Answer:
The speed of the water shoot out of the hole is 20 m/s.
(d) is correct option.
Explanation:
Given that,
Height = 20 m
We need to calculate the velocity
Using formula Bernoulli equation
![\dfrac{1}{2}\rho v_{1}^2+\rho gh_{1}=\dfrac{1}{2}\rho v_{2}^2+\rho gh_{2}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7D%5Crho%20v_%7B1%7D%5E2%2B%5Crho%20gh_%7B1%7D%3D%5Cdfrac%7B1%7D%7B2%7D%5Crho%20v_%7B2%7D%5E2%2B%5Crho%20gh_%7B2%7D)
Where,
v₁= initial velocity
v₂=final velocity
h₁=total height
h₂=height of the hole from the base
Put the value into the formula
![v_{1}^2=2g(h_{2}-h_{1})](https://tex.z-dn.net/?f=v_%7B1%7D%5E2%3D2g%28h_%7B2%7D-h_%7B1%7D%29)
![v_{1}=\sqrt{2g(h_{2}-h_{1})}](https://tex.z-dn.net/?f=v_%7B1%7D%3D%5Csqrt%7B2g%28h_%7B2%7D-h_%7B1%7D%29%7D)
![v_{1}=\sqrt{2\times9.8\times(20-0.005)}](https://tex.z-dn.net/?f=v_%7B1%7D%3D%5Csqrt%7B2%5Ctimes9.8%5Ctimes%2820-0.005%29%7D)
![v_{1}=19.7\ m/s= approximate\ 20\ m/s](https://tex.z-dn.net/?f=v_%7B1%7D%3D19.7%5C%20m%2Fs%3D%20approximate%5C%2020%5C%20m%2Fs)
Hence, The speed of the water shoot out of the hole is 20 m/s.
W, because as time is moving up at a consistent rate the speed is as well, creating the straight line.
Answer:
![\theta = 25.3^\circ](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2025.3%5E%5Ccirc)
Explanation:
The acceleration of the block can be found by the kinematics equations:
![v = v_0 + at\\5.88 = 1.26 + a(1.1)\\a = 4.2~m/s^2](https://tex.z-dn.net/?f=v%20%3D%20v_0%20%2B%20at%5C%5C5.88%20%3D%201.26%20%2B%20a%281.1%29%5C%5Ca%20%3D%204.2~m%2Fs%5E2)
Since the plane is frictionless, the only force acting on the block along the motion of the block is its weight.
![F = mg\sin(\theta) = ma\\g\sin(\theta) = a\\(9.8)\sin(\theta) = 4.2\\\theta = 25.3^\circ](https://tex.z-dn.net/?f=F%20%3D%20mg%5Csin%28%5Ctheta%29%20%3D%20ma%5C%5Cg%5Csin%28%5Ctheta%29%20%3D%20a%5C%5C%289.8%29%5Csin%28%5Ctheta%29%20%3D%204.2%5C%5C%5Ctheta%20%3D%2025.3%5E%5Ccirc)
To solve this problem we will apply the principle of buoyancy of Archimedes and the relationship given between density, mass and volume.
By balancing forces, the force of the weight must be counteracted by the buoyancy force, therefore
![\sum F = 0](https://tex.z-dn.net/?f=%5Csum%20F%20%3D%200)
![F_b -W = 0](https://tex.z-dn.net/?f=F_b%20-W%20%3D%200)
![F_b = W](https://tex.z-dn.net/?f=F_b%20%3D%20W)
![F_b = mg](https://tex.z-dn.net/?f=F_b%20%3D%20mg)
Here,
m = mass
g =Gravitational energy
The buoyancy force corresponds to that exerted by water, while the mass given there is that of the object, therefore
![\rho_w V_{displaced} g = mg](https://tex.z-dn.net/?f=%5Crho_w%20V_%7Bdisplaced%7D%20g%20%3D%20mg)
Remember the expression for which you can determine the relationship between mass, volume and density, in which
![\rho = \frac{m}{V} \rightarrow m = V\rho](https://tex.z-dn.net/?f=%5Crho%20%3D%20%5Cfrac%7Bm%7D%7BV%7D%20%5Crightarrow%20m%20%3D%20V%5Crho)
In this case the density would be that of the object, replacing
![\rho_w V_{displaced} g = V\rho g](https://tex.z-dn.net/?f=%5Crho_w%20V_%7Bdisplaced%7D%20g%20%3D%20V%5Crho%20g)
Since the displaced volume of water is 0.429 we will have to
![\rho_w (0.429V) = V \rho](https://tex.z-dn.net/?f=%5Crho_w%20%280.429V%29%20%3D%20V%20%5Crho)
![0.429\rho_w= \rho](https://tex.z-dn.net/?f=0.429%5Crho_w%3D%20%5Crho)
The density of water under normal conditions is
, so
![0.429(1000) = \rho](https://tex.z-dn.net/?f=0.429%281000%29%20%3D%20%5Crho)
![\rho = 429kg/m^3](https://tex.z-dn.net/?f=%5Crho%20%3D%20429kg%2Fm%5E3)
The density of the object is ![429kg / m ^ 3](https://tex.z-dn.net/?f=429kg%20%2F%20m%20%5E%203)
1. one-Half
2. Apogee
3. Any object that revolves around another object
4. Venus's gravitation pull