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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

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 box is being pulled to the right. What is the direction of the gravitational force?

ioda

Answer:

The correct answer would be downwards

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As beverages are produced what action will increase the solubility of co2 in them most?

Alla [95]

Answer: D. decreasing the temperature

Explanation:

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3 years ago

Read 2 more answers

Spitting cobras can defend themselves by squeezing muscles around their venom glands to squirt venom at an attacker. Suppose a s

Alenkasestr [34]

Answer: 1.289 m

Explanation:

The path the cobra's venom follows since it is spitted until it hits the ground, is described by a parabola. Hence, the equations for parabolic motion (which has two components) can be applied to solve this problem:

<u>x-component: </u>

$x=V_{o}cos\theta t$ (1)

Where:

$x$ is the horizontal distance traveled by the venom

$V_{o}=3.10 m/s$ is the venom's initial speed

$\theta=47\°$ is the angle

$t$ is the time since the venom is spitted until it hits the ground

<u>y-component: </u>

$y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=0.44 m$ is the initial height of the venom

$y=0$ is the final height of the venom (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity

Let's begin with (2) to find the time it takes the complete path:

$0=0.44 m+3.10 m/s sin\theta(47\°)+\frac{-9.8m/s^{2} t^{2}}{2}$ (3)

Rewritting (3):

$-4.9 m/s^{2} t^{2} + 2.267 m/s t + 0.44 m=0$ (4)

This is a quadratic equation (also called equation of the second degree) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ (5)

Where:

$a=-4.9 m/s^{2$

$b=2.267 m/s$

$c=0.44 m$

Substituting the known values:

$t=\frac{-2.267 \pm \sqrt{2.267^{2}-4(-4.9)(0.44)}}{2(-4.9)}$ (6)

Solving (6) we find the positive result is:

$t=0.609 s$ (7)

Substituting (7) in (1):

$x=(3.10 m/s)cos(47\°)(0.609 s)$ (8)

We finally find the horizontal distance traveled by the venom:

$x=1.289 m$

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3 years ago

Explain how to correctly add vectors in 2-D

Simora [160]

To add vectors we can use the head to tail method (Figure 1).
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2 years ago

If the crate shown here is moving at a constant speed in a straight line and the force applied is 310 N, what is the magnitude o

larisa86 [58]

Answer:

$f_k = 310N$

the answer is A.

Explanation:

Using the laws of newton:

∑F = ma

where ∑F is the sumatory of forces acting in the system, m the mass and a the acelertion of the system.

Then, if the block is moving with constant velocity, its aceleration is equal to 0, so:

∑F = m(0)

∑F = 0

It means that:

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where F is the force applied and $f_k$ is the friction force. Replacing the value of F, we get:

310N - $f_k$ = 0

Finally, solving for $f_k$ :

$f_k = 310N$

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3 years ago