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3 years ago

15

How much would an 80.0kg person weigh on Mars, where the acceleration of gravity is 3.93m/s and in Earths moon, where the accele

ration of gravity is 1.63m/s

1 answer:

LuckyWell [14K]3 years ago

7 0

Answer:

this is the answer

Explanation:

This is the solution, hope it helps

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Two objects of the same size are both perfect blackbodies. One has a temperature of 3000 K, so its frequency of maximum emission

bija089 [108]

Answer:

a) The colder body (3000k), b) hearter body c) 12000K body

Explanation:

This exercise should know the power emitted by the objects and the distribution of this emission in the energy spectrum, for this we will use Stefan's laws and that of Wien's displacement

Stefan's Law P = σ A e T⁴

Wien displacement law λ T = 2,898 10⁻³ m K

Let's calculate the power emitted for each object.

As they are perfect black bodies e = 1, they also indicate that they have the same area

T = 3000K

P₁ = σ A T₁⁴

T = 12000K

P₂ = σ A T₂⁴

P₂ / P₁ = T₂⁴ / T₁⁴

P₂ / P₁ = (12000/3000)⁴

P₂ / P₁ = 256

This indicates that the hottest body emission is 256 times the coldest body emission.

Let's calculate the maximum emission wavelength

Body 1

T = 3000K

λ T = 2,898 10-3

λ₁ = 2.89810-3 / T

λ₁ = 2,898 10-3 / 3000

λ₁ = 0.966 10-6 m

λ₁ = 966 nm

T = 12000K

λ₂ = 2,898 10-3 / 12000

λ₂ = 0.2415 10-6 m

λ₂ = 214 nm

a) The colder body (3000k) emits more light in the infrared, since the emission of the hot body is at a minimum (emission tail)

b) The two bodies have emission in this region, the body of 3000K in the part of rise of the emission and the body to 12000K in the descent of the emission even when this body emits 256 times more than the other, so this body should have the highest broadcast in this area

c) The emission of the hottest 12000K body is mainly in UV

d) The hottest body emits more energy in UV and visible

e) No body has greater emission in all zones

5 0

3 years ago

In a given system of units the ratio of the unit of volume to that of area gives the unit of

ch4aika [34]

Answer:

length

Explanation:

SI unit of volume = m^3

SI unit of area = m^2

volume unit / Area unit = m^3 / m^2

i.e, unit of length

8 0

3 years ago

A man pushing a mop across a floor causes it to undergo two displacements. The first has a magnitude of 152 om and makes an angl

aliya0001 [1]

Answer:

D₂= 167,21 cm : Magnitude of the second displacement

β= 21.8° , countercockwise from the positive x-axis: Direction of the second displacement

Explanation:

We find the x-y components for the given vectors:

i: unit vector in x direction

j:unit vector in y direction

D₁: Displacement Vector 1

D₂: Displacement Vector 2

R= resulta displacement vector

D₁= 152*cos110°(i)+152*sin110°(j)=-51.99i+142.83j

D₂= -D₂(i)-D₂(j)

R= 131*cos38°(i)+ 131*sin38°(j) = 103.23i+80.65j

We propose the vector equation for sum of vectors:

D₁+ D₂= R

-51.99i+142.83j+D₂x(i)-D₂y(j) = 103.23i+80.65j

-51.99i+D₂x(i)=103.23i

D₂x=103.23+51.99=155.22 cm

+142.83j-D₂y(j) =+80.65j

D₂y=142.83-80.65=62.18 cm

Magnitude and direction of the second displacement

$D_{2} =\sqrt{(D_{x})^{2} +(D_{y} )^{2} }$

$D_{2} =\sqrt{(155.22)^{2} +(62.18 )^{2} }$

D₂= 167.21 cm

Direction of the second displacement

$\beta = tan^{-1} \frac{D_{y}}{D_{x} }$

$\beta = tan^{-1} \frac{62.18}{155.22 }$

β= 21.8°

D₂= 167,21 cm : Magnitude of the second displacement

β= 21.8.° , countercockwise from the positive x-axis: Direction of the second displacement

6 0

3 years ago

21. Compare and contrast beta particles to the electrons that surround neutral atoms.

stiv31 [10]

Beta particles come from the nucleus. Electrons are found around the nucleus.
Beta particles normally travel very fast out of a nucleus in a straight line. Electrons normally orbit the nucleus of an atom.

They both have the same mass and the same charge.

6 0

3 years ago

A constant magnetic field passes through a single rectangular loop whose dimensions are 0.42 m x 0.54 m. The magnetic field has

olga_2 [115]

Answer:

$0.29887\ \text{V}$

Explanation:

A = Area = $0.42\times0.54\ \text{m}^2$

B = Magnetic field = 1.7 T

$\theta$ = Angle that magnetic field makes with the plane of the loop = $71^{\circ}$

t = Time = 0.42 s

EMF is given by

$\varepsilon=A\cos\theta\dfrac{dB}{dt}\\ =\dfrac{0.42\times 0.54\times\cos 71^{\circ}\times 1.7}{0.42}\\ =0.29887\ \text{V}$

The magnitude of the average emf induced in the loop is $0.29887\ \text{V}$ .

8 0

3 years ago