Answer:

Explanation:
Given data
Time t=2.5 minutes=150 seconds
Distance A=1600 ft=487.68 m........east
Distance B=2500 ft=762m ........north
To find
Average velocity
Solution
First we need to find the resultant distance magnitude.To find that we apply Pythagorean theorem to find hypotenuse
So


It should be 0.25kg because you converter from g to kg and since 1g<1kg so you move the decimal to the left
<span>When an electron is hit by a photon of lights,it absorbs the quanta of energy the photon was carrying and moves to a higher energy state.Electrons therefore have to jump around within the atom as they either gain or lose energy.</span>
Answer: It states that the BCD equivalent would be 0001000100000000000100010001000100010000000100000001000000000001.
Answer:
a = 3,0 m/s²
Explanation:
En este ejercicio se pide calcular la aceleracion del cuerpo, usemos las ecuaciones de cinematica en una dimensión.
v= v₀ + a t
como el corredor parte del reposo si velocidad inicial es cero
v = at
a = v/t
calculemos
a = 12 /4,0
a = 3,0 m/s²