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.
The magnetic field direction and direction of induced current in a wire are related by the right hand grip rule. Since the magnetic field was upwards, the thumb points upwards and the fingers curl around it. When viewed from above, it is seen as a current flowing in the counter clockwise direction.
The only graph that accurately depict the given motion is graph D.
The given parameters;
- initial position of the man = 0
- direction of the man's first displacement = backward
- time of first motion, t₁ = 6 seconds
- velocity of this first displacement = v₁
- time without any motion (<em>zero movement</em>) = 6 seconds
- direction of the second displacement = forward
- velocity of second displacement = 2v₁
Let the acceleration of the first displacement = a
Acceleration of the second displacement = 2a
From the given graphs we can eliminate every graph without initial decrease or motion towards the negative direction.
The only options with initial motion towards the negative direction are;
The difference between graph B and D;
- in graph B there is a uniform motion for 6 seconds
- in graph D there is no motion for 6 seconds (<em>this is obvious as the line fall directly on top of the horizontal axis maintaining a value of zero for 6 seconds</em>).
Thus, the only graph that accurately depict the given motion is graph D.
Learn more here: brainly.com/question/21095906
speed = 86km/hr
= 86000/3600
= 23.889m/s
time = 4.6s
acceleration = 23.889/4.6
a = 5.19m/s^2
