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vlabodo [156]
4 years ago
5

Three forces act on an object. If the object is in translational equilibrium, which of the following must be true? I. The vector

sum of the three forces must equal zero; II. The magnitude of the three forces must be equal; III. The three forces must be parallel.
Physics
1 answer:
Svet_ta [14]4 years ago
3 0

Answer:

Option I

Explanation:

When ever the system is in equilibrium, it means the net force on the system is zero.

If the number of forces acting on a system and then net force on the system is zero, it only shows that the vector sum of all the forces is zero.

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Particles of m1 and m2 (m2>m1) are connected by a line in extensible string passing over a smooth fixed pulley. Initially, bo
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Answer:

The velocity with which the mass will hit the floor is v_f = \sqrt{2(\dfrac{m_2-m_1}{m_2+m_1}) x.}

Explanation:

If the tension in the string is T, for m_1 we have

T- m_1g =m_1a,

and for the mass m_2

T -m_2g = -m_2a

From these equations we solve for a and get:

a =(\dfrac{m_2-m_1}{m_2+m_1}) g.

The kinematic equation

v_f^2 = v_0^2+2ax

gives the final velocity v_f of a particle, when its initial velocity was v_0, and has traveled a distance x while undergoing acceleration a.

In our case

v_0 = 0 (the initial velocity of the particles is zero)

a =(\dfrac{m_2-m_1}{m_2+m_1}) g.

which gives us

v_f^2 = 2ax

v_f^2 =2(\dfrac{m_2-m_1}{m_2+m_1}) g

\boxed{v_f = \sqrt{2(\dfrac{m_2-m_1}{m_2+m_1}) x.} }

which is the velocity with which the mass m_2 will hit the floor.

8 0
3 years ago
A 2500‐kg vehicle traveling at 25 m/s can be stopped by gently applying the breaks for 20 seconds. What is the average force sup
katen-ka-za [31]
Momentum = (mass) x (speed)

Change in momentum = (force) x (time)

The initial momentum is (mass) x (speed) = 2500x 25 = 62,500 kg-m/s.

Since you want to <u>stop</u> the vehicle, that number is also the required <em>change</em>
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62,500 = (force) x (time) = 20 x force

Divide each side by 20 :

force = 62,500 / 20 = <em>3,125 newtons </em>
3 0
3 years ago
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