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cestrela7 [59]
3 years ago
8

Block A, with mass mA, is initially at rest on a horizontal floor. Block B, with mass mB, is initially at rest on the horizontal

top surface of A. The coefficient of static friction between the two blocks is μs. Block A is pulled with a horizontal force. It begins to slide out from under B if the force is greater than:
(A) mA*g
(B) mB*g
(C) μs*mA*g
(D) μs*mB*g
(E) μs(mA +mB)g
Physics
1 answer:
nexus9112 [7]3 years ago
8 0

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

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Answer:

However, the disadvantages are:

1. Many atimes for some motion prolems, free-body diagrams has to be drawn many times so to have enough equations to solve for the unknowns. This is not the same with energy conservation principles.

2. In situations where we need to find the internal forces acting on an object, we can't truly solve such problems using free-body diagram as it captures external forces. This is not the same with energy conservation principles.

Explanation:

Often times the ideal method to use in solving motion problem related questions are mostly debated.

Energy conservation principles applies to isolated systems are useful when object changes their positions in moving upward or downward converts its potential energy due to gravity for kinetic energy, or the other way round. When energy in a system or motion remains constant that is energy is neither created nor destroyed, it can therefore be easier to calculate other unknown paramters like in the motion problem velocity, distance bearing it in mind that energy can only change from one type to another.

On the other hand, free body diagram which is a visual representation of all the forces acting on an object including their directions has so many advantages in solving motion related problems which include finding relationship between force and motion in identifying the force acting on a body.

5 0
3 years ago
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s344n2d4d5 [400]

Answer:

v = √2G M_{earth} / R

Explanation:

For this problem we use energy conservation, the energy initiated is potential and kinetic and the final energy is only potential (infinite r)

        Eo = K + U = ½ m1 v² - G m1 m2 / r1

        Ef = - G m1 m2 / r2

When the body is at a distance R> Re, for the furthest point (r2) let's call it Rinf

       Eo = Ef

       ½ m1v² - G m1 M_{earth} / R = - G m1 M_{earth} / R

      v² = 2G M_{earth} (1 / R - 1 / Rinf)

If we do Rinf = infinity     1 / Rinf = 0

       v = √2G M_{earth} / R

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The mechanical energy is conserved  

 

      Em = -G m1  M_{earth} / R

      Em = - G m1  M_{earth} / R

     R = int        ⇒  Em = 0

6 0
3 years ago
This property of waves is the only property where the relationship between energy and this property are indirect or inverse
yaroslaw [1]

Answer: I don't understand

Explanation:

study and pay attention

4 0
2 years ago
What is the magnitude of the electric force between a proton and an electron when they are at a distance of 4.09 angstrom from e
Zinaida [17]

Answer:

F=1.38*10^{-9}N

Explanation:

According to Coulomb's law, the magnitude of the electric force between two point charges is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them:

F=\frac{kq_1q_2}{d^2}

Here k is the Coulomb constant. In this case, we have q_1=-e, q_2=e and d=4.09*10^-10m. Replacing the values:

F=\frac{8.99*10^{9}\frac{N\cdot m^2}{C^2}(-1.6*10^{-19}C)(1.6*10^{-19}C)}{(4.09*10^{-10})^2}\\F=-1.38*10^{-9}N

The negative sign indicates that it is an attractive force. So, the magnitude of the electric force is:

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yarga [219]

Answer:

4000 kg.m/s

Explanation:

p=mv

m=800 kg

v=5 m/s

p=(800)(5)= 4000 kg.m/s

3 0
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