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liberstina [14]
3 years ago
13

John is a field researcher who studies social interaction within motorcycle groups. He is not an accomplished rider himself, but

he is able to obtain jobs at various motorcycle shops that allow him to observe members of motorcycle groups and to socialize with them after work. He lets members think that he is a regular employee and does not mention his research role to anyone connected with the groups. His note-taking is done after hours in his own apartment. His role as a field research is that of a:A. Complete observer
B. Partial observer
C. Participant and observer
D. Complete participant
Physics
1 answer:
uysha [10]3 years ago
7 0
His role as a field research is that of a: Complete participant!
(Option D.)

~Good luck!
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Over 4 seconds, a car's momentum decreases by 1000 kg m/s how much force did it take to make this happen?
kotykmax [81]

Answer:

250N

Explanation:

Given parameters:

Time  = 4s

Momentum  = 1000kgm/s

Unknown:

Force  = ?

Solution:

To solve this problem, we use Newton's second law of motion;

      Ft  = Momentum

F is the force

t is the time

So;

          F x 4 = 1000kgm/s

          F  = 250N

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3 years ago
John dragged a wooden block across a table with 4 different surfaces and the results
AnnyKZ [126]

If the surface is rough then the average force needed to move the wooden block will be more as the friction between the surfaces will be more.

Hence, the least force will be required on the smoothest surface and the greatest force will be required for the roughest surface.

a) Hence, the order will be: B > D > A > C

b) When the surface is glued then the force of friction will increase. Hence, it will require more force to move the wooden block on the surface. Hence, the more required will be more than 105 N.

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3 years ago
How would the photos set up change the mechanical advantage of lifting the block as compared with lifting the block without a le
lisabon 2012 [21]
<h3>Answer:</h3>

  The mechanical advantage would decrease, making the block more difficult to lift.

<h3>Explanation:</h3>

The mechanical advantage in such a setup is the ratio of distance from A to B to the distance from D to B. In this picture, that ratio is less than 1, meaning the advantage of having this setup is less than the advantage of no setup at all.

While the force required to lift the block is increased by this setup, the distance over which that force is applied will be smaller for raising the block to a given height. (Overall, for the same height, more work is required with the lever setup because you're raising part of the mass of the lever as well as the mass of the block.)

5 0
3 years ago
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In Anchorage, collisions of a vehicle with a moose are so common that they are referred to with the abbreviationMVC. Suppose a 1
lara31 [8.8K]

Answer:

Part a)

f = \frac{8}{9}

Part b)

f = \frac{120}{169}

Part c)

So from above discussion we have the result that energy loss will be more if the collision occurs with animal with more mass

Explanation:

Part a)

Let say the collision between Moose and the car is elastic collision

So here we can use momentum conservation

m_1v_{1i} = m_1v_{1f} + m_2v_{2f}

1000 v_o = 1000 v_{1f} + 500 v_{2f}

also by elastic collision condition we know that

v_{2f} - v_{1f} = v_o

now we have

2v_o = 2v_{1f} + v_o + v_{1f}

now we have

v_{1f} = \frac{v_o}{3}

Now loss in kinetic energy of the car is given as

\Delta K = \frac{1}{2}m(v_o^2 - v_{1f}^2)

\Delta K = \frac{1}{2}m(v_o^2 - \frac{v_o^2}{9})

so fractional loss in energy is given as

f = \frac{\Delta K}{K}

f = \frac{\frac{4}{9}mv_o^2}{\frac{1}{2}mv_o^2}

f = \frac{8}{9}

Part b)

Let say the collision between Camel and the car is elastic collision

So here we can use momentum conservation

m_1v_{1i} = m_1v_{1f} + m_2v_{2f}

1000 v_o = 1000 v_{1f} + 300 v_{2f}

also by elastic collision condition we know that

v_{2f} - v_{1f} = v_o

now we have

10v_o = 10v_{1f} + 3(v_o + v_{1f})

now we have

v_{1f} = \frac{7v_o}{13}

Now loss in kinetic energy of the car is given as

\Delta K = \frac{1}{2}m(v_o^2 - v_{1f}^2)

\Delta K = \frac{1}{2}m(v_o^2 - \frac{49v_o^2}{169})

so fractional loss in energy is given as

f = \frac{\Delta K}{K}

f = \frac{\frac{60}{169}mv_o^2}{\frac{1}{2}mv_o^2}

f = \frac{120}{169}

Part c)

So from above discussion we have the result that energy loss will be more if the collision occurs with animal with more mass

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You need to clean and sand wood of any ruff edges
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