Answer:
V = 10 km / 1 hr = 10 km/hr
V = -10 j km / hr if one were to use i, j, k as unit vectors with the usual orientation
Answer:
F₂= 210 pounds
Explanation:
Conceptual analysis
Hooke's law
Hooke's law establishes that the elongation (x) of a spring is directly proportional to the magnitude of force (F) applied to it, provided that said spring is not permanently deformed:
F= K*x Formula (1)
Where;
F is the magnitude of the force applied to the spring in Newtons (Pounds)
K is the elastic spring constant, which relates force and elongation. The higher its value, the more work it will cost to stretch the spring. (Pounds/inch)
x the elongation of the spring (inch)
Data
The data given is incorrect because if we apply them the answer would be illogical.
The correct data are as follows:
F₁ =80 pounds
x₁= 8 inches
x₂= 21 inches
Problem development
We replace data in formula 1 to calculate K :
F₁= K*x₁
K=( F₁) / (x₁)
K=( 80) / (8) = 10 pounds/ inche
We apply The formula 1 to calculate F₂
F₂= K*x₂
F₂= (10)*(21)
F₂= 210 pounds
Answer:
As light travels in a straight line at a constant speed, it's acceleration is <u>0 m/s²</u>.
There is no rate of change of speed, so there is no acceleration.
- <u>0 m/s²</u> is the right answer.
Answer:
The third particle should be at 0.0743 m from the origin on the negative x-axis.
Explanation:
Let's assume that the third charge is on the negative x-axis. So we have:

We know that the electric field is:

Where:
- k is the Coulomb constant
- q is the charge
- r is the distance from the charge to the point
So, we have:

Let's solve it for r(3).
Therefore, the third particle should be at 0.0743 m from the origin on the negative x-axis.
I hope it helps you!