-%2813%29%5E2%3D2a%288.5%29)
![132.25-169=17a](https://tex.z-dn.net/?f=132.25-169%3D17a)
![-36.75=17a](https://tex.z-dn.net/?f=-36.75%3D17a)
![a=\frac{-36.75}{17}=-2.2m/s^2](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B-36.75%7D%7B17%7D%3D-2.2m%2Fs%5E2)
We know that
Acceleration=![a=\frac{v-u}{t}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bv-u%7D%7Bt%7D)
Substitute the values
![-2.2=\frac{11.5-13}{t}](https://tex.z-dn.net/?f=-2.2%3D%5Cfrac%7B11.5-13%7D%7Bt%7D)
![-2.2=\frac{-1.5}{t}](https://tex.z-dn.net/?f=-2.2%3D%5Cfrac%7B-1.5%7D%7Bt%7D)
![t=\frac{1.5}{2.2}=0.68 s](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B1.5%7D%7B2.2%7D%3D0.68%20s)
Hence, the time taken by box to slow by this amount=0.68 s