<h2>
Answer: 13.61 N/m</h2>
Hooke's law establishes that the elongation of a spring is directly proportional to the modulus of the force
applied to it, <u>as long as the spring is not permanently deformed</u>:
(1)
Where:
is the elastic constant of the spring. The higher its value, the more work it will cost to stretch the spring.
is the length of the spring without applying force.
is the length of the spring with the force applied.
According to this, we have a spring where only the force due gravity is applied.
In other words, the force applied is the weigth
of the block:
(2)
Where
is the mass of the block and
is the gravity acceleration.
(3)
(4)
Knowing the force applied
and
and
, we can substitute the values in equation (1) and find
:
(5)
(6)
<u>Finally:</u>
Answer:
Same magnitude of the 10 nc charge cause the electric field is external.
Explanation:
To do a better explanation, let's go and suppose we have an electric field of, 1300 N/C with a 10 nC charge.
As the system we are talking about is really big, and the charge is small, we can assume always if the charge is sitting right in the same point where the electric field is, then, the electric field would not suffer any kind of alteration in it's value. Therefore, no matter what value of the charge is sitting here, the electric field is independent of the charge, so it would not feel any alteration. However, the force that the charge is feeling would be stronger than in the first case.
F = qE
If charge is doubled, then the force would be bigger in the second case than in the first case, but electric field remain the same value.
Answer:
a. El tiempo de recorrido es
segundos para un objeto localizado a un metro de distancia de la cámara fotográfica.
b. El tiempo de recorrido es 0.118 segundos para un objeto localizado a un metro de distancia de la cámara fotográfica.
Explanation:
El sonido es un tipo de onda mecánica, que es un tipo de onda que necesita de un medio material para propagarse. En este caso, entendemos que el sonido se propaga a través del aire atmosférico hasta llegar a su destino y devolverse a rapidez constante. Entonces, podemos estimar el tiempo (
), medido en segundos, a partir de la siguiente fórmula:

Donde:
- Distancia entre la cámara fotográfica y el objeto, medida en metros.
- Rapidez del sonido en el aire atmosférico, medida en metros por segundo.
A continuación, calculamos el tiempo de recorrido:
a. (
,
)


El tiempo de recorrido es
segundos para un objeto localizado a un metro de distancia de la cámara fotográfica.
b. (
,
)


El tiempo de recorrido es 0.118 segundos para un objeto localizado a un metro de distancia de la cámara fotográfica.
Answer:
The distance is 11 m.
Explanation:
Given that,
Friction coefficient = 0.24
Time = 3.0 s
Initial velocity = 0
We need to calculate the acceleration
Using newton's second law
...(I)
Using formula of friction force
....(II)
Put the value of F in the equation (II) from equation (I)
....(III)

Put the value in the equation (III)


We need to calculate the distance,
Using equation of motion



Hence, The distance is 11 m.