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Xelga [282]
3 years ago
5

A total Δν of 15 km/s is required to achieve an interplanetary mission. The proposed rocket has two stages. The first stage alon

e has a mass of 1000 tonnes and consists of 10 solid rocket boosters and a liquid rocket motor. The liquid rocket has a specific impulse of 300 s and a fuel mass flow rate of 1500 kg/s. The solid rocket boosters have a specific impulse of 250 s and a fuel mass flow rate of 200 kg/s. The first stage burn time is 1 minute. The first stage is jettisoned immediately after burn-out. The second stage has an initial mass of 150 tonnes. The second stage motor provides an effective exhaust velocity of 4 km/s and has a structural coefficient of 0.05. The rocket carries a 2 tonne payload. a. What is the jettisoned mass of the first stage?
Physics
1 answer:
AlexFokin [52]3 years ago
3 0

Answer:

102000 kg

Explanation:

Given:

A total Δν = 15 km/s

first stage mass = 1000 tonnes

specific impulse of liquid rocket =  300 s

Mass flow rate of liquid fuel = 1500 kg/s

specific impulse of solid fuel = 250 s

Mass flow of solid fuel = 200 kg/s

First stage burn time = 1 minute = 1 × 60 seconds = 60 seconds

Now,

Mass flow of liquid fuel in 1 minute = Mass flow rate × Burn time

or

Mass flow of liquid fuel in 1 minute = 1500 × 60 = 90000 kg

Also,

Mass flow of solid fuel in 1 minute = Mass flow rate × Burn time

or

Mass flow of solid fuel in 1 minute = 200 × 60 = 12000 kg

Therefore,

The total jettisoned mass flow of the fuel in first stage

= 90000 kg +  12000 kg

= 102000 kg

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An object is placed at O ona number line. It moves 3 units to the right, then 4 units to the left, and then 6 units to
kvv77 [185]

Answer:

You have a displacement of 5 units to the right.

Explanation:

First you go three to the right which lands on the 3 mark. Then you move it 4 to the left which substracts 4, landing the object at -1. Finally you move 6 to the right, and you finish at marker 5. Since displacement is not total distance but just final distance from the start point directly to end point, it is only a displacement of 5.

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3 years ago
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The following table lists the work functions of a few common metals, measured in electron volts. Metal Φ(eV) Cesium 1.9 Potassiu
Citrus2011 [14]

A. Lithium

The equation for the photoelectric effect is:

E=\phi + K

where

E=\frac{hc}{\lambda} is the energy of the incident light, with h being the Planck constant, c being the speed of light, and \lambda being the wavelength

\phi is the work function of the metal (the minimum energy needed to extract one photoelectron from the surface of the metal)

K is the maximum kinetic energy of the photoelectron

In this problem, we have

\lambda=190 nm=1.9\cdot 10^{-7}m, so the energy of the incident light is

E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.9\cdot 10^{-7} m}=1.05\cdot 10^{-18}J

Converting in electronvolts,

E=\frac{1.05\cdot 10^{-18}J}{1.6\cdot 10^{-19} J/eV}=6.5 eV

Since the electrons are emitted from the surface with a maximum kinetic energy of

K = 4.0 eV

The work function of this metal is

\phi = E-K=6.5 eV-4.0 eV=2.5 eV

So, the metal is Lithium.

B. cesium, potassium, sodium

The wavelength of green light is

\lambda=510 nm=5.1\cdot 10^{-7} m

So its energy is

E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5.1\cdot 10^{-7} m}=3.9\cdot 10^{-19}J

Converting in electronvolts,

E=\frac{3.9\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.4 eV

So, all the metals that have work function smaller than this value will be able to emit photoelectrons, so:

