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musickatia [10]
3 years ago
15

"For punishment to work (i.e., to weaken the frequency of undesirable behaviors without creating a backlash), the punishment mus

t be strong enough to stop the undesired behavior and must be administered"
Physics
1 answer:
Flauer [41]3 years ago
3 0

Answer: Consistently,

Contingently and quickly

Explanation:

Consistently: it must follow a prescribed pattern and procedure putting the legal provisions in check.

Contingently: It must be administered in such a way that it is seen as dependent on the for a change character to occur. It should be administered putting into consideration of the undesirable character which are to be corrected.

Quickly: It must be administered on time,this is to ensure that the punishment gives immediate corrective action and prevent a reoccurrence of such undesirable character.

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Phyics !!! Cannon ball question
natima [27]

Answer:

Zero; no force is required to keep it going

Explanation:

Since the cannon ball is fired into frictionless space, there will be nothing to stop it, so it will keep going and going.

3 0
3 years ago
True or false all waves need a medium in order to travel
Roman55 [17]
No light does not need medium to travel.Its an electromagnetic(EM) wave.All EM waves travel independent of medium. In contrast mechanical waves need a medium,sound waves,shock waves are mechanical and they need medium to transfer energy or travel.
3 0
3 years ago
Help asapp!!!!!!!!!!!!!!
IRINA_888 [86]
Sorry don’t know this one
3 0
3 years ago
1. A 2.5 kg led projector is launched as a projectile off a tall building. At one point, as it
spin [16.1K]

Answer:

Explanation:

I got everything but i. Don't know why but it's eluding me. So let's do everything but that.

a. PE = mgh so

   PE = (2.5)(98)(14) and

   PE = 340 J

b. KE=\frac{1}{2}mv^2 so

   KE=\frac{1}{2}(2.5)(14)^2 and

   KE = 250 J

c. TE = KE + PE so

   TE = 340 + 250 and

   TE = 590 J

d. PE at 8.7 m:

   PE = (2.5)(9.8)(8.7) and

   PE = 210 J

e. The KE at the same height:

   TE = KE + PE and

   590 = KE + 210 so

   KE = 380 J

f. The velocity at that height:

   380=\frac{1}{2}(2.5)v^2 and

   v=\sqrt{\frac{2(380)}{2.5} } so

   v = 17 m/s

g. The velocity at a height of 11.6 m (these get a bit more involed as we move forward!). First we need to find the PE at that height and then use it in the TE equation to solve for KE, then use the value for KE in the KE equation to solve for velocity:

   590 = KE + PE and

   PE = (2.5)(9.8)(11.6) so

   PE = 280 then

   590 = KE + 280 so

   KE = 310 then

   310=\frac{1}{2}(2.5)v^2 and

   v=\sqrt{\frac{2(310)}{2.5} } so

   v = 16 m/s

h. This one is a one-dimensional problem not using the TE. This one uses parabolic motion equations. We know that the initial velocity of this object was 0 since it started from the launcher. That allows us to find the time at which the object was at a velocity of 26 m/s. Let's do that first:

   v=v_0+at and

   26 = 0 + 9.8t and

   26 = 9.8t so the time at 26 m/s is

   t = 2.7 seconds. Now we use that in the equation for displacement:

   Δx = v_0t+\frac{1}{2}at^2 and filling in the time the object was at 26 m/s:

   Δx = 0t + \frac{1}{2}(-9.8)2.7)^2 so

   Δx = 36 m

i. ??? In order to find the velocity at which the object hits the ground we would need to know the initial height so we could find the time it takes to hit the ground, and then from there, sub all that in to find final velocity. In my estimations, we have 2 unknowns and I can't seem to see my way around that connundrum.

4 0
3 years ago
Compare the mass of block A, the mass of block B is:
Alexxx [7]

the answer would be b. it`s half as great

6 0
3 years ago
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