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myrzilka [38]
4 years ago
13

Explanation A 5000 kg rocket is at rest in deep space. The rocket burns fuel pushing 10kg of exhaust gases rearward at 4000 m/s.

This process takes 10 s. During the time interval what average force is applied to the rocket?
Physics
2 answers:
katovenus [111]4 years ago
4 0

Answer:

F = 4000 N

Explanation:

given,

mass of rocket (M)= 5000 Kg

10 Kg gas burns at speed (m)= 4000 m/s

time = 10 s

average force = ?

at the end the rocket is at rest

by conservation of momentum

 M v + m v' = 0

 5000 x v - 10 x 4000 = 0

 5000 v = 40000

    v = 8 m/s

speed of rocket = 8 m/s

now,

we know

change in momentum = F x Δ t

F = \dfrac{m(v_i-v_f)}{\Delta t}

F = \dfrac{5000(8-0)}{10}

      F = 4000 N

Hence, the average force applied to the rocket is equal to F = 4000 N

Yuri [45]4 years ago
3 0

Answer:

4000 N

Explanation:

mass of fuel, m = 10 kg

velocity of ejected fuel, v = 4000 m/s

time, t = 10 s

force, F = velocity x mass of fuel burning per second

F = 4000 x 10 / 10 = 4000 N

Thus, the force is 4000 N.

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Answer:

3335400 N/m² or 483.75889 lb/in²

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

A = Area = 1.5 cm²

m = Mass of woman = 51 kg

F = Force = mg

When we divide force by area we get pressure

P=\frac{F}{A}\\\Rightarrow P=\frac{mg}{A}\\\Rightarrow P=\frac{51\times 9.81}{1.5\times 10^{-4}}\\\Rightarrow P=3335400\ N/m^2

1\ N/m^2=\frac{1}{6894.757}\ lb/in^2

3335400\ N/m^2=3335400\times \frac{1}{6894.757}\ lb/in^2=483.75889\ lb/in^2

The pressure exerted on the floor is 3335400 N/m² or 483.75889 lb/in²

7 0
3 years ago
Question: Self-test 3.12 Calculate the change in G for ice at -10°C, with density 917 kg mº, when the pressure is increased from
Akimi4 [234]

The change in the Gibb's free energy per mole (G) is 1.96 J.

The given parameters:

  • Density of the ice, ρ = 917 kg/m³
  • Initial pressure, P₁ = 1.0 bar
  • Final pressure, P₂ = 2.0 bar
  • Temperature, T = - 10 C
  • Mass of water = 18 g

The change in the Gibb's free energy per mole (G) is calculated as follows;

\Delta G = V(P_2-P_1) \\\\

where;

V is the volume of the ice

Density = \frac{Mass}{Volume} \\\\Volume = \frac{Mass}{Density} \\\\Volume = \frac{18 \times 10^{-3} \ kg}{917 \ m^3} \\\\Volume = 1.96 \times 10^{-5} \ m^3\\\\Volume = 1.96 \times 10^{-5} \ m^3 \times \frac{1000 \ L}{m^3} \\\\Volume = 0.0196 \ L

Change in pressure;

P_2 - P_1 = 2.0 \ bar \ - \ 1.0 \ bar = 1.0 \ bar = 0.987 \ atm

The change in the Gibb's free energy per mole (G);

\Delta G= V(P_2-P_1)\\\\\Delta G = 0.0196\ L \times 0.987\ atm \\\\\Delta G = 0.0193 \ L.atm\\\\1 \ L.atm = 101.325 \ J\\\\\Delta G =  0.0193 \ L.atm \times \frac{101.325 \ J}{1 \ L.atm} \\\\\Delta G = 1.96 \ J

Thus, the change in the Gibb's free energy per mole (G) is 1.96 J.

Learn more about Gibb's free energy here: brainly.com/question/10012881

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\\ \sf\longmapsto E_P=25(100)

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