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bazaltina [42]
1 year ago
15

A special rocket can produce 7.66 ✕ 10^5 N of instantaneous thrust with an exhaust speed of 3.05 ✕ 103 m/s in vacuum. What mass

of fuel does the engine burn each second to produce this thrust? (Enter your answer in kg/s.)
Physics
1 answer:
adelina 88 [10]1 year ago
8 0

The mass of fuel the engine burn each second to produce a thrust of 7.66×10⁵ N is 2.5×10² kg/s.

<h3 /><h3>What is mass?</h3>

Mass can be defined as the quantity of matter contained in a body. The S.I unit of mass is kilogram(kg)

To calculate the mass the engine burns each seconds, we use the formula below.

Formual:

  • M = T/v............. Equation

Where:

  • M = Mass per seconds of the rocket
  • T = Thrust
  • v = Velocity

From the question,

Given:

  • T = 7.66×10⁵ N
  • v = 3.05×10³ m/s

Substitute these values into equation 1

  • M = (7.66×10⁵)/(3.05×10³)
  • M = 2.5×10² kg/s

Hence, the mass of fuel burned in each second is 2.5×10² kg/s.

Learn more about mass here: brainly.com/question/25121535

#SPJ1

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Heat is extracted from a certain quantity of steam at
vodomira [7]

Answer:v=2452.91 m/s

Explanation:

Given

initially steam is at 100^{\circ}C and converted to 0^{\circ} C ice

Let m be the mass of steam

latent heat of fusion and vaporization for water is

L_f=3.33\times 10^5 J/kg

L_v=2.26\times 10^6 J/kg

Heat required to convert steam in to water at 100^{\circ}C

Q_1=m\times L_v=m\cdot 2.26\times 10^6 J

Heat required to lower water temperature to 0^{\circ}C

Q_2=m\times c\times \Delta T

Q_2=m\times 4.184\times (100)

Q_2=4.184m\times 10^5 J

Heat required to convert 0^{\circ}C water to ice at 0^{\circ}C is

Q_3=m\times L_f

Q_3=m\times 3.33\times 10^5=3.33m\times 10^5 J

Q=Q_1+Q_2+Q_3

Q=(2.26+0.4184+0.33)m\times 10^6 J

Q=3.0084m\times 10^6 J

So this energy is equal to kinetic energy of  bullet of mass m moving with velocity v

Q=\frac{1}{2}mv^2

3.0084m\times 10^6=\frac{1}{2}mv^2

v^2=3.0084\times 2\times 10^6

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5 0
3 years ago
By what factor is the heat flow increased if a window 0.550 mm on a side is inserted in the door? The glass is 0.450 cmcm, and t
Andrei [34K]

This question is incomplete, the complete question is;

A carpenter builds a solid wood door with dimensions 1.95 m × 0.99 m × 4.5 cm . Its thermal conductivity is k=0.120W/(m.K). The air films on the inner and outer surfaces of the door have the same combined thermal resistance as an additional 1.6 cm thickness of solid wood. The inside air temperature is 19.0°C , and the outside air temperature is -6.50°C .

a) What is the rate of heat flow through the door?

b) By what factor is the heat flow increased if a window 0.550 m on a side is inserted in the door? The glass is 0.450 cm , and the glass has a thermal conductivity of 0.80 W/(m.K). The air films on the two sides of the glass have a total thermal resistance that is the same as an additional 12.0 cm of glass.

Answer:

a) the rate of heat flow through the door is 97 watts

b) The factor of increased heat flow is 1.353

Explanation:

Given that;

room dimension = 1.95m × 0.99m × 4.5cm,

thermal conductivity = 0.120 w/m.k

additional thickness of solid wood Δt = 1.6 cm

a)

first we determine the effective thickness of the door;

t = 4.5cm + 1.6 cm = 61 cm ≈ 0.061 m

Now rate of heat flow is given by the relation

Q = KA( (TH -TC)/L)

= 0.12 × (1.95 m × 0.99 m) × ( (19°C - (-6.50°C)) / 0.061m)

= 0.23166 × 418.0327

= 96.8414 watts

Q = 97 watts

therefore the rate of heat flow through the door is 97 watts

b)

by intensity the glass of thickness 0.450 cm

the effective thickness is

L = 0.45cm + 12 cm = 12.45 cm = 0.1245 m

additionally area of glass A = (0.550 m)²

A = 0.3025 m²

Now

Qglass = KA ((TH-TC)/L)  

= 0.80 w/m.k × 0.3025 m² × (19°C - (-6.50°C)) / 0.1245m)

= 0.242 × 204.819

Qglass = 49.57 watt  

Qwood = KA ((TH-TC)/L)  

area of wooden door = (1.95×0.99) - 0.3025 m² = 1.628m²

so Qwood = 0.12 × 1.628 × (19°C - (-6.50°C)) / 0.061m)

= 0.19536 × 418.0327

Qwood = 81.67 watt

Q = Qglass + Qwood

Q = 49.57 watt  + 81.67 watt

Q = 131.24 watt

The factor of increased heat flow is;

f = 131.24 watt / 97 watts

f = 1.353

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