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Georgia [21]
3 years ago
9

For an object like a planet, with a typical temperature of a few hundred kelvin, what kind of blackbody radiation would it princ

ipally emit
Physics
1 answer:
navik [9.2K]3 years ago
6 0

Answer:

Low-temperature blackbody

Explanation:

There are 3 types of blackbody temperatures.

Low-temperature blackbody

High temperature extended area blackbody

High-temperature cavity blackbody

A Low-temperature blackbody is a type of black body radiation that has the range of -40° C to 175° C, typically between 233 K and 448 K. A perfect fit for the temperature range mentioned in the question, "a few hundred Kelvin". Therefore, it's the kind of blackbody temperature that the object would emit.

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A plane is flying east when it drops some supplies to a designated target below. The supplies land after falling for 10 seconds.
Diano4ka-milaya [45]

solution:

As Given plane is flying in east direction.

It throws back some supplies to designated target.

Time taken by the supply to reach the target =10 seconds

g = Acceleration due to gravity = - 9.8 m/s²[Taken negative as object is falling Downwards]

As we have to find distance from the ground to plane which is given by d.

d = \frac{1}{2}\times g\times t^2

 = \frac{1}{2}\times (9.8) \times(100) =50\times 9.8=490 meters

Distance from the ground where supplies has to be land  to plane  =  Option B =490 meters

6 0
3 years ago
Read 2 more answers
An inventor claims to have invented a heat engine that receives 750kJ of heat from a source at 400K and produces 250kJ of net wo
IRISSAK [1]

Answer:

the claim is not valid or reasonable.

Explanation:

In order to test the claim we will find the maximum and actual efficiencies. maximum efficiency of a heat engine can be found as:

η(max) = 1 - T₁/T₂

where,

η(max) = maximum efficiency = ?

T₁ = Sink Temperature = 300 K

T₂ = Source Temperature = 400 K

Therefore,

η(max) = 1 - 300 K/400 K

η(max) = 0.25 = 25%

Now, we calculate the actual frequency of the engine:

η = W/Q

where,

W = Net Work = 250 KJ

Q = Heat Received = 750 KJ

Therefore,

η = 250 KJ/750 KJ

η = 0.333 = 33.3 %

η > η(max)

The actual efficiency of a heat engine can never be greater than its Carnot efficiency or the maximum efficiency.

<u>Therefore, the claim is not valid or reasonable.</u>

3 0
3 years ago
________ is the average kinetic energy of each atom.<br><br> radiation<br> temperature<br> potential
solong [7]
I think it's temperature
7 0
3 years ago
A helicopter is ascending vertically with a speed of 5.10m/s. At a height of 105m above the Earth, a package is dropped from a w
valkas [14]

Op here is another problem exactly like that. Just plug in your variables instead. And remember, time is never negative.

5 0
3 years ago
You and your friend live in different states and plant identical hydrangea bulbs in your garden. When they bloom, you send each
Whitepunk [10]

Answer:

sample. rewrite in your own words if you'd like.

Explanation:

Since the hydrangeas are changing color depending on whether the soil is acidic or basic, they must have an indicator. This means that any hydrangea planted in acidic soil will have blue flowers and any hydrangea planted in basic soil will have pink. One common indicator present in many plants is anthocyanin, so maybe hydrangeas contain that chemical.

8 0
3 years ago
Read 2 more answers
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