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3 years ago

Galileo’s most important contribution to astronomy was his

Physics

2 answers:

olchik [2.2K]3 years ago

6 0

The answer is D I just took the test.

ikadub [295]3 years ago

3 0

When Galileo got ahold of a little telescope ... no more than
a kid's toy today ... he soon turned it towards the night sky.
The story says that he was the first one to do that. Apparently
everyone else who had these things used them to look at their
neighbors' windows. Anyway, When Galileo looked at Jupiter,
he saw four tiny dots. As he watched them night after night, he
saw that the four dots always stayed with Jupiter, but got nearer
and farther from it. He eventually realized that they were small
bodies in orbit around Jupiter. And then ... because he was Galileo ...
he made the big intellectual jump: He reasoned that if there were
smaller bodies in orbit around Jupiter, then Jupiter and the other
planets might also be in orbit around something, and not necessarily
around the Earth. He understood this as evidence that the sun is
the center of the solar system.

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A rectangular barge, 5.2 m long and 2.4 m wide, floats in fresh water. Suppose that a 410-kg crate of auto parts is loaded onto

sesenic [268]

<h2>Answer:</h2><h2>The depth of barge float=3 cm</h2><h2>Explanation:</h2>

Length of rectangular barge=5.2 m

Width of rectangular barge=2.4m

Mass of crate=410 kg

Let h be the height of barge float

Volume of barge float= $l\times b\times h=5.2\times 2.4\times h=12.48h$

Density of water= $10^3kg/m^3$

Weight of water displaced by barge=Buoyant force=-Weight of horse

$Volume\;of\;water\times density\;of\;water\times g=410\times g$

$12.48h\times 1000=410$

$h=\frac{410}{12.48\times 1000}=0.03 m$

1 m=100 cm

$0.03 m=0.03\times 100=3$ cm

Hence, the depth of barge float=3 cm

<h2 />

4 0

4 years ago

Read 2 more answers

Two masses are attracted by a gravitational force of 7 N.

damaskus [11]

Answer:

The answer to your question is: F = 0.4375 N. The force will be 16 times lower than with the first conditions.

Explanation:

Data

F = 7 N

F = ? if the masses is quartered

Formula

$F = \frac{Km1m2}{r2}$

Process

Normal conditions F = Km₁m₂/r² = 7

When masses quartered F = K(m₁/4)(m₂/4)/r² = ?

F = K(m₁m₂/16)/r²

F = K(m₁m₂/16r² = 7/16 = 0.4375 N

3 0

3 years ago

Except for the nodes on a standing wave, what is the frequency f of the points executing simple harmonic motion?

Katen [24]

Take into account that in a standing wave, the frequency f of the points executing simple harmonic motion, is simply a multiple of the fundamental harmonic fo, that is:

f = n·fo

where n is an integer and fo is the first harmonic or fundamental.

fo is given by the length L of a string, in the following way:

fo = v/λ = v/(L/2) = 2v/L

becasue in the fundamental harmonic, the length of th string coincides with one hal of the wavelength of the wave.

6 0

1 year ago

Before colliding, the momentum of block A is -100 kg*m/, and block B is -150 kg*m/s. After, block A has a momentum -200 kg*m/s.

rjkz [21]

Answer:

Momentum of block B after collision = $-50\ kg\ ms^{-1}$