Ep = mgh
= (35)(9.8)(3.5)
= 1200.5 J
Ek = (1/2)(mv^2)
= (0.5)(35)(5^2)
= 437.5 J
Answer:
94.1 m
Explanation:
From Coulombs law,
F = Gm1m2/r²................... Equation 1
where F = force, m1 = first mass, m2 = second mass, G = universal constant, r = distance of separation.
Make r the subject of the equation,
r = √(Gm1m2/F)................. Equation 2
Given: F = 7×10² N, m1 = 15×10⁷ kg, m2 = 62×10⁷ kg,
Constant: G = 6.67×10⁻¹¹ Nm²/kg²
Substitute into equation 2
r = √( 6.67×10⁻¹¹×15×10⁷×62×10⁷/7×10²)
r √(886.16×10)
r √(88.616×10²)
r = 9.41×10
r = 94.1 m.
Hence the distance of separation = 94.1 m
I believe it is A) 2(1/2mv^2)=mgh <===cross out masses
2(1/2V^2)=gh
Solve for V
Answer:
c.
Explanation:
In positron emission, also called positive beta decay (β+-decay), a proton in the parent nucleus decays into a neutron that remains in the daughter nucleus, and the nucleus emits a neutrino and a positron, which is a positive particle like an ordinary electron in mass but of opposite charge.
Does this help? Sorry it's wordy that's just how my teachers taught me :'(
Just took the quiz on k12 and the answer is A. compressional waves