To solve this problem it is necessary to apply Boyle's law in which it is specified that

Where,
and
are the initial pressure and volume values
and
are the final pressure volume values
The final pressure here is the atmosphere, then




Pressure at the water is given by,


Using Boyle equation we have,



Therefore the volume of the lungs at the surface is 5.9L
<span>When the particles of a medium move with simple harmonic motion, this means the wave is a sinusoidal wave.
Know that a sinusoidal curve can describe either sine or cosine functions (remember your cofunction identities for sine and cosine).</span>
Answer:
The volume of the larger cube is 5.08 g/cm³.
Explanation:
Given that,
Mass of smaller cube = 20 g
Density of smaller cube 
Dylan has two cubes of iron.
The larger cube has twice the mass of the smaller cube.

Density is same for both cubes because both cubes are same material.
The density is equal to the mass divided by the volume.


Where, V = volume
m = mass

We need to calculate the volume of smaller mass
The volume of smaller mass



Now, We need to calculate the volume of large cube



Hence, The volume of the larger cube is 5.08 g/cm³.
Answer: 1 m/s
Explanation:
We have an object whose position
is given by a vector, where the components X and Y are identified by the unit vectors
and
(where each unit vector is defined to have a magnitude of exactly one):
![r=[2 m + (2 m/s) t] i + [3 m - (1 m/s^{2})t^{2}] j](https://tex.z-dn.net/?f=r%3D%5B2%20m%20%2B%20%282%20m%2Fs%29%20t%5D%20i%20%2B%20%5B3%20m%20-%20%281%20m%2Fs%5E%7B2%7D%29t%5E%7B2%7D%5D%20j)
On the other hand, velocity is defined as the variation of the position in time:

This means we have to derive
:
![\frac{dr}{dt}=\frac{d}{dt}[2 m + (2 m/s) t] i + \frac{d}{dt}[3 m - (1 m/s^{2})t^{2}] j](https://tex.z-dn.net/?f=%5Cfrac%7Bdr%7D%7Bdt%7D%3D%5Cfrac%7Bd%7D%7Bdt%7D%5B2%20m%20%2B%20%282%20m%2Fs%29%20t%5D%20i%20%2B%20%5Cfrac%7Bd%7D%7Bdt%7D%5B3%20m%20-%20%281%20m%2Fs%5E%7B2%7D%29t%5E%7B2%7D%5D%20j)
This is the velocity vector
And when
the velocity vector is:

This is the velocity vector at 2 seconds
However, the solution is not complete yet, we have to find the module of this velocity vector, which is the speed
:


Finally:
This is the speed of the object at 2 seconds
Answer:
a rock of 50kg should be placed =drock=0.5m from the pivot point of see saw
Explanation:
τchild=τrock
Use the equation for torque in this equation.
(F)child(d)child)=(F)rock(d)rock)
The force of each object will be equal to the force of gravity.
(m)childg(d)child)=(m)rockg(d)rock)
Gravity can be canceled from each side of the equation. for simplicity.
(m)child(d)child)=(m)rock(d)rock)
Now we can use the mass of the rock and the mass of the child. The total length of the seesaw is two meters, and the child sits at one end. The child's distance from the center of the seesaw will be one meter.
(25kg)(1m)=(50kg)drock
Solve for the distance between the rock and the center of the seesaw.
drock=25kg⋅m50kg
drock=0.5m