The cheetah is one of the fastest accelerating animals, for it can go from rest to 28.0 m/s in 5.20 s. If its mass is 100 kg, de
1 answer:
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Answer:
I₁ = (7.78 i ^ - 6.71 j ^) 10⁻³ J s , I₂ = (-12.5 i ^ -14.6 j ^) 10⁻³ J s , I₃ = (19.1i ^ + 18.6 j ^) 10⁻³ J s and I₄ = (-9.14i ^ + 7.24 j ^) 10⁻³ J s
Explanation:
The impulse is equal to the variation of the moment, to apply this relationship to our case, we will assume that initially the mouse was at rest
I = Δp = m -m v₀
I = m ( -v₀)
Bold indicates vector quantities, let's calculate the momentum of each mouse in for the x and y axes
We recommend bringing all units to the SI system
Mouse 1.
It has a mass of 22.3 g = 22.3 10⁻³ kg, a final velocity of (v = 0.349 i ^ - 0.301 j ^) m / s with an initial velocity of zero
Iₓ = m ( - v₀ₓ)
Iₓ = 22.3 10⁻³ (0.349 -0)
Iₓ = 7.78 10⁻³ J s
= m (
-
)
= 22.3 10⁻³ (-0.301)
= -6.71 10⁻³ J s
I₁ = (7.78 i ^ - 6.71 j ^) 10⁻³ J s
Mouse 2
Mass 17.9 g = 17.9 10⁻³ kg
Speed (-0.699 i ^ - 0.815 j ^) m / s
Iₓ = m ( - v₀ₓ)
Iₓ = 17.9 10⁻³ (-0.699 -0)
Iₓ = -12.5 10⁻³ J s
= 17.9 10⁻³ (-0.815 - 0)
= -14.6 10⁻³ J s
I₂ = (-12.5 i ^ -14.6 j ^) 10⁻³ J s
Mouse 3
Mass 19.1 g = 19.1 10⁻³ kg
Speed (0.745i ^ + 0.975 j ^) m / s
Iₓ = 19.1 10⁻³ (0.745 -0)
Iₓ = 14.2 10⁻³ J s
= 19.1 10⁻³(0.975 -0)
= 18.6 10⁻³ J s
I₃ = (19.1i ^ + 18.6 j ^) 10⁻³ J s
Mouse 4
Mass 10.1 g = 10.1 10⁻³ kg
Speed (-0.905i ^ + 0.717j ^) m / s
Iₓ = 10.1 10⁻³ (-0.905 -0)
Iₓ = -9.14 10⁻³ J s
= 10.1 10⁻³ (0.717 -0)
= 7.24 10⁻³ J s
I₄ = (-9.14i ^ + 7.24 j ^) 10⁻³ J s
Answer:
The increase in temperature of the bullet is 351.1 kelvin
Explanation:
First, we should find the kinetic energy of the bullet is:
with m the mass and v the velocity.
Now we know that half of the kinetic energy of the bullet is transformed into internal energy, by second's law of thermodynamics that means heat (Q) to raise bullet temperature (T), so:
To know what the increase in temperature is, we should use specific heat of lead:
The equation that relates specific heat, change in temperature and mass is:
solving for :