The initial mass fraction of the spacecraft that must be burned and ejected to achieve an increase in speed is 0,00219 m/s
<h3>What fraction of the initial mass of the spacecraft?</h3>
Increase the speed: Vf-Vi = 2.2 m/s
Speed of aircraft: Vr = 400 m/s
Speed of ejected products: Vrel = 1000 m/s
The answer is:
So, the initial mass fraction of the spacecraft that must be burned and ejected to achieve an increase in speed is 0,00219 m/s
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M = 10.0 g, the mass of the iron sample
ΔT = 75 - 25.2 = 49.5°C, the decrease in temperature
c = 0.449 J/(g-°C), the specific heat of iron
The heat released is
Q = m*c*ΔT
= (10.0 g)*(0.449 J/(g-°C))*(49.5 C)
= 222.255 J
Answer: 222.3 J (nearest tenth)
Answer:
0.00016 kg
Explanation:
Given:
Power = P = 1.2 × 10⁹ Watts
Power = work done / Time
efficiency = 0.30
Input power = 1.2 × 10⁹ / 0.30 = 4 × 10⁹ W
Energy = 4 × 10⁹ x 60 x 60 = 1.44 x 10¹³ joules
E = m c² , where c is the speed of light and m is the mass.
⇒ mass = m = E / c² = (1.44 x 10¹³) / (3 × 10⁸ )²
= 0.00016 kg
<span>1/3
The key thing to remember about an elastic collision is that it preserves both momentum and kinetic energy. For this problem I will assume the more massive particle has a mass of 1 and that the initial velocities are 1 and -1. The ratio of the masses will be represented by the less massive particle and will have the value "r"
The equation for kinetic energy is
E = 1/2MV^2.
So the energy for the system prior to collision is
0.5r(-1)^2 + 0.5(1)^2 = 0.5r + 0.5
The energy after the collision is
0.5rv^2
Setting the two equations equal to each other
0.5r + 0.5 = 0.5rv^2
r + 1 = rv^2
(r + 1)/r = v^2
sqrt((r + 1)/r) = v
The momentum prior to collision is
-1r + 1
Momentum after collision is
rv
Setting the equations equal to each other
rv = -1r + 1
rv +1r = 1
r(v+1) = 1
Now we have 2 equations with 2 unknowns.
sqrt((r + 1)/r) = v
r(v+1) = 1
Substitute the value v in the 2nd equation with sqrt((r+1)/r) and solve for r.
r(sqrt((r + 1)/r)+1) = 1
r*sqrt((r + 1)/r) + r = 1
r*sqrt(1+1/r) + r = 1
r*sqrt(1+1/r) = 1 - r
r^2*(1+1/r) = 1 - 2r + r^2
r^2 + r = 1 - 2r + r^2
r = 1 - 2r
3r = 1
r = 1/3
So the less massive particle is 1/3 the mass of the more massive particle.</span>
What Is A Molecule? B) NaCi
