Answer:
"How does the volume of a gas kept at constant pressure change as its temperature is increased?"
Explanation:
One possible question can be:
"How does the volume of a gas kept at constant pressure change as its temperature is increased?"
The answer to this question is contained in Charle's law, which states that for a gas at constant pressure, the volume of the gas is proportional to its absolute temperature:
Or also written as
By looking at this equation, we can find immediately the answer to our question: as the (absolute) temperature of the gas increases, the volume increases as well, by the same proportion.
the answer is (A) the movement of the magnet relative to the coil
Answer:
Speed of both blocks after collision is 2 m/s
Explanation:
It is given that,
Mass of both blocks, m₁ = m₂ = 1 kg
Velocity of first block, u₁ = 3 m/s
Velocity of other block, u₂ = 1 m/s
Since, both blocks stick after collision. So, it is a case of inelastic collision. The momentum remains conserved while the kinetic energy energy gets reduced after the collision. Let v is the common velocity of both blocks. Using the conservation of momentum as :
v = 2 m/s
Hence, their speed after collision is 2 m/s.
Explanation:
Given Data
Total mass=93.5 kg
Rock mass=0.310 kg
Initially wagon speed=0.540 m/s
rock speed=16.5 m/s
To Find
The speed of the wagon
Solution
As the wagon rolls, momentum is given as
P=mv
where
m is mass
v is speed
put the values
P=93.5kg × 0.540 m/s
P =50.49 kg×m/s
Now we have to find the momentum of rock
momentum of rock = mv
momentum of rock = (0.310kg)×(16.5 m/s)
momentum of rock =5.115 kg×m/s
From the conservation of momentum we can find the wagons momentum So
wagon momentum=50.49 -5.115 = 45.375 kg×m/s
Speed of wagon = wagon momentum/(total mass-rock mass)
Speed of wagon=45.375/(93.5-0.310)
Speed of wagon= 0.487 m/s
Throwing rock backward,
momentum of wagon = 50.49+5.115 = 55.605 kg×m/s
Speed of wagon = wagon momentum/(total mass-rock mass)
speed of wagon = 55.605 kg×m/s/(93.5kg-0.310kg)
speed of wagon= 0.5967 m/s
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.
