Answer:
the volume is 0.253 cm³
Explanation:
The pressure underwater is related with the pressure in the surface through Pascal's law:
P(h)= Po + ρgh
where Po= pressure at a depth h under the surface (we assume = 1atm=101325 Pa) , ρ= density of water ,g= gravity , h= depth at h meters)
replacing values
P(h)= Po + ρgh = 101325 Pa + 1025 Kg/m³ * 9.8 m/s² * 20 m = 302225 Pa
Also assuming that the bubble behaves as an ideal gas
PV=nRT
where
P= absolute pressure, V= gas volume ,n= number of moles of gas, R= ideal gas constant , T= absolute temperature
therefore assuming that the mass of the bubble is the same ( it does not absorb other bubbles, divides into smaller ones or allow significant diffusion over its surface) we have
at the surface) PoVo=nRTo
at the depth h) PV=nRT
dividing both equations
(P/Po)(V/Vo)=(T/To)
or
V=Vo*(Po/P)(T/To) = 0.80 cm³ * (101325 Pa/302225 Pa)*(277K/293K) = 0.253 cm³
V = 0.253 cm³
B) gravitational to kinetic
Explanation:
The skydiver, when he is located at a certain height h above the ground, possesses gravitational potential energy, equal to:

where m is the mass of the skydiver, g is the gravitational acceleration and h is the height above the ground. As he falls, its height h decreases, while his speed v increases, so part of the gravitational potential energy is converted into kinetic energy, which is given by

so, we see that as v increases, the kinetic energy increases. Therefore the correct answer is
B) gravitational to kinetic
Answer:
B. The same on the moon.
Explanation:
The density of an object is the ratio of the mass contained by the object to the volume occupied by that mass.

When the object is taken from the earth to anywhere in the universe, its mass remains constant. The dimensions of the object and hence its volume also remains constant anywhere in the universe.
Therefore, the density of the object will also remain the same as it depends upon the mass and the volume of the object.
So, the correct option is:
<u>B. The same on the moon.</u>
Answer:
16.12
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To solve this problem we will apply the concepts related to the conservation of kinetic energy and elastic potential energy. Thus we will have that the kinetic energy is

And the potential energy is

Here,
m = mass
v = Velocity
x = Displacement
k = Spring constant
There is equilibrium, then,
KE = PE

Our values are given as,

Replacing we have that


Therefore the speed of the cart is 2.19m/s