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adoni [48]
2 years ago
12

A wall (of thermal conductivity 0.98 W/m · ◦ C) of a building has dimensions of 3.7 m by 15 m. The average inside and outside

temperatures are, respectively, 12â—¦C and 8â—¦C. If the heat transfer rate through the wall is 1540 W, what is the thickness of the wall in centimeters? Answer in units of cm. ________
Physics
1 answer:
disa [49]2 years ago
4 0

Answer:

Answer:

the amount of energy flowing is 1.008x10⁹J

Explanation:

To calculate how much heat flows, the expression is the following:

Where

K=thermal conductivity=0.81W/m°C

A=area=6.2*12=74.4m²

ΔT=30-8=22°C

L=thickness=8cm=0.08m

t=time=16.9h=60840s

Replacing:

Explanation:

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When testing a home for radon, where is the most likely place for the highest detected level? A) the basement, due to ground see
german

Answer:

A

Explanation:

ive been buying loui i been shoppin

5 0
3 years ago
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1.) I’m bored, so I decide to play catch by myself. I throw a 1.5kg ball in the air at 25 m/s. How long do I have to wait to cat
navik [9.2K]

Answer:

technically yes

Explanation:

with a gun depending on how fast it shoots so when you fire at something you shoot in front of it a little bit so you hit it but a plane that fast you shoot like 100 feet infront of it...

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3 years ago
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At an altitude of 5000 m the rocket's acceleration has increased to 6.9 m/s2 . What mass of fuel has it burned?
sergey [27]

1) Initial upward acceleration: 6.0 m/s^2

2) Mass of burned fuel: 0.10\cdot 10^4 kg

Explanation:

1)

There are two forces acting on the rocket at the beginning:

- The force of gravity, of magnitude F_g = mg, in the downward direction, where

m=1.9\cdot 10^4 kg is the rocket's mass

g=9.8 m/s^2 is the acceleration of gravity

- The thrust of the motor, T, in the upward direction, of magnitude

T=3.0\cdot 10^5 N

According to Newton's second law of motion, the net force on the rocket must be equal to the product between its mass and its acceleration, so we can write:

T-mg=ma (1)

where a is the acceleration of the rocket.

Solving for a, we find the initial acceleration:

a=\frac{T-mg}{m}=\frac{3.0\cdot 10^5-(1.9\cdot 10^4)(9.8)}{1.9\cdot 10^4}=6.0 m/s^2

2)

When the rocket reaches an altitude of 5000 m, its acceleration has increased to

a'=6.9 m/s^2

The reason for this increase is that the mass of the rocket has decreased, because the rocket has burned some fuel.

We can therefore rewrite eq.(1) as

T-m'g=m'a'

where

m' is the new mass of the rocket

Re-arranging the equation and solving for m', we find

m'=\frac{T}{g+a}=\frac{3.0\cdot 10^5}{9.8+6.9}=1.8\cdot 10^4 kg

And since the initial mass of the rocket was

m=1.9 \cdot 10^4 kg

This means that the mass of fuel burned is

\Delta m = m-m'=1.9\cdot 10^4 - 1.80\cdot 10^4 = 0.10\cdot 10^4 kg

3 0
3 years ago
Robert has just bought a new model rocket, and is trying to measure its flight characteristics. The rocket engine package claims
777dan777 [17]

Answer:

The work done by the drag force is given by 29.96 J

Explanation:

Given :

Thrust force F = 12.3 N

Displacement d = 10.2 m

Mass of rocket m = 0.663 Kg

From work energy theorem,

  W = \Delta K

 W_{t} - Wd - W_{g} = KE

Where W_{t} = thrust work W_{g} = gravitational work

KE = 12.3 \times 10.2 -Wd - 0.663 \times 9.8 \times 10.2

KE = 59.2 -Wd

After cutoff kinetic energy is converted into potential energy,

KE = Wd' + mg\Delta h

Put value of KE

59.2 -Wd = Wd' + 0.663 \times 9.8 \times 4.5

Work done by drag force is given by,

Wd'+Wd =  59.2 -29.23

                 = 29.96 J

Therefore, the work done by the drag force is given by 29.96 J

5 0
3 years ago
NASA launches a probe with a mass of 15,000 kg to another planet more massive than Earth. Which statement is true about the prob
vivado [14]

Answer: The weight of the probe will increase.

8 0
3 years ago
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