To convert minutes per second into kilometre per hour we multiply the speed with

Galileo's observational contributionsGalileo Galilei was the first scientist to perform experiments in order to test his ideas.

ozzi

<h2>Answer: </h2><h2>- Jupiter has orbiting moons.</h2><h2>- The Sun has sunspots and rotates on its axis.</h2><h2>- The Moon has mountains, valleys, and craters.</h2><h2>- Venus goes through a full set of phases.</h2>

Explanation:

In 1609 Galileo built a telescope, with which he observed mountains and craters on the Moon, discovered Jupiter’s major satellites and the next year he published these discoveries in his book <em>The Sidereal Messenger</em>.

In addition, Galileo observed that Venus presented phases (such as those of the moon) together with a variation in size; observations that are only compatible with the fact that Venus rotates around the Sun and not around Earth. This is because <u>Venus presented its smaller size when it was in full phase and the largest size when it was in the new one, when it is between the Sun and the Earth. </u>

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On the other hand, <u>although Galileo was not the first to observe sunspots</u>, he gave the correct explanation of their existence, which supported the idea that planets revolve around the Sun.

These observations and discoveries were presented by Galileo to the Catholic Church (which supported the geocentric theory at that time) as a proof that completely refuted Ptolemy's geocentric system and affirmed Copernicus' heliocentric theory.

4 0

3 years ago

The Z0 boson, discovered in 1985, is the mediator of the weak nuclear force, and it typically decays very quickly. Its average r

Dvinal [7]

Answer:

The lifetime of the particle is $\Delta t = 2.6*10^{-25} \ s$

Explanation:

From the question we are told that

The average rest energy is $E = 91.19 \ GeV = 91.19GeV * \frac{1.60 *10^{-10} J }{1GeV} = 1.46 *10^{-8}J$

The intrinsic width is $\Delta E =2.5eV = 2.5GeV * \frac{1.60 *10^{-10}J }{1GeV} = 4*10^{-10} J$

The lifetime is mathematically represented as

$\Delta t = \frac{h}{\Delta E}$

Where h is the Planck's constant with a value of $1.055*10^{-34} \ J\cdot s$

substituting values

$\Delta t = \frac{1.055*10^{-34}}{4 *10^{-10}}$

$\Delta t = 2.6*10^{-25} \ s$

6 0

2 years ago

2. A 1750 kg car accelerates at a rate of 4.0 m/s2. How much force is the cars engine producing?

8090 [49]

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 1750 × 4

We have the final answer as

Hope this helps you

6 0

2 years ago

When a physical change in a sample occurs what does not change

Lelu [443]

Answer:

When a physical change in a sample occurs, composition of the sample does not change. It stays the same. It stays the same. Also, the properties of the sample will still be the same

5 0

2 years ago

An extremely long wire laying parallel to the x -axis and passing through the origin carries a current of 250A running in the po

viktelen [127]

Answer:

$1.232\times 10^{-5}\ T.$

Explanation:

<u>Given:</u>

Current through the wire, passing through the origin, $I_1 = 250\ A.$

Current through the wire, passing through the y axis, $r_y=1.8\ m.$ , $I_2 = 50\ A.$

According to Ampere's circuital law, the line integral of magnetic field over a closed loop, called Amperian loop, is equal to $\mu_o$ times the net current threading the loop.

$\oint \vec B \cdot d\vec l=\mu_o I.$

In case of a circular loop, the directions of magnetic field and the line element $d\vec l$ , both are along the tangent of the loop at that point, therefore, $\vec B\cdot d\vec l = B\ dl$ .

$\oint \vec B \cdot d\vec l = \oint B\ dl = B\oint dl.\\$

$\oint dl$ is the circumference of the Amperian loop = $2\pi r$

Therefore,

$B\ 2\pi r=\mu_o I\\B=\dfrac{\mu_o I}{2\pi r}.$

It is the magnetic field due to a current carrying wire at a distance r from it.

For the first wire, passing through the origin:

Consider an Amperian loop of radius 3.510 m, concentric with the axis of the wire, such that it passes through the point where magnetic field is to be found, therefore, $r_1 = 3.510\ m.$

The magnetic field at the given point due to this wire is given by:

$B_1 = \dfrac{\mu_o I_1}{2\pi r_1}\\=\dfrac{4\pi \times 10^{-7}\times 250}{2\pi \times 3.510}=1.42\times 10^{-5}\ T.$

For the first wire, passing through the y-axis:

Consider an Amperian loop of radius (3.510+1.8) m = 5.310 m, concentric with the axis of the wire, such that it passes through the point where magnetic field is to be found, therefore, $r_2 = 5.310\ m.$

The magnetic field at the given point due to this wire is given by:

$B_2 = \dfrac{\mu_o I_2}{2\pi r_2}\\=\dfrac{4\pi \times 10^{-7}\times 50}{2\pi \times 5.310}=1.88\times 10^{-6}\ T.$

The directions of current in both the wires are opposite therefore, the directions of the magnetic field due to both the wires are also opposite to that of each other.

Thus, the net magnetic field at $r_r=-3.510\ m$ is given by

$B=B_1-B_2 = 1.42\times 10^{-5}-1.88\times 10^{-6}=1.232\times 10^{-5}\ T.$

6 0

2 years ago

To convert minutes per second into kilometre per hour we multiply the speed with​

To convert minutes per second into kilometre per hour we multiply the speed with