Answer:
Las fuerzas se representan dibujando flechas sobre el cuerpo en el que actúan, por ejemplo, la fuerza que se aplica al empujar un barril. ... Para obtener la fuerza resultante, puede aprovecharse la representación gráfica mediante flechas.
Explanation:
Gracias!!!
Answer:
True
Explanation:
I kind of didn’t understand your question, but if is a true and false question is true!
Answer:
magnitude v0x = v0 * cos θ = 31 m/s * cos 16° = 30 m/s
Explanation:
Please see the figure for a graphical description of the problem. As you can see from the figure, the vector v0 (initial velocity) is the sum of its horizontal and vertical components v0x + v0y.
Using the trigonometric rule for right triangles:
cos θ = adjacent / hypotenuse
We can calculate the magnitude of the v0x vector.
Seeing the figure, notice that the vectors v0, its horizontal component, v0x, and its vertical component, v0y, form a right triangle.
v0x is the side adjacent to the angle θ, and v0 is the hypotenuse of the triangle. Then:
cos θ = magnitude v0x / magnitude v0
magnitude v0x = magnitude v0 * cos θ = 31 m/s * cos 16° = 30 m/s
Thank you for posting your question here at brainly. I think your question is incomplete. Below is the complete question, it can be found elsewhere:
What is the probability of finding an electron within one Bohr radius of the nucleus?<span>Consider an electron within the 1s orbital of a hydrogen atom. The normalized probability of finding the electron within a sphere of a radius R centered at the nucleus is given by 1-a0^2[a0^2-e^(-2R/a0)(a0^2+2a0R+2R2)]. Where a0 is the Bohr radius (for a hydrogen atom, a0 = 0.529 Å.). What is the probability of finding an electron within one Bohr radius of the nucleus? What is the probability of finding an electron of the hydrogen atom within a 2.30a0 radius of the hydrogen nucleus?
Below is the answer:
</span><span>you plug the values for A0 and R into your formula</span>
Answer:
2.35 x 10^{5} J
Explanation:
force (f) = 1080 N
distance (d) = 218 m
find the work done
work is said to be done when force applied to an object moves the object, therefore work done = force x distance
work done = 1080 x 218 = 235,440 J = 2.35 x 10^{5} J