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saveliy_v [14]
3 years ago
8

A sphere has surface area 1.25 m2, emissivity 1.0, and temperature 100.0°C. What is the rate at which it radiates heat into empt

y space? The Stefan-Boltzmann constant is 5.67 × 10-8 W/(m2 ∙ K4).
A.) 9.9 mW

B.) 3.7 W

C.) 7.1 W

D.) 1.4 kW
Physics
1 answer:
Yuri [45]3 years ago
6 0
The total power emitted by an object via radiation is:
P=A\epsilon \sigma T^4
where:
A is the surface of the object (in our problem, A=1.25 m^2
\epsilon is the emissivity of the object (in our problem, \epsilon=1)
\sigma = 5.67 \cdot 10^{-8} W/(m^2 K^4) is the Stefan-Boltzmann constant
T is the absolute temperature of the object, which in our case is T=100^{\circ} C=373 K

Substituting these values, we find the power emitted by radiation:
P=(1.25 m^2)(1.0)(5.67 \cdot 10^{-8}W/(m^2K^4)})(373 K)^4=1371 W = 1.4 kW
So, the correct answer is D.
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iren [92.7K]

Answer:

372.3 J/^{\circ}C

Explanation:

First of all, we need to calculate the total energy supplied to the calorimeter.

We know that:

V = 3.6 V is the voltage applied

I = 2.6 A is the current

So, the power delivered is

P=VI=(3.6)(2.6)=9.36 W

Then, this power is delivered for a time of

t = 350 s

Therefore, the energy supplied is

E=Pt=(9.36)(350)=3276 J

Finally, the change in temperature of an object is related to the energy supplied by

E=C\Delta T

where in this problem:

E = 3276 J is the energy supplied

C is the heat capacity of the object

\Delta T =29.1^{\circ}-20.3^{\circ}=8.8^{\circ}C is the change in temperature

Solving for C, we find:

C=\frac{E}{\Delta T}=\frac{3276}{8.8}=372.3 J/^{\circ}C

5 0
3 years ago
Help me with this question Please
zysi [14]

Answer:

W has the lowest density and Y has the greatest density

Explanation:

Density of W = mass/volume = 11/24 = 0.45

Density of X = mass/volume = 11/12 = 0.91

Density of Y = m/v = 5.5/4 = 1.375

Density of Z = m/v = 5.5/11 = 0.5

From these we can find the answer......

Hope this answer is useful......

3 0
3 years ago
A 13-kg sled is moving at a speed of 3.0 m/s. At which of the following speeds will the sled have twice as much kinetic energy?
AysviL [449]
K.E. = 1/2 mv²
K.E. is directly proportional to v^2
So, when K.E. increase by 2, K.E. increase by root. 2
v' = 1.41v
original v value was 3 so, final would be:
v' = 1.41*3 = 4.23
After round-off to it's tenth value, it will be:
v' = 4.2

So, option B is your answer!

Hope this helps!

7 0
3 years ago
On a distant planet, a rock falls in 48.4 s from the top of a 1.10e+02 m cliff to the planet surface below. What is the accelera
pav-90 [236]

this can be solve using the formala of free fall

t = sqrt( 2y/ g)

where t is the time of fall

y is the height

g is the acceleration due to gravity

48.4 s = sqrt (2 (1.10e+02 m)/ g)

G = 0.0930 m/s2

The velocity at impact

V = sqrt(2gy)

= sqrt( 2 ( 0.0930 m/s2)( 1.10e+02 m)

V = 4.523 m/s

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8 0
4 years ago
What are the characteristics of high energy waves?
sladkih [1.3K]
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4 0
3 years ago
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