Hello!
Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton force is applied to it.
Data:
Hooke represented mathematically his theory with the equation:
F = K * Δx
On what:
F (elastic force) = 2 N
K (elastic constant) = 4 N/cm
Δx (deformation or elongation of the elastic medium or distance from a spring) = ?
Solving:




simplify by 2


Answer:
B.) 1/2 cm
_______________________
I Hope this helps, greetings ... Dexteright02! =)
Answer: 9/10
Explanation:
because it's really important and makes you energetic
False, Compounds can contain more than 2 elements.
Answer:
Explanation:
Magnetic field due to a long current carrying conductor
μ₀ / 4π x 2i / r ( i = current , r = distance of point from wire )
= 10⁻⁷ x 2 x 120 / 6.4 ( i = 120 A , r = 6.4m )
= 37.5 x 10⁻⁷ T .
= 3. 75 X 10⁻⁶ T .
= 3.75 µT.
b )
The direction of this field will be horizontal hence it will affect magnetic needle.
Answer is B! Hope you have a good day