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ryzh [129]
3 years ago
14

The box part of the box plot contains all the values between which numbers?

Mathematics
1 answer:
Valentin [98]3 years ago
5 0

Answer:c between 32 and 37

Step-by-step explanation:

if you look at where the box starts and where it ends you can find the numbers that you need

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Marcus is making homemade barbecue sauce.The recipe caps for 3 cups of ketchup for every 1/2 cup of mustard
SSSSS [86.1K]

Answer:

3 divided  by 1/2

Step-by-step explanation:

turn 3 into 3/1  then flip 1/2 into 2/1 and multiply which is <u><em>6</em></u>

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3 0
3 years ago
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What is<br><img src="https://tex.z-dn.net/?f=1895%20%5Ctimes%20906" id="TexFormula1" title="1895 \times 906" alt="1895 \times 90
netineya [11]

Answer:

1895 x 906=

1,716,870

4 0
3 years ago
Need help with a math question
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Using PYTHAGORAS THEOREM answer the question A vertical pillar ab bent at c at a height of 2.4m and its upper end b touches the
Alisiya [41]

Answer:

  5.4 m

Step-by-step explanation:

The Pythagorean theorem tells you the relationship between the various lengths is ...

  cd² = ac² +ad²

  cd² = (2.4 m)² + (1.8 m)² = 9.00 m²

  cd = √(9.00 m²) = 3.0 m

Since cd = cb and ab = ac+cb, we have ...

  ab = 2.4 m + 3.0 m = 5.4 m

_____

You may recognize this as a 3-4-5 right triangle with a scale factor of 0.6. We are asked for the sum of the lengths of the "4" and "5" legs, which is ...

  0.6·(4+5) = 5.4

7 0
3 years ago
A light bulb of 200W emits 1.5μm.How many photons are emmited per second.​
Vinvika [58]

Answer:

Step-by-step explanation:

An incandescent light bulb filament is approximated by a black body radiator, and in the case of a 60W bulb the filament temperature is around 2500˚C which is 2870˚K.

There are a couple of standard black body results that we can use. Firstly the total irradiance emitted per unit area of black body is equal to:

Ed=σT4

where σ=5.67×10−8 Wm−2K−4 is the Stephan-Boltzmann constant. For our bulb we get:

Ed=5.67×10−8⋅28704=3.85×106 Wm−2

As this is a 60W bulb then it has a total irradiance of 60W. Therefore the equivalent black body surface area is:

603.85×106=1.56×10−5 m2

which is 15.6mm2 of filament area.

Secondly we have that the total number of photons emitted per second per unit area of black body is equal to:[1]

Qd=σQT3

where σQ=1.52×1015 photons.sec−1m−2K−3. For our bulb this is:

Qd=1.52×1015⋅28703=3.59×1025 photons.sec−1m−2

Multiplying by the bulb’s equivalent black body surface area gives the result we require:

3.59×1025⋅1.56×10−5=5.6×1020 photons/sec

As a sanity check we know that these photons have a total energy of 60 joules per second, so the average energy per photon is:

605.6×1020=1.1×10−19 joules

A photon of wavelength λ has energy E=hcλ, and so the average energy corresponds with a photon of wavelength:

λ=hc1.1×10−19=1.9μm

Here’s a chart of the power distribution by wavelength, with the average photon wavelength shown as the dashed line, and visible wavelengths highlighted:

emember that there are a higher proportion of photons for the longer, lower energy wavelengths, so the average is weighted to the right.

We can also see from the original calculation that the general case is:

QdWEd=σQT3WσT4=2.68×1022WT photons/sec

for a bulb of wattage W watts and filament temperature T ˚K. So the photon emission rate is inversely proportional to the filament temperature. As a somewhat counter-intuitive example, a 60W halogen bulb with 3200˚K filament only emits photons at 90% of the rate of the standard 60W bulb, despite being visibly brighter.

The reason is that as the temperature increases then an increased proportion of shorter wavelength photos are emitted and therefore the average energy per photon increases, decreasing the number emitted per second. However at the same time an increased proportion of the photons are visible rather than infra-red, making the bulb appear brighter. Here’s the power distribution chart with the 60W halogen curve added for comparison:

3 0
3 years ago
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