Answer:
coefficient of friction =0.268
magnitude of force P=289.78N
Explanation:
The coefficient of friction is obtained by mgsinФ/mgcosФ=tanФ=tan15=0.268
force P is horizontal as stated in the question, horizontal component of P=mgcosФ=30*10*cos15=289.78
Answer: B. It demonstrates a behavior or particles.
Explanation: I took the test and got it right
Labels that belong in regions marked X and Y are;
A) X: Maintains magnetic properties in the presence of another magnet
Y: Is made from magnetically hard ferromagnetic material
Answer:
i think number 4 :/ i hope its right...