3 years ago

When was the copernican treatise published

Physics

1 answer:

LiRa [457]3 years ago

5 0

They were published in 1542.

You might be interested in

The main part of the nuclear power plant is the ____. Where nuclear ____ reactions take place.

emmasim [6.3K]

Turbine an unclean fission

7 0

3 years ago

Read 2 more answers

What is the kinetic energy of an object that has a mass of 12 kilograms and moves with a velocity of 10 m/s?

Alik [6]

1/2mv^2

1/2x12x10^2=600J

The kinetic energy is 600J

3 0

2 years ago

Read 2 more answers

Please help ill give brainly and heart !!!

Snezhnost [94]

Answer:

July

Explanation:

In july you would expect least hurricane

4 0

2 years ago

Read 2 more answers

We know today that atoms cannot be divided into smaller parts true or false

djverab [1.8K]

Hi , the answer is false ,atoms can be divided into smaller parts , electrons , protons and neutrons.

3 0

2 years ago

If rho(x,y) is the density of a wire (mass per unit length), then

lidiya [134]

Answer:

See description

Explanation:

With the given information we have:

$x(t) = 1 + cos(t)\\ y(t)=sin(t)\\ \rho(x,y) = 3x$

the interval is $[0,\pi ]$

now the mass $m$ has the given expression:

$m = \int \rho(x,y) dS$

we will use the formula for a line integral and let:

$dS=\sqrt{x'(t)^2 + y'(t)^2}=\sqrt{cos(t)^2 + sin(t)^2}dt=dt$

therefore we have:

$m=\int \rho(x,y)dS=\int\limits^\pi_0 {3*x}dS=\int\limits^\pi _0{3*(1+cos(t))dS\\=\int\limits^\pi _0{3*(1+cos(t))dt$

we solve the integral:

$m=3*\int\limits^\pi _0{(1+cos(t))dt= 3*(t+sin(t))\limits^\pi _0=3*\pi=9.42$

7 0

2 years ago