Which model could represent a neutral atom of boron?
2 answers:
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Answer:
29,7 m
Explanation:
We need to devide the problem in two parts:
A) Energy
B) MRUV
<u>Energy:</u>
Since no friction between pint (1) and (2), then the energy conservatets:
Energy = constant ----> Ek(cinética) + Ep(potencial) = constant
Ek1 + Ep1 = Ek2 + Ep2
Ek1 = 0 ; because V1 is zero (the ball is "dropped")
Ep1 = m*g*H1
Ep2= m*g*H2
Then:
Ek2 = m*g*(H1-H2)
By definition of cinetic energy:
m*(V2)²/2 = m*g*(H1-H2) --->
Replaced values: V2 = 14,0 m/s
<u>MRUV:</u>
The decomposition of the velocity (V2), gives a for the horizontal component:
V2x = V2*cos(α)
Then the traveled distance is:
X = V2*cos(α)*t.... but what time?
The time what takes the ball hit the ground.
Since: Y3 - Y2 = V2*t + (1/2)*(-g)*t²
In the vertical axis:
Y3 = 0 ; Y2 = H2 = 2 m
Reeplacing:
-2 = 14*t + (1/2)*(-9,81)*t²
solving the ecuation, the only positive solution is:
t = 2,99 sec ≈ 3 sec
Then, for the distance:
X = V2*cos(α)*t = (14 m/s)*(cos45°)*(3sec) ≈ 29,7 m
The question is incomplete. Here is the complete question.
Calculate the maximum deceleration of a car that is heading down a 14° slope (one that makes an anlge of 14° with the horizontal) under the following road conditions. You may assum that the weight of the car is evenlydistributed on all four tires and that the sttic coefficient of friction is involved - that is, the tires are not allowed to slip during the deceleration. (Ignore rolling) Calculate for a car: (a) On a dry concrete. (b) On a wet concrete. (c) On ice, assuming that μs = 0.100, the same as for shoes on ice.
Answer: (a) a = - 11.05 m/s²; (b) a = - 10.64 m/s²; (c) a = - 9.84m/s²
Explanation: The image in the attachment describe the forces acting on the car. Observing that, we know that:
= -
-
The is a x-component of force due to gravity (W) and, in this case, is given by:
= W.sin(14)
W is described as: W = m.g
Force due to friction () is given by:
= μs.N
N is the normal force and, in the system, is equivalent of , so:
= m.g.cos(14)
Therefore, the formula will be:
= -
-
m.a = - (m.g.sin14) - (μs.mg.cos14)
a = - g (sin14 + μscos 14)
a) For dry concrete, μs = 1:
a = - g (sin14 + μscos 14)
a = - 9.8 (sin14 + 1.cos14)
a = - 11.05 m/s²
b) For wet concrete, μs = 0.7:
a = - g (sin14 + μscos 14)
a = - 9.8 (sin 14 + 0.7.cos14)
a = - 10.64 m/s²
c) For ice, μs = 0.1:
a = - g (sin14 + μscos 14)
a = - 9.8 (sin14 + 0.1cos14)
a = - 9.84 m/s²
