vector is the answer of this blank
Answer:
The water level rises more when the cube is located above the raft before submerging.
Explanation:
These kinds of problems are based on the principle of Archimedes, who says that by immersing a body in a volume of water, the initial water level will be increased, raising the water level. That is, the height in the container with water will rise in level. The difference between the new volume and the initial volume of the water will be the volume of the submerged body.
Now we have two moments when the steel cube is held by the raft and when it is at the bottom of the pool.
When the cube is at the bottom of the water we know that the volume will increase, and we can calculate this volume using the volume of the cube.
Vc = 0.45*0.45*0.45 = 0.0911 [m^3]
Now when a body floats it is because a balance is established in the densities, the density of the body and the density of the water.
![Ro_{H2O}=R_{c+r}\\where:\\Ro_{H2O}= water density = 1000 [kg/m^3]\\Ro_{c+r}= combined density cube + raft [kg/m^3]](https://tex.z-dn.net/?f=Ro_%7BH2O%7D%3DR_%7Bc%2Br%7D%5C%5Cwhere%3A%5C%5CRo_%7BH2O%7D%3D%20water%20density%20%3D%201000%20%5Bkg%2Fm%5E3%5D%5C%5CRo_%7Bc%2Br%7D%3D%20combined%20density%20cube%20%2B%20raft%20%5Bkg%2Fm%5E3%5D)
Density is given by:
Ro = m/V
where:
m= mass [kg]
V = volume [m^3]
The buoyancy force can be calculated using the following equation:
![F_{B}=W=Ro_{H20}*g*Vs\\W = (200+730)*9.81\\W=9123.3[N]\\\\9123=1000*9.81*Vs\\Vs = 0.93 [m^3]](https://tex.z-dn.net/?f=F_%7BB%7D%3DW%3DRo_%7BH20%7D%2Ag%2AVs%5C%5CW%20%3D%20%28200%2B730%29%2A9.81%5C%5CW%3D9123.3%5BN%5D%5C%5C%5C%5C9123%3D1000%2A9.81%2AVs%5C%5CVs%20%3D%200.93%20%5Bm%5E3%5D)
Vs > Vc, What it means is that the combined volume of the raft and the cube is greater than that of the cube at the bottom of the pool. Therefore the water level rises more when the cube is located above the raft before submerging.
Answer:
2.5m/s^2
Explanation:
Step one:
given
distance = 20meters
time = 2 seconds
initial velocity u= 0m/s
let us solve for the final velocity
velocity = distance/time
velocity= 20/2
velocity= 10m/s

divide both sides by 40

<span>31.3 m/s
Since the water balloon is being launched at a 45 degree angle, the horizontal and vertical speeds will be identical. Also the time the balloon takes to reach its peak altitude will match the time it takes to fall. So let's create a few expressions about what we know.
Distance the water balloon travels at velocity v for time t
d = vt
Total time required for the entire trip is double since the balloon goes up, then goes down
t = 2v/a
Now let's plug in the numbers we have, assuming the acceleration due to gravity is 9.8 m/s^2
t = 2v/9.8
100 = vt
Substitute 2v/9.8 for t in the 2nd formula
100 = v(2v/9.8)
Solve for v.
100 = v(2v/9.8)
100 = 2v^2/9.8
980. = 2v^2
490 = v^2
22.13594 = v
So we now know that both the horizontal velocity and vertical velocity needed is 22.13594 m/s. Let's verify that
2*22.13594 / 9.8 = 4.51754
So it will take 4.51754 second for the balloon to hit the ground after being launched.
4.51754 * 22.13594 = 100
And during that time it will travel 100 meters horizontally.
But we need to know the total velocity. And the Pythagorean theorem comes to the rescue. Just square the 2 velocities, add them together, and take the square root. We already know the square is 490 from the work above, so
sqrt(490+490) = sqrt(980) = 31.30495 m/s</span>