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4 years ago

Someone who likes to fix things without talking or writing about what he is doing is most closely associated with

1 answer:

6 0

The answer is technical reasoning.
Hope that helps.

Also, maybe add the actual choices for the question so people don't have to do a lot of research...

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50 POINTS! A Boy throws a ball horizontally a distance of 22m downrange from the top of a tower that is 20.0m tall. What is his

DerKrebs [107]

The ball's horizontal and vertical velocities at time $t$ are

$v_x=v_{xi}$

$v_y=v_{yi}-gt$

but the ball is thrown horizontally, so $v_{yi}=0$ . Its horizontal and vertical positions at time $t$ are

$x=v_{xi}t$

$y=20.0\,\mathrm m-\dfrac g2t^2$

The ball travels 22 m horizontally from where it was thrown, so

$22\,\mathrm m=v_{xi}t$

from which we find the time it takes for the ball to land on the ground is

$t=\dfrac{22\,\rm m}{v_{xi}}$

When it lands, $y=0$ and

$0=20.0\,\mathrm m-\dfrac{9.8\frac{\rm m}{\mathrm s^2}}2\left(\dfrac{22\,\rm m}{v_{xi}}\right)^2$

$\implies v_i=v_{xi}=11\dfrac{\rm m}{\rm s}$

7 0

3 years ago

A positive or negative outcome of a decision or action can be called a

Mandarinka [93]

The answer to this would be :
<span>consequence
</span>hope that this helps you!
Source: I had done a research about this. =)

7 0

3 years ago

Two blocks on a frictionless horizontal surface are on a collision course. one block with mass 0.6 kg moves at 0.8 m/s to the ri

Elanso [62]

First we can say that since there is no external force on this system so momentum is always conserved.

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

$0.6*0.8 + 1.2*0 = 0.6*v_{1f} + 1.2*v_{2f}$

$0.48= 0.6*v_{1f} + 1.2*v_{2f}$

$0.8 = v_{1f} + 2v_{2f}$

now by the condition of elastic collision

$v_{2f} - v_{1f} = 0.8 - 0[\tex]now add two equations[tex]3*v_{2f} = 1.6$

$v_{2f} = 0.533 m/s$

also from above equation we have

$v_{1f} = -0.267 m/s$

So ball of mass 0.6 kg will rebound back with speed 0.267 m/s and ball of mass 1.2 kg will go forwards with speed 0.533 m/s.

8 0

4 years ago

Q|C Review. A particle of mass 4.00kg is attached to a spring with a force constant of 100 N/m . It is oscillating on a friction

Nezavi [6.7K]

The change in energy after the collision is <u>0.5</u>

<u />

<h3>What is change in energy?</h3>

This refers to the difference in the energy where energy is the capacity to do work. There different forms of energy they include mechanical energy, solar energy, electrical energy and so on.

The energy described in the problem is mechanical energy and it is of two types kinetic energy and potential energy

<h3>solving for the change in energy as a result of the collision</h3>

where mass of particle mp = 4 kg

mass of object mb = 6 kg

force constant of spring k = 100 N/m

amplitude A = 2 m

kinetic energy = 1/2 mv^2

initial velocity u = Aω

ω = sqrt ( 100/ 4 )

u = 2 sqrt ( 100/ 4 )

u = 10m/s

final velocity v = 5 m/s

change in energy

= - 0.5 * ( 4 + 4 ) * 5^2 + 0.5 * 4 * 10^2 ) / 0.5 * 4 * 10^2

= 0.5

Read more on change in energy here: brainly.com/question/26066414

#SPJ4

8 0

2 years ago

An angry mob lynches a phsical teacher after receiving their grades. They throw the physics teacher off a tall building . They t

babunello [35]

Answer:

5010 m

Explanation:

The vertical position of the teacher at time t can be found by using the equation:

$y(t) = h + ut+ \frac{1}{2}gt^2$

where

h is the initial height

$u = -20 m/s$ is the initial velocity (negative since it's downward)

$g=-9.8 m/s^2$ is the acceleration of gravity

t is the time

The teacher reaches the ground when y = 0, so the equation becomes

$h=-ut-\frac{1}{2}gt^2$

Substituting t = 30 s, we find the initial height:

$h=-(-20)(30)-\frac{1}{2}(-9.8)(30)^2=5010 m$

5 0

3 years ago