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3 years ago

Fill in this pretty little blank.

Physics

2 answers:

IgorLugansk [536]3 years ago

6 0

Freaky girl , it from my

Trava [24]3 years ago

3 0

Answer:

freaky girl

it from my

Explanation:

lol

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Use the periodic table to answer the question

True [87]

Answer: I hope this helps you

Explanation:

8 0

3 years ago

Read 2 more answers

A satellite orbits the Earth in an elliptical orbit. At perigee its distance from the center of the Earth is 22500 km and its sp

aleksley [76]

Answer:

$6.09294\times 10^{24}\ kg$

Explanation:

K = Kinetic energy

$v_p$ = Perigee speed = 4280 m/s

$v_a$ = Apogee speed = 3990 m/s

$r_p$ = Perigee Distance = 22500000 m

$r_a$ = Apogee Distance = 24100000 m

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

M = Mass of Earth

m = Mass of satellite

In this system the kinetic and potential energies are conserved

$K_p+P_p=K_a+P_a\\\Rightarrow \frac{1}{2}mv_p^2-\frac{GMm}{r_p}=\frac{1}{2}mv_a^2-\frac{GMm}{r_a}\\\Rightarrow \frac{1}{2}m(v_p^2-v_a^2)+GMm\left(\frac{1}{r_a}-\frac{1}{r_p}\right)=0\\\Rightarrow M=\frac{v_a^2-v_p^2}{2G}\times \left(\frac{1}{r_a}-\frac{1}{r_p}\right)^{-1}\\\Rightarrow M=\frac{3990^2-4280^2}{2\times 6.67\times 10^{-11}}\times \left(\frac{1}{24100000}-\frac{1}{22500000}\right)^{-1}\\\Rightarrow M=6.09294\times 10^{24}\ kg$

The mass of the Earth is $6.09294\times 10^{24}\ kg$

3 0

4 years ago

a. What is the peak wavelength for AM0? What temperature T corresponds to this peak wavelength for a blackbody source? Assuming

atroni [7]

Answer:

A) T = 5510 K , B) I = 5,226 10⁷ W / m² , C) I₂ = 1128 W / m²

Explanation:

A) The mass of air is defined as the ratio between the shortest path that sunlight must pass to reach the planet's surface and the length of the beam, is called AM

Am = 1 / cos θ

The AM0 value corresponds to solar radiation in the outer part of the Earth's atmosphere.

The peak of this emission is the peak that emitted from the sun

λ = 526 nm

To find the temperature that corresponds to this emission we use the Wien displacement law

λ T = 2,898 10⁻³

T = 2,898 10⁻³ / 526 10⁻⁹

T = 5510 K

i) The radiance on the surface of the sun is

I = P / A

We can calculate the potency by Stefan's law, for a black body

P = σ A e T⁴

P / A = σ e T⁴

The σ constant is value 5,670 10⁻⁸ W / m²K⁴, we will assume that the Sun emits as a black body, so e = 1

I = sig T⁴

I = 5,670 10⁻⁸ 5510⁴

I = 5,226 10⁷ W / m²

ii) the irradiation at a distance of 1 ua (1,496 1011 m)

Let's use the relationship

P = I A

I₁ A₁ = I₂ A₂

I₂ = I₁ A₁ / A₂

The area of a sphere is

A = 4π R²

Let's replace

I₂ = I₁ (r₁ / r₂)²

Index 1 corresponds to the sun and the index to Earth that is an astronomical unit

r₁ = 6.96 10⁸m (Sun radius)

r₂ = 1,498 1011 m (Earth-Sun distance)

Calculous

I₂ = 5,226 10⁷ (6.96 10⁸ / 1,498 10¹¹)²

I₂ = 1.1281 10³ W / m²

I₂ = 1128 W / m²

8 0

3 years ago

Precision

Schach [20]

Answer:

how close a true value a measurement is

6 0

3 years ago

Read 2 more answers

A 0.140-kg baseball is dropped from rest from a height of 1.8 m above the ground. It rebounds to a height of 1.4 m. What change

aliina [53]

Answer:

$\Delta p=-1.56\ kg-m/s$

Explanation:

It is given that,

Mass of the baseball, m = 0.14 kg

It is dropped form a height of 1.8 m above the ground. Let u is the velocity when it hits the ground. Using the conservation of energy as :

$u=\sqrt{2gh}$

h = 1.8 m

$u=\sqrt{2\times 9.8\times 1.8}$

u = 5.93 m/s

Let v is the speed of the ball when it rebounds. Again using the conservation of energy to find it :

$v=\sqrt{2gh'}$

h' = 1.4 m

$v=-\sqrt{2\times 9.8\times 1.4}$

v = -5.23 m/s

The change in the momentum of the ball is given by :

$\Delta p=m(v-u)$

$\Delta p=0.14(-5.23-5.93)$

$\Delta p=-1.56\ kg-m/s$

So, the change in the ball's momentum occurs when the ball hits the ground is 1.56 kg-m/s. Hence, this is the required solution.

4 0

4 years ago