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nevsk [136]
3 years ago
15

Can someone help me with the exercise 3 only c,d,e please!!!

Physics
2 answers:
saul85 [17]3 years ago
4 0
I'm not sure...
I feel ya.
fgiga [73]3 years ago
3 0
Im not sure, but i think you should try C
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An unstable atomic nucleus of mass 1.80 10-26 kg initially at rest disintegrates into three particles. One of the particles, of
beks73 [17]

Answer:

a)   v = (-7.73 i ^ - 6.95j ^) 10⁶ m/s  and b)  ΔK = 3.96 10⁻¹³ J

Explanation:

We can work this process of disintegration as a conservation problem of the moment.

We create a system formed by the initial nucleus and the three final particles, in this case all the forces involved are internal and the amount of movement is conserved. We will write the equations each axis

X axis

      p₀ = 0

      pf = m1 v1x + m vfx

      m₁ = 8.50 10-27 kg

      vf₁ = 4.00 106 m / s

     p₀ = pf

     0 = m₁ v₁ₓ + m₃ vfₓ

     vfₓ = -m₁ / m₃ v1ₓ

Y Axis

     P₀ = 0

     Pf = m₂ v₂ + m₃ vfy

     m₂ = 5.10 10⁻²⁷ kg

     v₂ = 6.00 10⁶ m / s

     p₀ = pf

     0 = m₂ v₂ + m₃ vfy

     vfy = -m₂ / m3₃v₂

We have the initial particle mass and it decomposes into three parts after disintegration

     m = m₁ + m₂ + m₃

     m₃ = m-m₁-m₂

     m₃ = 1.80 10⁻²⁶ - 5.10 10⁻²⁷ - 8.50 10⁻²⁷ = (1.80 -0.510 -0.850) 10⁻²⁶

     m3 = 0.44 10⁻²⁶ kg = 4.4 10⁻²⁷ kg

Let's replace and calculate

    vfₓ = -m₁ / m₃ v₁ₓ

    vfₓ = - 8.50 10⁻²⁷/4.4 10⁻²⁷   4.00 10⁶

    vfₓ = -7.73 10⁶ m / s

    vfy = -m₂ / m₃ v₂

    vfy = - 5.10 10⁻²⁷ /4.4 10⁻²⁷   6.00 10⁶

    vfy = -6.95 10⁶ m/s

We set the speed vector

     v = (-7.73 i ^ - 6.95j ^) 10⁶ m/s

b) Let's calculate the kinetic energy

Initial

   K₀ = 0

Final

   Kf = K₁ + K₂ + K₃

   Kf = ½ m₁ v₁² + ½ m₂ v₂² + ½ m₃ v₃²

   v₃² = (7.73 10⁶)²+ (6.95 10⁶)²  = 108.5 10⁻¹²

  Kf = ½ 8.50 10⁻²⁷(4 10⁶) 2 + ½ 5.1 10⁻²⁷ (6 10⁶) 2 + ½ 4.4 10⁻²⁷ 108.05 10⁻¹²

  Kf = 68 10⁻¹⁵ + 91.8 10⁻¹⁵ + 237.7 10⁻¹⁵

  Kf = 396.8 10⁻¹⁵ J

   Kf = 3.96 10⁻¹³ J

  ΔK = Kf -K₀

  ΔK = 3.96 10⁻¹³ J

7 0
3 years ago
A firearms company is testing a new model of rifle by firing a 7.50-g lead bullet into a block of wood having a mass of 17.5 kg.
Anestetic [448]

Answer:

Explanation:

Let the bullets speed be V .

Kinetic energy = 1/2 mV² where m is mass of bullet

This energy is converted into heat Q which raises the temperature of target by Δ T .

Q = mc Δ T  , m is mass , c is specific heat and Δ T is rise in temperature .

heat absobed by bullet

= .0075 x 130 x .040

= .039 J

heat absorbed by block of wood

= 17.5 x 1700 x .04

= 1190 J

Total heat absorbed

= 1190.039 J

So kinetic energy = heat absobed

= 1/2 x .0075 x V² = 1190.039

V² = 317343.73

V  = 563.33 m /s

8 0
3 years ago
5 ohm resistor is in series with a bulb, a switch and a 12V battery.
Nikitich [7]

Answer: yo light bill finna be turnt off  im just kidding  1`

Explanation:

6 0
4 years ago
An object that’s
KonstantinChe [14]
An object thats NEGATIVELY charged has more electrons than protons, An object thats POSITIVELY charged has less electrons than protons, An object that NOT charged has the same number of electrons and protons.
5 0
3 years ago
Read 2 more answers
Two blocks in contact with each other are pushed to the right across a rough horizontal surface by the two forces shown. If the
ale4655 [162]

I assume the blocks are pushed together at constant speed, and it's not so important but I'll also assume it's the smaller block being pushed up against the larger one. (The opposite arrangement works out much the same way.)

Consider the forces acting on either block. Let the direction in which the blocks are being pushed by the positive direction.

The 2.0-kg block feels

• the downward pull of its own weight, (2.0 kg) <em>g</em>

• the upward normal force of the surface, magnitude <em>n₁</em>

• kinetic friction, mag. <em>f₁</em> = 0.30<em>n₁</em>, pointing in the negative horizontal direction

• the contact force of the larger block, mag. <em>c₁</em>, also pointing in the negative horizontal direction

• the applied force, mag. <em>F</em>, pointing in the positive horizontal direction

Meanwhile the 3.0-kg block feels

• its own weight, (3.0 kg) <em>g</em>, pointing downward

• normal force, mag. <em>n₂</em>, pointing upward

• kinetic friction, mag. <em>f₂</em> = 0.30<em>n₂</em>, pointing in the negative horizontal direction

• contact force from the smaller block, mag. <em>c₂</em>, pointing in the <u>positive</u> horizontal direction (this is the force that is causing the larger block to move)

Notice the contact forces form an action-reaction pair, so that <em>c₁</em> = <em>c₂</em>, so we only need to find one of these, and we can get it right away from the net forces acting on the 3.0-kg block in the vertical and horizontal directions:

• net vertical force:

<em>n₂</em> - (3.0 kg) <em>g</em> = 0   ==>   <em>n₂</em> = (3.0 kg) <em>g</em>   ==>   <em>f₂</em> = 0.30 (3.0 kg) <em>g</em>

• net horizontal force:

<em>c₂</em> - <em>f₂</em> = 0   ==>   <em>c₂</em> = 0.30 (3.0 kg) <em>g</em> ≈ 8.8 N

4 0
3 years ago
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