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Answer:
(E) μs(mA +mB)g
Explanation:
We can apply for mB:
∑ Fx = mB*a (→)
⇒ Ffriction = mB*a ⇒ a = Ffriction / mB = μs*N / mB
⇒ a = μs*(mB*g) / mB ⇒ a = μs*g (acceleration of the system)
Now, for mA we have
∑ Fx = mA*a (→)
F - Ffriction = mA*a ⇒ F = mA*a + Ffriction
⇒ F = mA*(μs*g) + μs*(mB*g) ⇒ F = μs*g*(mA + mB)
We must know that the friction acts only between the two blocks
Alkali metals alkaline earth metals and halogens and noble gases. I’m pretty sure
Answer:
the child is 1.581 m far from the fence
Explanation:
The diagrammatic illustration that give a better view of what the question denote can be seen in the image attached below.
From the image attached below, let assume that the release point is the origin, then equation of the motion (x) is as follows:
---- (1)
the equation of the motion y is :
By using the quadratic formula, we have;
where;
a = 4.9, b = -5.14 c = 1
In as much as the ball is traveling upward, then we consider t= 0.258sec.
From equation (1)
Thus, the child is 1.581 m far from the fence
You can't change time to a measurement of distance