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NNADVOKAT [17]
3 years ago
8

What is the total resistance in this circuit?

Physics
1 answer:
saul85 [17]3 years ago
6 0

Answer:B

Explanation:

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A body which has surface area 5cm² and temperature of 727°C radiates 300J of energy in one minute. Calculate it's emissivity giv
cestrela7 [59]
<h2>Answer: 0.17</h2>

Explanation:

The Stefan-Boltzmann law establishes that a black body (an ideal body that absorbs or emits all the radiation that incides on it) "emits thermal radiation with a total hemispheric emissive power proportional to the fourth power of its temperature":  

P=\sigma A T^{4} (1)  

Where:  

P=300J/min=5J/s=5W is the energy radiated by a blackbody radiator per second, per unit area (in Watts). Knowing 1W=\frac{1Joule}{second}=1\frac{J}{s}

\sigma=5.6703(10)^{-8}\frac{W}{m^{2} K^{4}} is the Stefan-Boltzmann's constant.  

A=5cm^{2}=0.0005m^{2} is the Surface area of the body  

T=727\°C=1000.15K is the effective temperature of the body (its surface absolute temperature) in Kelvin.

However, there is no ideal black body (ideal radiator) although the radiation of stars like our Sun is quite close.  So, in the case of this body, we will use the Stefan-Boltzmann law for real radiator bodies:

P=\sigma A \epsilon T^{4} (2)  

Where \epsilon is the body's emissivity

(the value we want to find)

Isolating \epsilon from (2):

\epsilon=\frac{P}{\sigma A T^{4}} (3)  

Solving:

\epsilon=\frac{5W}{(5.6703(10)^{-8}\frac{W}{m^{2} K^{4}})(0.0005m^{2})(1000.15K)^{4}} (4)  

Finally:

\epsilon=0.17 (5)  This is the body's emissivity

3 0
3 years ago
A satellite omass1000 kg moves in a circular orbit of radius 8000 km round the earth,assumed to be a sphere of radius 6400 km. C
lubasha [3.4K]

Answer:

ΔE = 37.8 x 10^9 J

Explanation:

The energy required will increased the potential energy and increase the kinetic energy.

As the altitude change is fairly small compared to the earth radius, we can ASSUME that the average gravity will be a good representative

Gravity acceleration at altitude would be 9.8(6400²/8000²) = 6.272 m/s²

G(avg) = (9.8 + 6.272)/2 = 8.036 m/s²

ΔPE = mG(avg)Δh = 1000(8.036)(8e6 - 6.4e6) = 12.857e9 J

The centripetal force at orbit must be equal to the gravity force

mv²/R = mg'

v²/8.0e6 = 6.272

v² = (6.272(8.0e6)) = 50.2e6 m²/s²

The maximum velocity when resting on earth at the equator is about 460 m/s.

The change in kinetic energy is

ΔKE = ½m(vf² - vi²)(1000)

ΔKE = ½(1000)(50.2e6 - 460²) = 25e9 J

Total energy increase is

25e9 + 12.857e9 = 37.8e9 J

3 0
3 years ago
The largest hailstone every measured fell in Vivian, Nebraska in 2010. The circumference of that hailstone was 19 inches. Using
Crazy boy [7]

Answer:

6.04788 in

115.82654\ in^3

78.38779 m/s

0.88159 kg

34.55294 J

Explanation:

Circumference is given by

c=2\pi r\\\Rightarrow r=\dfrac{c}{2\pi}\\\Rightarrow r=\dfrac{19}{2\pi}\\\Rightarrow r=3.02394\ in

Diameter is given by

d=2r\\\Rightarrow d=2\times 3.02394\\\Rightarrow d=6.04788\ in

The diameter is 6.04788 in

6.04788\times 2.54=15.3616152\ cm

Volume of sphere is given by

v=\dfrac{4}{3}\pi r^3\\\Rightarrow v=\dfrac{4}{3}\pi 3.02394^3\\\Rightarrow v=115.82654\ in^3

