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svlad2 [7]
3 years ago
13

A bullet has a mass of 8 grams and a muzzle velocity of 340m/sec. A baseball has a mass of 0.2kg and is thrown by the pitcher at

40m/sec. What is the momentum of the baseball? What is the momentum of the bullet?
Physics
1 answer:
Thepotemich [5.8K]3 years ago
3 0

Answer:

Momentum of bullet

P = 2.72 kg m/s

momentum of baseball

P = 8 kg m/s

Explanation:

As we know that momentum is defined as the product of mass and velocity

here we know that

mass of the bullet = 8 gram

velocity of bullet = 340 m/s

momentum of the bullet is given as

P = mv

P = (\frac{8}{1000})(340)

P = 2.72 kg m/s

Now we have

mass of baseball = 0.2 kg

velocity of baseball = 40 m/s[/tex]

momentum of baseball is given as

P = (0.2)(40)

P = 8 kg m/s

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A gray kangaroo can bound across a flat stretch of ground with each jump carrying it 10 m from the takeoff point. you may want t
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<u>Answer:</u>

Takeoff speed of Kangaroo = 12.91 m/s

<u>Explanation:</u>

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   Range of projectile = Time taken for the projectile to reach ground* Horizontal velocity

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       Time taken for the projectile to reach maximum height = Vertical speed / Acceleration = u sin θ/g

      Time taken for the projectile to reach ground = 2 u sin θ/g

  So Range of projectile = ucos\theta*\frac{2usin\theta}{g} =\frac{u^2sin2\theta}{g}

 We have Range = 10 meter, θ = 18⁰

    Substituting

        10=\frac{u^2sin(2*18)}{9.8}\\ \\ u^2= 166.73\\ \\ u=12.91 m/s

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The initial temperature of the metal bolt is 80.8 °C

We'll begin by calculating the heat absorbed by the water.

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  • Initial temperature (T₁) = 21 °C
  • Final temperature (T₂) = 25 °C
  • Change in temperature (ΔT)  = T₂ – T₁ = 25 – 21 = 4 °C
  • Specific heat capacity of water (C) = 4184 J/KgºC
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Q = MCΔT

Q = 0.15 × 4184 × 4

Q = 2510.4 J

Finally, we shall determine the initial temperature of the metal bolt.

  • Heat absorbed by water = 2510.4 J
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Q = MC(T₂ – T₁)

–2510.4 = 0.050 × 899 (25 – T₁)

–2510.4 = 44.95 (25 – T₁)

Clear bracket

–2510.4 = 1123.75 – 44.95T₁

Collect like terms

–2510.4 – 1123.75 = –44.95T₁

–3634.15 = –44.95T₁

Divide both side by –44.95

T₁ = –3634.15 / –44.95

T₁ = 80.8 °C

Thus, the initial temperature of the metal is 80.8 °C.

Learn more about heat tranfer:

brainly.com/question/26034272

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