Answer:
Final velocity at the top of the ramp is 6.58m/s
Explanation
Check the attachment
If the pressure exerted on a 300.0 mL sample of hydrogen gas at constant temperature is increased from 0.500 kPa to 0.750 kPa, w
1 answer:
Answer:
200 mL
Explanation:
Given that,
Initial volume, V₁ = 300 mL
Initial pressure, P₁ = 0.5 kPa
Final pressure, P₂ = 0.75 kPa
We need to find the final volume of the sample if pressure is increased at constant temperature. It is based on Boyle's law. Its mathematical form is given by :
V₂ is the final volume
So, the final volume of the sample is 200 mL.
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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.
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:
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:
4.75 m/s
Explanation:
The computation of the velocity of the existing water is shown below:
Data provided in the question
Tall = 2 m
Inside diameter tank = 2m
Hole opened = 10 cm
Bottom of the tank = 0.75 m
Based on the above information, first we have to determine the height which is
= 2 - 0.75 - 0.10
= 2 - 0.85
= 1.15 m
We assume the following things
1. Compressible flow
2. Stream line followed
Now applied the Bernoulli equation to section 1 and 2
So we get
where,
P_1 = P_2 = hydrostatic
z_1 = 0
z_2 = h
Now
= 4.7476 m/sec
= 4.75 m/s
