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2 years ago

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A two-stage rocket is traveling at 1210m/s with respect to the earth when the first stage runs out of fuel. Explosive bolts rele

ase the first stage and push it backward with a speed of 40m/s relative to the second stage after the explosion. The first stage is three times as massive as the second stage.
What is the speed of the second stage after the separation?

1 answer:

d1i1m1o1n [39]2 years ago

7 0

Answer:

R

Explanation:

Given that,

Two stage rocket traveling at

V = 1210 m/s with respect to earth

First stage

When fuel Is run out , explosive bolts releases and push rocket backward at speed is

V1 = 40m/s relative to second stage

Therefore

V1 = 40 - V2

The first stage is 3 times as massive as the second stage

I.e Mass of first stage is 3 times the second stage

Let Mass of second stage be

M2 = M

Then, M1 = 3M

Velocity of second stage V2?

Applying conservation of linear momentum

Momentum before explosion = momentum after explosion

Momentum p=mv

Then,

(M1+M2)V = —M1•V1 + M2•V2

(3M+M)•1210 = —3M•(40-V2) +M•V2

4M × 1210 = —120M + 3M•V2 + MV2

4840M = —120M + 4M•V2

4840M + 120M = 4M•V2

4960M = 4M•V2

Then, V2 = 4960M / 4M

V2 = 1240 m/s

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: Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at

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Complete Question

Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at sea level. One container is in Denver at an altitude of about 6,000 ft and the other is in New Orleans (at sea level). The surface area of the container lid is A=0.0155 m. The air pressure in Denver is PD = 79000 Pa. and in New Orleans is PNo = 100250 Pa. Assume the lid is weightless.

Part (a) Write an expression for the force FNo required to remove the container lid in New Orleans.

Part (b) Calculate the force FNo required to lift off the container lid in New Orleans, in newtons.

Part (c) Calculate the force Fp required to lift off the container lid in Denver, in newtons.

Part (d) is more force required to lift the lid in Denver (higher altitude, lower pressure) or New Orleans (lower altitude, higher pressure)?

Answer:

a

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b

$F_{No}= 7771.125 \ N$

c

$F_p = 2.2*10^{6} N$

d

From the value obtained we can say the that the force required to open the lid is higher at Denver

Explanation:

The altitude of container in Denver is $d_D = 6000 \ ft = 6000 * 0.3048 = 1828.8m$

The surface area of the container lid is $A = 0.0155m^2$

The altitude of container in New Orleans is sea-level

The air pressure in Denver is $P_D = 79000 \ Pa$

The air pressure in new Orleans is $P_{ro} = 100250 \ Pa$

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$F_{No} = \Delta P A$

So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would

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$\Delta P = P_{No} - \frac{P_{sea}}{2}$

Therefore

$F_{No} = A [P_{No} - \frac{P_{sea}}{2}]$

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$F_{No} = 0.0155 [100250 - \frac{101000}{2}]$

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$F_p = \Delta P_d A$

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$\frac{P_D}{\rho h} = \frac{P_d_D}{\rho d_D}$

$P_d_D = P_D * d_D$

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$P_d__{D}} = 1828.2 * 79000$

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