Cesium

Potassium

Sodium

C. 4.9 eV

In this case, we have

- Copper work function: \phi = 4.5 eV

- Maximum kinetic energy of the emitted electrons: K = 2.7 eV

So, the energy of the incident light is

E=\phi+K=4.5 eV+2.7 eV=7.2 eV

Then the copper is replaced with sodium, which has work function of

\phi = 2.3 eV

So, if the same light shine on sodium, then the maximum kinetic energy of the emitted electrons will be

K=E-\phi = 7.2 eV-2.3 eV=4.9 eV

7 0
4 years ago
A tennis ball was thrown straight up with initial velocity of 22.5 m/s. How high does the ball rise
kiruha [24]

Answer:

h_m=25.8\ m

Explanation:

<u>Vertical Launch Upwards</u>

In a vertical launch upwards, an object is launched vertically up without taking into consideration friction with the air.

If vo is the initial speed and g is the acceleration of gravity, the maximum height reached by the object is given by:

\displaystyle h_m=\frac{v_o^2}{2g}

The tennis ball was thrown straight up with a speed of v0=22.5 m/s. The acceleration of gravity is g=9.81\ m/s^2, thus:

\displaystyle h_m=\frac{22.5^2}{2\cdot 9.81}

\mathbf{h_m=25.8\ m}

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sveticcg [70]
False, it is not the theory
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A balloon is rising vertically upwards at a velocity of 10m/s. When it is at a height of 45m from the ground, a parachute bails
harina [27]

(a) 30.9 m

Let's analyze the motion of the parachutist. Its vertical position above the ground is given by

y=h+ut+\frac{1}{2}gt^2

where

h = 45 m is the initial height

u = 10 m/s is the initial velocity (upward)

t is the time

g = -9.8 m/s^2 is the acceleration of gravity (downward)

Substituting t=3 s , we find the height of the parachutist when it opens the parachute:

y=45 m+(10 m/s)(3 s)+\frac{1}{2}(-9.8 m/s^2)(3 s)^2=30.9 m

(b) 44.1 m

Here we have to find first the height of the balloon 3 seconds after the parachutist has jumped off from it. The vertical position of the balloon is given by

y = h + ut

where

h = 45 m is the initial height

u = 10 m/s is the initial velocity (upward)

t is the time

Substituting t = 3 s, we find

y = 45 m + (10 m/s)(3 s) = 75 m

So the distance between the balloon and the parachutist after 3 s is

d = 75 m - 30.9 m = 44.1 m

(c) 8.2 m/s downward

The velocity of the parachutist at the moment he opens the parachute is:

v = u +gt

where

u = 10 m/s is the initial velocity (upward)

t is the time

g = -9.8 m/s^2 is the acceleration of gravity (downward)

Substituting t = 3 s,

v = 10 m/s + (-9.8 m/s^2)(3 s)= -19.4 m/s

where the negative sign means it is downward

After t=3 s, the parachutist open the parachute and it starts moving with a deceleration of

a =+5 m/s^2

where we put a positive sign since this time the acceleration is upward.

The total distance he still has to cover till the ground is

d = 30.9 m

So we can find the final velocity by using

v^2-u^2 = 2ad

where this time we have u = 19.4 m/s as initial velocity. Taking the downward direction as positive, the deceleration must be considered as negative:

a = -5 m/s^2

Solving for v,

v=\sqrt{u^2 +2ad}=\sqrt{(19.4 m/s)^2+2(-5 m/s^2)(30.9 m)}=8.2 m/s

(d) 5.24 s

We can find the duration of the second part of the motion of the parachutist (after he has opened the parachute) by using

a=\frac{v-u}{t}

where

a = -5 m/s^2 is the deceleration

v = 8.2 m/s is the final velocity

u = 19.4 m/s is the initial velocity

t is the time

Solving for t, we find

t=\frac{v-u}{a}=\frac{8.2 m/s-19.4 m/s}{-5 m/s^2}=2.24 s

And added to the 3 seconds between the instant of the jump and the moment he opens the parachute, the total time is

t = 3 s + 2.24 s = 5.24 s

8 0
3 years ago
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