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Ierofanga [76]
3 years ago
5

Insight therapies involve verbal interactions between a therapist and a client that are used to promote positive changes in one'

s life.
Please select the best answer from the choices provided

T
F​
Physics
2 answers:
Furkat [3]3 years ago
8 0

Answer:

True

Explanation:

Insight therapies are very powerful psychological tools used by therapist to help them unlock their past and provide a better well being and a more sound future.

  • It involves verbal communication between the client and the therapist.
  • The stages are grouped into sessions where both parties frequently meet.
  • Issues in the past that might be the premise of current behaviors are focused on.
  • The goal is to be able to annul the effect such issues currently have on the present lifestyle of the client.
Mumz [18]3 years ago
8 0

Answer:

true is the answer

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Suppose you push a hockey puck of mass m across frictionless ice for a time \Delta t, starting from rest, giving thepuck speed v
bazaltina [42]

Answer:

1. t_2 = 2t_1

2. t_2 = t_1\sqrt{2}

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1. According to Newton's law of motion, the puck motion is affected by the acceleration, which is generated by the push force F.

In Newton's 2nd law: F = ma

where m is the mass of the object and a is the resulted acceleration. So in the 2nd experiment, if we double the mass, a would be reduced by half.

a_1 = 2a_2

Since the puck start from rest, in the 1st experiment, to achieve speed of v it would take t time

t = v / a_1

Now that acceleration is halved:

t = \frac{v}{2a_2}

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You would need to push for twice amount of time t_2 = 2t_1

2. The distance traveled by the puck is as the following equation:

d = at^2

So if the acceleration is halved while maintaining the same d:

\frac{d_1}{d_2} = \frac{a_1t_1^2/2}{a_2t_2^2/2}

As d_1 = d_2, then d_1/d_2 = 1. Also a_1 = 2a_2

1 = \frac{2a_2t_1^2}{a_2t_2^2}

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3 years ago
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A wedge with an inclination of angle θ rests next to a wall. A block of mass m is sliding down the plane. There is no friction b
Softa [21]

Answer:

  The net force on the block  F(net)  = mgsinθ).

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Here, m is the mass of the block, n is the normal force, \thetaθ is the wedge angle, and Fw  is the force exerted by the wall on the wedge.

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(b)

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The horizontal component of normal force on the block is equal to force

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Substitute mgcosθ for n in the above equation;

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Since, there is no friction between the wedge and the wall, there is component force acting on the wall to restrict the motion of the wedge on the surface and that force is arises from the horizontal component for normal force on the block.

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