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Vesnalui [34]
3 years ago
13

A pump jack scaffold must be fitted with two ______ gripping mechanisms to prevent slippage.

Physics
1 answer:
FromTheMoon [43]3 years ago
5 0
A pump jack scaffold must be fitted with two positive gripping mechanisms to prevent slippage. Pump jacks are a uniquely designed scaffold consisting of a platform supported by movable brackets on vertical poles. The brackets are designed to be raised and lowered in a manner similar to an automobile jack. It is important to make sure that pump jack brackets have two positive gripping mechanisms to prevent any failure or slippage. 
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Another engineer is running a series of experiments with hydraulic systems. If she doubles the area of the input piston, what ha
Zepler [3.9K]

Answer:

The pressure remains the same.

F = P * A  the amount of force transmitted would double if only the input area is changed. Pressure is the same at all points in the system.

7 0
3 years ago
Crystallization is a separation technique used to produce rock candy from a sugar solution. true or false
Zinaida [17]
This is true due to the reaction that happens from water evaporating and leaving the sugar crystals behind to form.
3 0
3 years ago
The quotient of 8.4x10° and a number n results in 5.6x 1027. What is the value of n?
konstantin123 [22]
Answer: n = 1.5 x 10^-18

5 0
2 years ago
Two circular coils are concentric and lie in the same plane. The inner coil contains 170 turns of wire, has a radius of 0.012 m,
GREYUIT [131]

Answer:

The current of the outer coil is  I_o   =   3.99 \ A

Explanation:

From the question we are told that

    The number of turns of the inner coil is  N_i  = 170 \ turn

    The radius of the  inner coil is R_i =  0.012 \ m

     The current of the inner coil is  I_i  =  6.2 \ A

      The number of turns of the outer coil is N_o  =  220 \ turns

      The radius of the  outer coil is R_o  =  0.02 0 \ m

       

Generally the net magnetic field is mathematically represented as

              B  =  \frac{N \ mu I }{ 2 * R  }

Now from told that the net magnetic field is common

So  

           \frac{N_i  \mu I_i} {2 * R _i} =  \frac{N_o  \mu I_o} {2 * R _o}

Here  \mu is the permeability of free space

making  I_o the subject

            I_o   =   \frac{ N_i  I_i *2 * R _o}{N_o  *2 * R _i}

substituting values

           I_o   =   \frac{ 170 *6.2 *2 * 0.020}{220   *2 * 0.012}

         I_o   =   3.99 \ A

5 0
4 years ago
A runner of mass 56.0kg runs around the edge of a horizontal turntable mounted on a vertical, frictionless axis through its cent
SCORPION-xisa [38]

Answer: -0.84 rad/sec (clockwise)

Explanation:

Assuming no external torques act on the system (man + turntable), total angular momentum must be conserved:

L1 = L2

L1 = It ω + mm. v . r = 81.0 kg . m2 .21 rad/s – 56.0 kg. 3.1m/s . 3.1 m  

L1 = -521.15 kg.m2/sec (1)

(Considering to the man as a particle that is moving opposite to the rotation of  the turntable, so the sign is negative).

Once at rest, the runner is only a point mass with a given rotational inertia respect from the axis of rotation, that can be expressed as follows:

Im = m. r2 = 56.0 kg. (3.1m)2 = 538.16 kg.m2

The total angular momentum, once the runner has come to an stop, can be written as follows:

L2= (It + Im) ωf = -521.15 kg.m2/sec  

L2= (81.0 kg.m2 + 538.16 kg.m2) ωf = -521.15 kg.m2/sec  

Solving for ωf, we get:

ωf = -0.84 rad/sec  (clockwise)

5 0
3 years ago
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