The blank in the question can be filled with the word, “Graph”. Therefore, Graphs are the pictures which are in relationships.
<u>Explanation:
</u>
Graph usually represents a set of data which is nonlinear in occurrence and has some relationship between the two given data. And as graph are pictorial representation, it is simply assumed as the pictures of relationships.
For example, a graph can be drawn for the set of data for the presence of number of students of all the sections of the particular class of a school, as they are relative. But making the graph for number of students in all section of all class but different school cannot be done as non-relative.
Well if the child weighs 44 kg then 44 kg. Do you mean if they weigh 44 kg in space to earth?
Answer:
The type of decay illustrated by the equation is a Beta decay.
Option B is correct.
Explanation:
Complete Question
What type of decay is illustrated by the equation below?
²¹⁴₈₃Bi → ²¹⁴₈₄Po + ⁰₋₁e
- alpha decay
- beta decay
- positron emission
- electron capture
Solution
The type of radioactive decay is usually discernable from studying the products and reactants (the parent nucleus/atom, the daughter nucleus and the emitted or absorbed particles) of the radioactive decay.
The changes in atomic and mass numbers is accounted for in the particular particle that that is emitted and subsequently termed the type of radioactive decay that is going on.
For example, this equation represented shows that the Bi atom splits into a Po and a Beta particle. Hence, it is easy and straight forward to see that this radioactive decay is to relaese a Beta particle and the decay is a Beta decay.
(Note that, just as presented in the equation, a Beta particle has a mass number of 0 and an atomic number -1, kind of similar to an electron).
- An alpha decay releases an alpha particle.
- A beta decay releases a Beta particle.
- A positron emission releases a positron.
- An electron capture involves an electron attacking the parent atom/nucleus or the reactant.
Hope this Helps!!!
I believe the answer is C: For objects at extremely fast speeds.
Hope this helps!
Answer:
150 km/hr²
Explanation:
The data for the motion of the train includes;
The train starts from rest, therefore, the initial velocity of the train, u = 0 km/hr
The speed of the train after 0.5 hours, v = 75 km/hr
The change in time during the motion of the train, Δt = 0.5 hours
The acceleration, 'a', is the rate of change of velocity with time;
Therefore, we have;
a = (75 km/hr - 0 km/hr)/(0.5 hr) = 150 km/hr²
a = 150 km/hr² = 150 km/hr² × 1,000 m/km × hr²/(3,600 s)² = 5/432 m/s²
The acceleration of the train, a = 150 km/hr² = 5/432 m/s².
