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:
C
Explanation:
- Let acceleration due to gravity @ massive planet be a = 30 m/s^2
- Let acceleration due to gravity @ earth be g = 30 m/s^2
Solution:
- The average time taken for the ball to cover a distance h from chin to ground with acceleration a on massive planet is:
t = v / a
t = v / 30
- The average time taken for the ball to cover a distance h from chin to ground with acceleration g on earth is:
t = v / g
t = v / 9.81
- Hence, we can see the average time taken by the ball on massive planet is less than that on earth to reach back to its initial position. Hence, option C
If someone is underground, then therefore there is less planet/ground underneath them, so there would be less gravity. Gravity directly affects weight.
Answer:
T=183.21K
Explanation:
We have to take into account that the system is a ideal gas. Hence, we have the expression

where P is the pressure, V is the volume, n is the number of moles, T is the temperature and R is the ideal gas constant.
Thus, it is necessary to calculate n and V
V is the volume of a sphere

V=8.86*10^{50}L
and for n

Hence, we have (1 Pa = 9.85*10^{-9}atm)

hope this helps!!
In my text book for social studies it says a canal