Answer:
D. 12.4 m
Explanation:
Given that,
The initial velocity of the ball, u = 18 m/s
The angle at which the ball is projected, θ = 60°
The maximum height of the ball is given by the formula
h = u² sin²θ/2g m
Where,
g - acceleration due to gravity. (9.8 m/s)
Substituting the values in the above equation
h = 18² · sin²60 / 2 x 9.8
= 18² x 0.75 / 2 x 9.8
= 12.4 m
Hence, the maximum height of the ball attained, h = 12.4 m
Answer:
4.75 m/s
Explanation:
The computation of the velocity of the existing water is shown below:
Data provided in the question
Tall = 2 m
Inside diameter tank = 2m
Hole opened = 10 cm
Bottom of the tank = 0.75 m
Based on the above information, first we have to determine the height which is
= 2 - 0.75 - 0.10
= 2 - 0.85
= 1.15 m
We assume the following things
1. Compressible flow
2. Stream line followed
Now applied the Bernoulli equation to section 1 and 2
So we get

where,
P_1 = P_2 = hydrostatic
z_1 = 0
z_2 = h
Now

= 4.7476 m/sec
= 4.75 m/s
Answer:
D. two positively charged objects
120 is your answer. 120 mL = 120 cm^3