Answer:
Explanation:
In a beta (minus) decay, a neutron in a nucleus turns into a proton, emitting a fast-moving electron (called beta particle) alongside with an antineutrino.
The general equation for a beta decay is:
(1)
where
X is the original nucleus
Y is the daughter nucleus
e is the electron
is the antineutrino
We observe that:
- The mass number (A), which is the sum of protons and neutrons in the nucleus, remains the same in the decay
- The atomic number (Z), which is the number of protons in the nucleus, increases by 1 unit
In this problem, the original nucles that we are considering is iodine-131, which is
where
Z = 53 (atomic number of iodine)
A = 131 (mass number)
Using the rule for the general equation (1), the dauther nucleus must have same mass number (131) and atomic number increased by 1 (54, which corresponds to Xenon, Xe), therefore the equation will be:
Answer: 10000 kg
Explanation:
The momentum 
is given by the following equation:
(1)
Where:
is the mass of the object
is the velocity of the object
Now, in this case and according the conservation of momentum:
(2)
Where:
is the mass of the spacecraft
is the mass of the asteroid
is the initial velocity of the spacecraft
is the initial velocity of the asteroid (because we are told the asteroid is stationary, as the spacracft is on the sateroid it remains stationary as well)
is the final velocity of the spacecraft
is the final velocity of the asteroid
Rewritting (2):
(3)
(4)
Finding :
This is the mass of the asteroid
(2)(3)(2)=12⇒ D=M/V ⇒ D=94/12=7.8333(repeated) g/m³
Well, the answer is obviously D, because it's the only one that starts with 3.134.
Moving the decimal place 9 times to the right will give us:
3,134,000,000
I think the correct answer from the choices listed above is option B. The very high voltage needed to create a spark across the spark plug is produced at the transformer's secondary winding. <span>The secondary coil is engulfed by a powerful and changing magnetic field. This field induces a current in the coils -- a very high-voltage current.</span>
