Answer:
Explanation:
For this problem, we can use Boyle's law, which states that for a gas at constant temperature, the product between pressure and volume remains constant:
which can also be rewritten as
In our case, we have:
is the initial pressure
is the initial volume
is the final pressure
Solving for V2, we find the final volume:
Answer:
pu = 1260.9kg/m^3
the density of the unknown liquid is 1260.9kg/m^3
Explanation:
The density of a liquid is inversely proportional to the volume (height) of object submerged in it.
High density liquid possess higher buoyant force preventing objects from submerging.
p ∝ 1/V ∝ 1/h
since V = Ah
pu/pw = hw/hu
pu = pwhw/hu
Where;
p = density
h = height submerged
pu and pw is the density of unknown liquid and water respectively
hu and hw is the height of object submerged in unknown liquid and water respectively
pw = 1000kg/m^3
hu = 4.6cm = 0.046m
hw = 5.8cm = 0.058m
Substituting the given values;
pu = 1000×0.058/0.046
pu = 1260.9kg/m^3
the density of the unknown liquid is 1260.9kg/m^3
Let us first know the given: Tennis ball has a mass of 0.003 kg, Soccer ball has a mass of 0.43 kg. Having the same velocity at 16 m/s. First the equation for momentum is P=MV P=Momentum M=Mass V=Velocity. Now let us have the solution for the momentum of tennis ball. Pt=0.003 x 16 m/s= ( kg-m/s ) I use the subscript "t" for tennis. Momentum of Soccer ball Ps= 0.43 x 13m/s = ( km-m/s). If we going to compare the momentum of both balls, the heavier object will surely have a greater momentum because it has a larger mass, unless otherwise the tennis ball with a lesser mass will have a greater velocity to be equal or greater than the momentum of a soccer ball.
Answer:
T=1022.42 N
Explanation:
Given that
l = 32 cm ,μ = 1.5 g/cm
L =2 m ,V= 344 m/s
The pipe is closed so n= 3 ,for first over tone
f= 129 Hz
The tension in the string given as
T = f²(4l²) μ
Now by putting the values
T = f²(4l²) μ
T = 129² x (4 x 0.32²) x 1.5 x 10⁻³ x 100
T=1022.42 N
As per the third law of Newton, the force exerted by the boat over the student is equal in magnitude to the force that the student exerted on the boat.
So, calculate the force on the student using the second law of Newton, Force = mass * acceleration.
Force on the student = 60 kg * 2.0 m/s^2 = 120 N.
=> horizontal force exerted by the student on the boat = 120 N
Answer: option d. 120 N. toward the back of the boat.
Of course it is toward the back because that is where the student jumped from..
