I dont understand exactly what the question is but the main things in science will be a hypothesis, or theory things like that
Answer:
526.57 Pa
Explanation:
P ( pressure at the bottom of the container) = 1.049 × 10^5 pa
Using the formula of pressure in an open liquid
Pw ( pressure due to water) = ρhg where ρ is density of water in kg/m³, h is the height in meters, and g is acceleration due to gravity in m/s²
Pw = 1000 × 9.81 ×0.209 = 2050.29 Pa
P( atmospheric pressure) = 1.013 × 10^5 Pa
Pl ( pressure due to the liquid) = ρ(density of the liquid) × h (depth of the liquid) × g
Subtract each of the pressure from the absolute pressure at the bottom
P(bottom) - atmospheric pressure
(1.049 × 10^5) - (1.013 × 10^5) = 0.036 × 10^5 = 3600 Pa
subtract pressure due to water from the remainder
3600 - 2050.29 = 1549.71 Pa
1549.71 = ρ(density of the liquid) × h (depth of the liquid) × g
ρ (density of the liquid) = 1549.71 / (h × g) = 1549.71 / (0.3 × 9.81) =526.57 Pa
Answer:
9.6 m
Explanation:
This is a case of motion under variable acceleration . So no law of motion formula will be applicable here. We shall have to integrate the given equation .
a = 3.6 t + 5.6
d²x / dt² = 3.6 t + 5.6
Integrating on both sides
dx /dt = 3.6 t² / 2 + 5.6 t + c
where c is a constant.
dx /dt = 1.8 t² + 5.6 t + c
when t = 0 , velocity dx /dt is zero
Putting these values in the equation above
0 = 0 +0 + c
c = 0
dx /dt = 1.8 t² + 5.6 t
Again integrating on both sides
x = 1.8 t³ / 3 + 5.6 x t² /2 + c₁
x = 0.6 t³ + 2.8 t² + c₁
when t =0, x = 0
c₁ = 0
x = 0.6 t³ + 2.8 t²
when t = 1.6
x = .6 x 1.6³ + 2.8 x 1.6²
= 2.4576 + 7.168
= 9.6256
9.6 m
To find the maximum height you have to calculate the initial velocity in the y axis..
to do that resolve the velocity vector in the y direction
which is equal to
[email protected]27*√3/2 = 23.4 m/s
thus using kinematics.
v^2 = u^2 + 2as
23.4*23.4 /19.6 = s
27.96 that's 28 m approximately
