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3 years ago

Quando e como surgiu a ginastica rítmica no mundo e no Brasil?

1 answer:

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A ginástica rítmica no Brasil começou a ser praticada na década de 1960. Teve várias denominações, entre elas foi chamada de ginástica moderna e ginástica rítmica moderna e, sendo praticada essencialmente por mulheres, passou a ser chamada de ginástica feminina moderna. Depois, por decisão da Federação Internacional de Ginástica, passou à denominação de ginástica rítmica desportiva e, finalmente, ginástica rítmica.

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What does this same experiment tell you about light waves? Explain the evidence that supports your claim.

Oliga [24]

Answer:

Light wave is an EM wave that can only be seen by humans New questions in Physics Engineers at the Space Centre must determine the net force needed for a rockets engine to achieve an acceleration of 70 m/s2.

Explanation:

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3 years ago

A force of 385 N is applied in pushing a stalled automobile at a constant speed for a distance of 150 m. How much work (in J) wa

Helga [31]

Answer:57,750J

Explanation:

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2 years ago

Which wavelength produces fluorescence? Why do you think this wavelength produces fluorescence while the other does not?

Maurinko [17]

Answer:

Long wavelength

Explanation:

Wavelengths that corresponds to the bands of blue and red are strongly absorbed whereas the wavelengths that lie in the mid-range corresponds to green light that are absorbed weakly.

Fluorescence produced is always directed towards longer wavelengths of the spectra as compared to the corresponding spectra for absorption.

4 0

3 years ago

You obtain a 100-W light bulb and a 50-W light bulb. Instead of connecting them in the normal way, you devise a circuit that pla

lesantik [10]

Answer:

When they are connected in series

The 50 W bulb glow more than the 100 W bulb

Explanation:

From the question we are told that

The power rating of the first bulb is $P_1 = 100 \ W$

The power rating of the second bulb is $P_2 = 50 \ W$

Generally the power rating of the first bulb is mathematically represented as

$P_1 = V^2 R$

Where $V$ is the normal household voltage which is constant for both bulbs

$R_1 = \frac{V^2}{P_1 }$

substituting values

$R_1 = \frac{V^2}{100}$

Thus the resistance of the second bulb would be evaluated as

$R_2 = \frac{V^2}{50}$

From the above calculation we see that

$R_2 > R_1$

This power rating of the first bulb can also be represented mathematically as

$P_ 1 = I^2_1 R_1$

This power rating of the first bulb can also be represented mathematically as

$P_ 2 = I^2_2 R_2$

Now given that they are connected in series which implies that the same current flow through them so

$I_1^2 = I_2^2$

This means that

$P \ \alpha \ R$

So when they are connected in series

$P_2 > P_1$

This means that the 50 W bulb glows more than the 100 \ W bulb

3 0

3 years ago

Select the correct answer.

Stels [109]

Answer:

An electric bell is placed inside a transparent glass jar. The bell can be turned on and off using a switch on the outside of the jar. A vacuum is created inside the jar by sucking out the air. Then the bell is rung using the switch. What will we see and hear?

We’ll see the bell move, but we won’t hear it ring.

We won’t see the bell move, but we’ll hear it ring.

We’ll see the bell move and hear it ring.

We won’t see the bell move or hear it ring.

We’ll see the sound waves exit the vacuum pump.

Explanation:

so, the answer to the question is

We'll see the bell move, but we won’t hear it ring.

5 0

3 years ago

Read 2 more answers