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coldgirl [10]
3 years ago
13

A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.

Physics
1 answer:
Bas_tet [7]3 years ago
5 0

Answer:

Explanation:

Total momentum of the system before the collision

.5 x 3 - 1.5 x 1.5 = -0.75 kg m/s towards the left

If v be the velocity of the stuck pucks

momentum after the collision = 2 v

Applying conservation of momentum

2 v = -  .75

v =  - .375 m /s

Let after the collision v be the velocity of .5 kg puck

total momentum after the collision

.5 v + 1.5 x .231 = .5v +.3465

Applying conservation of momentum law

.5 v +.3465 = - .75

v = - 2.193 m/s

2 ) To verify whether the collision is elastic or not , we verify whether the kinetic energy is conserved or not.

Kinetic energy before the collision

= 2.25 + 1.6875

=3.9375 J

kinetic energy after the collision

= .04 + 1.2 =1.24 J

So kinetic energy is not conserved . Hence collision is not elastic.

3 ) Change in the momentum of .5 kg

1.5 - (-1.0965 )

= 2.5965

Average force applied = change in momentum / time

= 2.5965 / 25 x 10⁻³

= 103.86 N

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Answer:

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Explanation:

Iron and gadlinium are both very easily made into magnetic substances.  Cobalt is also capable of being magnetized. Aluminum, put in an alloy, can make a magnetic substance, but

Aluminum by itself is not able to be magnetized.

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Why does Energy appear continuous in everyday life?
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2 years ago
A +71 nC charge is positioned 1.9 m from a +42 nC charge. What is the magnitude of the electric field at the midpoint of these c
viva [34]

Answer:

The net Electric field at the mid point is 289.19 N/C

Given:

Q = + 71 nC = 71\times 10^{- 9} C

Q' = + 42 nC = 42\times 10^{- 9} C

Separation distance, d = 1.9 m

Solution:

To find the magnitude of electric field at the mid point,

Electric field at the mid-point due to charge Q is given by:

\vec{E} = \frac{Q}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}

\vec{E} = \frac{71\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}

\vec{E} = 708.03 N/C

Now,

Electric field at the mid-point due to charge Q' is given by:

\vec{E'} = \frac{Q'}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}

\vec{E'} = \frac{42\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}

\vec{E'} = 418.84 N/C

Now,

The net Electric field is given by:

\vec{E_{net}} = \vec{E} - \vec{E'}

\vec{E_{net}} = 708.03 - 418.84 = 289.19 N/C

5 0
3 years ago
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