The greatest height the ball will attain is 3.27 m
<h3>Data obtained from the question</h3>
- Initial velocity (u) = 8 m/s
- Final velocity (v) = 0 m/s (at maximum height)
- Acceleration due to gravity (g) = 9.8 m/s²
The maximum height to which the ball can attain can be obtained as follow:
v² = u² – 2gh (since the ball is going against gravity)
0² = 8² – (2 × 9.8 × h)
0 = 64 – 19.6h
Collect like terms
0 – 64 = –19.6h
–64 = –19.6h
Divide both side by –19.6
h = –64 / –19.6h
h = 3.27 m
Thus, the greatest height the ball can attain is 3.27 m
Learn more about motion under gravity:
brainly.com/question/13914606
Initial volume of mercury is
V = 0.1 cm³
The temperature rise is 35 - 5 = 30 ⁰C = 30 ⁰K.
Because the coefficient of volume expansion is 1.8x10⁻⁴ 1/K, the change in volume of the mercury is
ΔV = (1.8x10⁻⁴ 1/K)*(30 ⁰K)(0.1 cm³) = 5.4x10⁻⁴ cm³
The cross sectional area of the tube is
A = 0.012 mm² = (0.012x10⁻² cm²).
Therefore the rise of mercury in the tube is
h = ΔV/A
= (5.4x10⁻⁴ cm³)/(0.012x10⁻² cm²)
= 4.5 cm
Answer: 4.5 cm
After x seconds, an object will fall
where a is acceleration due to gravity and t is time
so when t=3.3
the distance it will fall is
=53.361m
it will fal 53.361 meters
