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Butoxors [25]
2 years ago
15

The principle of conservation of linear momentum can be strictly applied during a collision between two particles provided the t

ime of impact is​
Physics
1 answer:
Amanda [17]2 years ago
7 0

Answer:

The principle of conservation of linear momentum can be strictly applied during a collision between two particles provided the time of impact is​ extremely small.

Explanation:

During the collision between two particles, a large force acts between the two colliding bodies for a relatively short time, also known as impulse.

Therefore, the principle of conservation of linear momentum can be strictly applied during a collision between two particles provided the time of impact is​ extremely small.

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How does the input distance of a single fixed pulley compare to the out- put distance?
ololo11 [35]

A pulley is another sort of basic machine in the lever family. We may have utilized a pulley to lift things, for example, a banner on a flagpole.

<u>Explanation:</u>

The point in a fixed pulley resembles the support of a lever. The remainder of the pulley behaves like the fixed arm of a first-class lever, since it rotates around a point. The distance from the fulcrum is the equivalent on the two sides of a fixed pulley. A fixed pulley has a mechanical advantage of one. Hence, a fixed pulley doesn't increase the force.

It essentially alters the direction of the force. A moveable pulley or a mix of pulleys can deliver a mechanical advantage of more than one. Moveable pulleys are appended to the item being moved. Fixed and moveable pulleys can be consolidated into a solitary unit to create a greater mechanical advantage.

4 0
2 years ago
How much force does it take to bring a 1,375 N car from rest to a velocity of 26 m/s in 6 seconds?
V125BC [204]

Explanation:

<u>Mass of car</u> = 137.5 kg

<u>Acceleration</u> = v - u / t = 26 - 0 / 6 = 4.33 m/sec^2

Force = m * a = 137.5 * 4.33 = 595.3 N

6 0
3 years ago
The persistence of vision for normal eye is
attashe74 [19]

Answer:

The answer is 1/16

Explanation:

1. Persistence of vision refers to the optical illusion that occurs when visual perception of an object does not cease for some time after the rays of light proceeding from it have ceased to enter the eye. 2. The persistence of vision for normal eye is 1/16 if a second.

6 0
3 years ago
Read 2 more answers
A vehicle starts to move from the rest gets an acceleration of 5 m/s2 within 2
likoan [24]

Answer:

When an object moves in a straight line with a constant acceleration, you can calculate its acceleration if you know how much its velocity changes and how long this takes.

The formula is,

Acceleration = change in velocity / time taken

The equation for acceleration can also be represented as:

a = (v-u) \ t

The change in velocity v – u = 5 – 0 = 5 m/s.

The acceleration = change in velocity ÷ time = 5 m/s ÷ 2 s = 2.5 m/s^2

3 0
2 years ago
Read 2 more answers
A uniform beam with mass M and length L is attached to the wall by a hinge, and supported by a cable. A mass of value 3M is susp
Jobisdone [24]

Answer:

The tension is  T= \frac{11}{2\sqrt{3} } Mg

The horizontal force provided by hinge   Fx= \frac{11}{4\sqrt{3} } Mg

Explanation:

   From the question we are told that

          The mass of the beam  is   m_b =M

          The length of the beam is  l = L

           The hanging mass is  m_h = 3M

            The length of the hannging mass is l_h = \frac{3}{4} l

            The angle the cable makes with the wall is \theta = 60^o

The free body diagram of this setup is shown on the first uploaded image

The force F_x \ \ and \ \ F_y are the forces experienced by the beam due to the hinges

      Looking at the diagram we ca see that the moment of the force about the fixed end of the beam along both the x-axis and the y- axis is zero

     So

           \sum F =0

Now about the x-axis the moment is

              F_x -T cos \theta  = 0

     =>     F_x = Tcos \theta

Substituting values

            F_x =T cos (60)

                 F_x= \frac{T}{2} ---(1)

Now about the y-axis the moment is  

           F_y  + Tsin \theta  = M *g + 3M *g ----(2)

Now the torque on the system is zero because their is no rotation  

   So  the torque above point 0 is

          M* g * \frac{L}{2}  + 3M * g \frac{3L}{2} - T sin(60) * L = 0

            \frac{Mg}{2} + \frac{9 Mg}{4} -  T * \frac{\sqrt{3} }{2}    = 0

               \frac{2Mg + 9Mg}{4} = T * \frac{\sqrt{3} }{2}

               T = \frac{11Mg}{4} * \frac{2}{\sqrt{3} }

                   T= \frac{11}{2\sqrt{3} } Mg

The horizontal force provided by the hinge is

             F_x= \frac{T}{2} ---(1)

Now substituting for T

              F_{x} = \frac{11}{2\sqrt{3} } * \frac{1}{2}

                  Fx= \frac{11}{4\sqrt{3} } Mg

4 0
3 years ago
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