The volume is 115.82654\ in^3

115.82654\times \dfrac{1}{1728}=0.06702\ ft^3

Fall velocity is given by

V=k\sqrt{d}\\\Rightarrow V=20\sqrt{15.3616152}\\\Rightarrow V=78.38779\ m/s

The velocity of the fall will be 78.38779 m/s

Mass is given by

m=\rho v\\\Rightarrow m=29\times 0.06702\\\Rightarrow m=1.94358\ lb

1.94358\ lb=1.94358\times \dfrac{1}{2.20462}=0.88159\ kg

The mass is 0.88159 kg

Kinetic energy is given by

K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}0.88159\times 78.38779\\\Rightarrow K=34.55294\ J

The kinetic energy is 34.55294 J

4 0
3 years ago
A resistor is connected in series with an AC source that provides a sinusoidal voltage of v of t is equal to V times cosine of b
nekit [7.7K]
<h2>Answer:</h2>

In circuits, the average power is defined as the average of the instantaneous power  over one period. The instantaneous power can be found as:

p(t)=v(t)i(t)

So the average power is:

P=\frac{1}{T}\intop_{0}^{T}p(t)dt

But:

v(t)=v_{m}cos(\omega t) \\ \\ i(t)=i_{m}cos(\omega t)

So:

P=\frac{1}{T}\intop_{0}^{T}v_{m}cos(\omega t)i_{m}cos(\omega t)dt \\ \\ P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}cos^{2}(\omega t)dt \\ \\ But: cos^{2}(\omega t)=\frac{1+cos(2\omega t)}{2}

P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}(\frac{1+cos(2\omega t)}{2} )dt \\\\P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}[\frac{1}{2}+\frac{cos(2\omega t)}{2}]dt \\\\P=\frac{v_{m}i_{m}}{T}[\frac{1}{2}(t)\right|_0^T +\frac{sin(2\omega t)}{4\omega} \right|_0^T] \\ \\ P=\frac{v_{m}i_{m}}{2T}[(t)\right|_0^T +\frac{sin(2\omega t)}{2\omega} \right|_0^T] \\ \\ P=\frac{v_{m}i_{m}}{2}

In terms of RMS values:

V_{RMS}=V=\frac{v_{m}}{\sqrt{2}} \\ \\ I_{RMS}=I=\frac{i_{m}}{\sqrt{2}} \\ \\ Then: \\ \\ P=VI

7 0
3 years ago
2NaOH+H2SO4 Na2SO4+H2O the equation is balanced/unbalanced because the number of hydrogen atoms and oxygen/sodium is equal/not e
RUDIKE [14]
Your equation is:
2NaOH + H_{2}SO_{4}  ==\ \textgreater \  Na_{2}SO_{4} + H_{2}O

An equation is balanced only if there are the same number of atoms of each element on both sides of the arrow - aka same number of atoms of each element in both reactants (left of the arrow) and products (right of the arrow). 

It'll be easiest to tackle this by counting up the number of atoms of each element on the left and on the right and comparing those numbers. If there is a number in front of the entire compound, that means that number applies to all elements in the compound. If the number is a subscript (little number to the right of the element), that means that number only applies to the element that the subscript is attached to:
1) On the left, you have:
2NaOH + H_{2}SO_{4}\\&#10;\\&#10;Na:  \: 2 \: atoms\\&#10;O: \: 2 + 4 = 6 \: atoms\\&#10;H: \: 2 + 2 = 4 \: atoms\\&#10;S:  \: 1 \: atom

2) On the right, you have:
Na_{2}SO_{4} + H_{2}O\\&#10;\\ &#10;Na: \: 2 \: atoms\\ &#10;O: \: 4 + 1 = 5 \: atoms\\ &#10;H: \: 2 \: atoms\\ &#10;S: \: 1 \: atom

You can see that the number of oxygen and hydrogen atoms aren't equal on both the left (reactants) and the right (products), so the equation is unbalanced.

Your final answer is "T<span>he equation is unbalanced because the number of hydrogen atoms and oxygen is not equal in the reactants and in the products."</span>


5 0
3 years ago
Read 2 more answers
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