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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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The following diagram represents a cart with an initial velocity of 1.0 m/s sliding along a frictionless track from point A
ki77a [65]

Answer:

point b

Explanation:

7 0
3 years ago
How deep is the outer core beneath the surface
sukhopar [10]
Precisely around 1,800 miles below.
6 0
3 years ago
A 40-W lightbulb is 1.7 m from a screen. What is the intensity of light incident on the screen? Assume that a lightbulb emits ra
Sonja [21]

Answer:

Intensity, I=1.101\ W/m^2

Explanation:

Power of the light bulb, P  = 40 W

Distance from screen, r = 1.7 m

Let I is the intensity of light incident on the screen. The power acting per unit area is called the intensity of the light. Its formula is given by :

I=\dfrac{P}{A}

I=\dfrac{P}{4\pi r^2}

I=\dfrac{40\ W}{4\pi (1.7\ m)^2}

I=1.101\ W/m^2

So, the intensity of light is 1.101\ W/m^2.

6 0
3 years ago
A 20 ft ladder leans against a wall. The bottom of the ladder is 3 ft from the wall at time t=0 and slides away from the wall at
stellarik [79]

Answer: 0.516 ft/s

Explanation:

Given

Length of ladder L=20 ft

The speed at which the ladder moving away is v=2 ft/s

after 1 sec, the ladder is 5 ft away from the wall

So, the other end of the ladder is at

\Rightarrow y=\sqrt{20^2-5^2}=19.36\ ft

Also, at any instant t

\Rightarrow l^2=x^2+y^2

differentiate w.r.t.

\Rightarrow 0=2xv+2yv_y\\\\\Rightarrow v_y=-\dfrac{x}{y}\times v\\\\\Rightarrow v_y=-\dfrac{5}{19.36}\times 2=0.516\ ft/s

5 0
3 years ago
A 1-kg rock is suspended by a massless string from one end of a
maxonik [38]

Answer:

The weight of measuring stick is 9.8 N

Explanation:

given information:

the mass of the rock, m_{r} = 1 kg

measuring stick, x =1 m

d = 0.25 m

to find the weight of measuring stick, we can use the following equation:

τ = Fd

τ = 0

F_{r} d - F_{s}d = 0

F_{r} = the force of the rock

F_{s} = the force of measuring stick

F_{s} =F_{r}

    = m g

    = 1 kg x 9.8 m/s

    = 9.8 N

thus, the weight of measuring stick is 9.8 N

6 0
3 years ago
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