Answer:_COC1\/2+_H\/2O>_HC1+CO\/2
Explanation:
Need help asap
U=10 m/s
v=30 m/s
t=6 sec
therefore, a=(v-u)/t
=(30-10)/6
=(10/3) ms^-2
now, displacement=ut+0.5*a*t^2
=60+ 0.5*(10/3)*36
=120 m
And you can solve it in another way:
v^2=u^2+2as
or, s=(v^2-u^2)/2a
=(900-100)/6.6666666.......
=120 m
The answer is B because you are not using force or energy as if you were if you were to lift or push something.
Answer:
Explanation:
The cannonball goes a horizontal distance of 275 m . It travels a vertical distance of 100 m
Time taken to cover vertical distance = t ,
Initial velocity u = 0
distance s = 100 m
acceleration a = 9.8 m /s²
s = ut + 1/2 g t²
100 = .5 x 9.8 x t²
t = 4.51 s
During this time it travels horizontally also uniformly so
horizontal velocity Vx = horizontal displacement / time
= 275 / 4.51 = 60.97 m /s
Vertical velocity Vy
Vy = u + gt
= 0 + 9.8 x 4.51
= 44.2 m /s
Resultant velocity
V = √ ( 44.2² + 60.97² )
= √ ( 1953.64 + 3717.34 )
= 75.3 m /s
Angle with horizontal Ф
TanФ = Vy / Vx
= 44.2 / 60.97
= .725
Ф = 36⁰ .
