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irinina [24]
4 years ago
9

At what net rate does heat radiate from a 300-m^2 black roof on a night when the roof's temperature is 33.0°C and the surroundin

g temperature is 18.0°C? The emissivity of the roof is 0.900. (Enter the magnitude.)
Physics
1 answer:
Fynjy0 [20]4 years ago
7 0

Answer:

24445.85 J/s

Explanation:

Area, A = 300 m^2

T = 33° C = 33 + 273 = 306 k

To = 18° C = 18 + 273 = 291 k

emissivity, e = 0.9

Use the Stefan's Boltzman law

E = \sigma  \times e \times A\times\left ( T^4 -T_{0}^{4}\right )

Where, e be the energy radiated per unit time, σ be the Stefan's constant, e be the emissivity, T be the temperature of the body and To be the absolute temperature of surroundings.

The value of Stefan's constant, σ = 5.67 x 10^-8 W/m^2k^4

By substituting the values

E = 5.64 \times 10^{-8}\times 0.9 \times 300 \times  (306^{4}-291^{4})

E = 24445.85 J/s

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A particle moves in a straight line and has acceleration given by a(t) = 12t + 10. Its initial velocity is v(0) = −5 cm/s and it
soldi70 [24.7K]

Answer:

The position function is s_{t}=2t^3+5t^2-5t+9.

Explanation:

Given that,

Acceleration a =12t+10

Initial velocity v_{0} = -5\ cm/s

Initial displacement s_{0}=9\ cm

We know that,

The acceleration is the rate of change of velocity of the particle.

a = \dfrac{dv}{dt}

The velocity is the rate of change of position of the particle

v=\dfrac{dx}{dt}

We need to calculate the the position

The acceleration is

a_{t} = 12t+10

\dfrac{dv}{dt} = 12t+10

a_{t}=dv=(12t+10)dt

On integration both side

\int{dv}=\int{(12t+10)}dt

v_{t}=6t^2+10t+C

At t = 0

v_{0}=0+0+C

C=-5

Now, On integration again both side

v_{t}=\int{ds_{t}}=\int{(6t^2+10t-5)}dt

s_{t}=2t^{3}+5t^2-5t+C

At t = 0

s_{0}=0+0+0+C

C=9

s_{t}=2t^3+5t^2-5t+9

Hence, The position function is s_{t}=2t^3+5t^2-5t+9.

7 0
3 years ago
A tennis ball is dropped from 1.43 m above the
Rudiy27

Answer:

-5.29 m/s

Explanation:

Given:

y₀ = 1.43 m

y = 0 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: v

v² = v₀² + 2a (y − y₀)

v² = (0 m/s)² + 2(-9.8 m/s²) (0 m − 1.43 m)

v = -5.29 m/s

4 0
4 years ago
The metric system is now known as the international system of units.<br> a. T<br> b. F
inn [45]
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6 0
3 years ago
A 42.6kg lamp is hanging from wires as shown in figure.The ring has negligible mass. Find tensionsT1, T2,T3 if the object is in
IRISSAK [1]

Answer:

T1 = 417.48N

T2 = 361.54N

T3 = 208.74N

Explanation:

Using the sin rule to fine the tension in the strings;

Given

amass = 42.6kg

Weight = 42.6 * 9.8 = 417.48N

The third angle will be 180-(60+30)= 90 degrees

Using the sine rule

W/Sin 90 = T3/sin 30 = T2/sin 60

Get T3;

W/Sin 90 = T3/sin 30

417.48/1 = T3/sin30

T3 = 417.48sin30

T3 = 417.48(0.5)

T3 = 208.74N

Also;

W/sin90 = T2/sin 60

417.48/1 = T2/sin60

T2 = 417.48sin60

T2 = 417.48(0.8660)

T2 = 361.54N

The Tension T1 = Weight of the object = 417.48N

8 0
3 years ago
Photochemical smog is the result of nitrogen and carbon, often from car exhaust, reacting
BabaBlast [244]

Answer:

g

Explanation:

3 0
3 years ago
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