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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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kenneth warmed up a 100- ml sample of air in and expandable container. he displayed the resluts in the following graph. the resl

shepuryov [24]

Answer:

300k

Explanation:

3 0

3 years ago

A go-cart and rider have a mass of 14 kg. If the cart accelerates at 6m/s^2 during a 40 m sprint in 100 seconds, how much power

lara [203]

Answer:

P = 33.6 [N]

Explanation:

To solve this problem we must use Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

∑F = m*a

where:

F = forces [N]

m = mass = 14 [kg]

a = acceleration = 6 [m/s²]

$F = 14*6\\F = 84 [N]$

In the second part of this problem we must find the work done, where the work in physics is known as the product of force by distance, it is important to make it clear that force must be applied in the direction of movement.

$W = F*d$

where:

W = work [J]

F = force = 84 [N]

d = displaciment = 40 [m]

$W = 84*40\\W = 3360 [J]$

Finally, the power can be calculated by the relationship between the work performed in a given time interval.

$P=W/t\\$

where:

P = power [W]

W = work = 3360 [J]

t = time = 100 [s]

Now replacing:

$P=3360/100\\P=33.6[W]$

The power is given in watts

3 0

3 years ago

A 3-m-high, 7-m-wide rectangular gate is hinged at the top edge and is restrained by a fixed ridge. Determine the hydrostatic fo

Shalnov [3]

Answer:

The Hydrostatic force is $F = 137.2 kN$

The location of pressure center is $Z = 1.333 \ m$

Explanation:

From the question we are told that

The height of the gate is $h = 3 \ m$

The weight of the gate is $w = 7 \ m$

The height of the water is $h_w = 2 \ m$

The density of water is $\rho_w = 1000 \ kg/m^3$

Note used $h_w$ for height of water and height of gate immersed by water since both have the same value

The area of the gate immersed in water is mathematically represented as

$A = h_w * w$

substituting values

$A = 2* 7$

$A = 14 \ m^2$

The hydrostatic force is mathematically represented as

$F = \rho_w * g * h_f * A$

Where

$h_f =h- h_w$

$h_f =3 -2$

$h_f = 1\ m$

So

$F = 1000 * 9.8 * 1 * 14$

$F = 137.2 kN$

The center of pressure is mathematically represented as

$Z = h_f + \frac{I_g}{h_f * A}$

Where $I_g$ is the moment of inertia of the gate which mathematically represented as

$I_g = \frac{w * h_w^2}{12}$

The $h_w$ is the height of gate immersed in water

$I_g = \frac{7 * 2^2 }{12}$

$I_g = 4.667\ kg m^2$

Thus

$Z = 1 + \frac{4.66}{1 * 14}$

$Z = 1.333 \ m$

3 0

3 years ago

Which scientist described an atom made of a solid positively charged substance with electrons dispersed throughout it?.

swat32

The scientist that described an atom made a solid positively charged substance with electrons dispersed throughout it was: Ernest Rutherford

In 1911 Ernest Rutherford proposed his atomic model in which he considered the atom as a positively, densely charged center called a nucleus in which the electrons circulate around the core with a negative charge.

<h3>What is an atom?</h3>

The atom is the smallest part of the composition of matter, it is indivisible and is composed of a nucleus that has protons and neutrons, and around the nucleus there are the electrons.

Learn more about the atom at: brainly.com/question/17545314

#SPJ4

3 0

2 years ago

You are traveling on an interstate highway at the posted speed limit of 70 mph when you see that the traffic in front of you has

vovikov84 [41]

Answer:8.75 s,

136.89 m

Explanation:

Given

Initial velocity $=70 mph\approx 31.29 m/s$

velocity after 5 s is $30 mph\approx 13.41 m/s$

Therefore acceleration during these 5 s

$a=\frac{v-u}{t}$

$a=\frac{13.41-31.29}{5}=-3.576 m/s^2$

therefore time required to stop

v=u+at

here v=final velocity =0 m/s

initial velocity =31.29 m/s

$0=31.29-3.576\times t$

$t=\frac{31.29}{3.576}=8.75 s$

(b)total distance traveled before stoppage

$v^2-u^2=2as$

$0^2-31.29^2=2\times (-3.576)\cdot s$

s=136.89 m

3 0

3 years ago