Answer:
Weight of the fluid that the object displaces.
Explanation:
When the fluid is completely immersed in a fluid, it experiences pressure from all the direction. While the object is immersed in the fluid a force acts on it in the opposite direction, i.e., upwards. This force is termed as buoyant force.
Also, as per the Archimedes' Principle, the force experience by the object is the same as the weight of the fluid that gets displaced by the object.
Thus on complete immersion of the object in the fluid, it experiences the force same as the weight of the fluid that gets displaced
For each load, Work = (mass) x (gravity) x (distance .
Bigger load: Work = (10 kg) x (9.8 m/s²) x (2 m) = 196 joules .
Smaller load: Work = (5 kg) x (9.8 m/s²) x (4 m) = 196 joules.
The work required is equal in both cases.
The mass ratio of 2:1 is exactly balanced by
the height ratio of 1:2 .
I think this is the answer for the first line(Cooling ,Heating or mixing ) and for the second line is(broken down,cooled,mixed)
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.
Here, we are required to find the relationship between balls of different mass(a measure of weight) and different volumes.
- 1. Ball A will have the greater density
- 2. Ball C and Ball D have the same density.
- 3. Ball Q will have the greater density.
- 4. Ball X and Y will have the same density
The density of an object is given as its mass per unit volume of the object.
Mathematically;.
For Case 1:
- Va = Vb and Ma = 2Mb
- D(b) = (Mb)/(Vb) and D(a) = 2(Mb)/Vb
- Therefore, the density of ball A,
- D(a) = 2D(b).
- Therefore, ball A has the greater density.
For Case 2:
- D(c) = (Mc)/(Vc) and D(d) = (1/3)Md/(1/3)Vd
- Therefore, ball C and D have the same density
For Case 3:
- Vp = 2Vq and Mp = Mq
- D(p) = (Mq)/2(Vq) and D(q) = (Mq)/Vq
- Therefore, the density of ball P is half the density of ball Q
- Therefore, ball Q has the greater density.
For case 4:
Therefore, Ball X and Ball Y have the same density.
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