Answer:
The bubble gum experiment demonstrates the law of conservation of mass in that even though the bubble gum has lost mass, this loss in mass is not because some matter present in the gum has been destroyed, but it has changed form and has been removed from the gum.
Explanation:
In this science experiment, students investigate whether or not chewing gum should be considered eating. During the process of chewing the gum, the gum loses mass. The experiment is used to demonstrate the law of conservation of mass which states that matter can neither be created nor destroyed but may change from one form to another.
The loss in mass of the gum is due to the fact that the sugar present in the gum has changed form and has been removed from the gum. During the process of chewing the gum, the sugar in solid form present in the gum is dissolved in the saliva found in the mouth. The dissolved sugar is then swallowed and passes into the digestive tract for digestion. This shows that even though the bubble gum has lost mass, this loss in mass is not because some matter present in the gum has been destroyed, but it has changed form and has been removed from the gum, This demonstrates the law of conservation of mass.
Answer:
e) 0.099
Explanation:
For an adiabatic expansion:

where
P1 is the initial pressure
P2 is the final pressure
V1 is the initial volume
V2 is the final volume
is the adiabatic index (which is
for an ideal monoatomic gas
In this problem, we have
since the volume increases by a factor 4
We can re-write the equation to find by what factor the pressure changes:

<span>The answer is transformer. They utilize
electromagnetic induction to generate current. This is only possible in
alternating current due to the differential increase and decrease of electrical
current that induces changes in magnetic flux in the coil. This varies the
magnetic flux of the primary coil that generates current in the secondary coil.</span>
Answer:
Vo = 4.5 [m/s]
Explanation:
In order to solve this problem, we must use the following equation of kinematics.

where:
Vf = final velocity = 12 [m/s]
Vo = initial velocity [m/s]
a = acceleration = 1.5 [m/s²]
t = time = 5 [s]
Now replacing:
![12=v_{o}+1.5*5\\v_{o}=12- (7.5)\\v_{o}= 4.5[m/s]](https://tex.z-dn.net/?f=12%3Dv_%7Bo%7D%2B1.5%2A5%5C%5Cv_%7Bo%7D%3D12-%20%287.5%29%5C%5Cv_%7Bo%7D%3D%204.5%5Bm%2Fs%5D)
D) It isn't accelerating
This is because if both sides are pushing with the exact same force, than one force is not overpowering the other, therefor the wheelbarrow would not be moving.