Answer:
Speed of the helium after collision = 246 m/s
Explanation:
Given that
Mass of helium ,m₁ = 4 u
u₁=598 m/s
Mass of oxygen ,m₂ = 32 u
u₂ = 401 m/s
v₂ =445 m/s
Given that initially both are moving in the same direction and lets take they are moving in the right direction.
Speed of the helium after collision = v₁
There is no any external force on the masses that is why the linear momentum will be conserve.
Initial linear momentum = Final linear momentum
P = m v
m₁u₁+m₂u₂ = m₁v₁+m₂v₂
598 x 4 + 32 x 401 = 4 x v₁+ 32 x 445
v₁ = 246 m/s
Speed of the helium after collision = 246 m/s
The air pressure in the pressurized tank will be 24014.88 N/m²,196.2 N/m²,2084.625 N/m².
<h3 /><h3>What is pressure?</h3>
The force applied perpendicular to the surface of an item per unit area across which that force is spread is known as pressure.
It is denoted by P. The pressure relative to the ambient pressure is known as gauge pressure.
Pressure is found as the product of the density,acceleraton due to gravity and the height.
P₁=ρ₁gh₁
P₁=13,600 kg/m³×9.81 (m/s²)×0.18 m
P₁=24014.88 N/m²
P₂=ρ₂gh₂
P₂= 1000 kg/m³×9.81 (m/s²)×00.2 m
P₂=196.2 N/m²
P₃=ρ₃gh₃
P₃=850 kg/m³×9.81 (m/s²)×0.25
P₃=2084.625 N/m²
Hence,the air pressure in the pressurized tank will be 24014.88 N/m²,196.2 N/m²,2084.625 N/m².
To learn more about the pressure refer to the link;
brainly.com/question/356585
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Answer: Formula for Acceleration Due to Gravity
These two laws lead to the most useful form of the formula for calculating acceleration due to gravity: g = G*M/R^2, where g is the acceleration due to gravity, G is the universal gravitational constant, M is mass, and R is distance.please mark as brainliest
Explanation:
An output for is the force a person applies to a simple machine.
Refer to the diagram shown below.
μ = the coefficient of dynamic friction between the crate and the ramp.
1. The applied force of F acts over a distance, d.
The work done is F*d.
2. The component of the weight of the crate acting down the ramp is
mg sin(30) = 0.5mg.
The work done by this force is 0.5mgd.
3. The normal force is N = mgcos(30) = 0.866mg.
This force is perpendicular to the ramp, therefore the work done is zero.
4. The frictional force is μN = μmgcos(30) = 0.866μmg.
The work done by the frictional force is 0.866μmgd.
5. The total force acting on the crate up the ramp is
F - mgsin(30) - μmgcos(30) = F - mg(0.5 - 0.866μ)
6. The work done on the crate by the total force is
d*(F - 0.5mg - 0.866μmg)