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As scientists research, they often find information that does not fit with current theories. What happens when new, contradictor

timofeeve [1]

The correct answer is "C". 'Old theories are adjusted to incorporate all old new information.' This makes the most sense, regarded the old and new information should be taken into consideration.

I hope this helped you!

Brainliest answer is always appreciated!

8 0

2 years ago

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A sinusoidal wave traveling on a string has a period of 0.20 s, a wavelength of 32 cm, and an amplitude of 3 cm. The speed of th

Finger [1]

Answer:

$v = 1.6 \frac{m}{s} *\frac{100cm}{1m}= 160 \frac{cm}{s}$

Explanation:

If we have a periodic wave we need to satisfy the following basic relationship:

$v = \lambda f$

From the last formula we see that the velocity is proportional fo the frequency.

For this case we have the following info given by the problem:

$T= 0.2 s, \lambda =32 cm* \frac{1m}{100cm} =0.32 m, A= 3cm*\frac{1m}{100 cm}=0.03 m$

We know that the frequency is the reciprocal of the period so we have this formula:

$f = \frac{1}{T}$

And if we replace we got:

$f =\frac{1}{0.2 s}= 5Hz$

Now since we have the value for the wavelength we can find the velocity like this:

$v = 0.32 m * 5Hz = 1.6 \frac{m}{s}$

And if we convert this into cm/s we got:

$v = 1.6 \frac{m}{s} *\frac{100cm}{1m}= 160 \frac{cm}{s}$

6 0

2 years ago

A spring that has a spring constant of 1400 N/m is stretched to a length of 2.5 m. If the normal length of the spring is 1.0 m,

Elena L [17]

The answer is C. 1575 J. I just took this review.

5 0

3 years ago

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Simplified. The absorption spectra of ions have been used to identify the presence of the elements in the atmospheres of the sun

Nookie1986 [14]

Answer:

The answer is " $3.83 \times 10^9 \ m$ "

Explanation:

Z=2, so the equation is $E= \frac{-4B}{n^2}$

Calculate the value for E when:

n=2 and n=9

The energy is the difference in transformation, name the energy delta E Deduct these two energies

In this transition, the wavelength of the photon emitted is:

$\Delta E=2.18 \times 10^{-18} ( \frac{1}{4}- \frac{1}{81})$

$\lambda = \frac{h c}{\Delta E}$

$h ( Planck's\ constant) = 6.62 \times 10^{(-34)} \ Js \\\\ speed \ of \ light = 3 \times 10^{8} \ \frac{m}{s}\\\\= \frac{6.62 \times 10^{(-34)} \times 3 \times 10^ {8}}{2.18 \times 10^{-18}} (\frac{1}{4}- \frac{1}{81}) \\\\=3.83 \times 10^9 \ m\\\\$

6 0

2 years ago

A spaceship hovering over the surface of Venus drops an object from a height of 17 m. How much longer does it take to reach the

Paraphin [41]

1.96s and 1.86s. The time it takes to a spaceship hovering the surface of Venus to drop an object from a height of 17m is 1.96s, and the time it takes to the same spaceship hovering the surface of the Earth to drop and object from the same height is 1.86s.

In order to solve this problem, we are going to use the motion equation to calculate the time of flight of an object on Venus surface and the Earth. There is an equation of motion that relates the height as follow:

$h=v_{0} t+\frac{gt^{2}}{2}$

The initial velocity of the object before the dropping is 0, so we can reduce the equation to:

$h=\frac{gt^{2}}{2}$

We know the height h of the spaceship hovering, and the gravity of Venus is $g=8.87\frac{m}{s^{2}}$ . Substituting this values in the equation $h=\frac{gt^{2}}{2}$ :

$17m=\frac{8.87\frac{m}{s^{2} } t^{2}}{2}$

To calculate the time it takes to an object to reach the surface of Venus dropped by a spaceship hovering from a height of 17m, we have to clear t from the equation above, resulting:

$t=\sqrt{\frac{2(17m)}{8.87\frac{m}{s^{2} } }} =\sqrt{\frac{34m}{8.87\frac{m}{s^{2} } } }=1.96s$

Similarly, to calculate the time it takes to an object to reach the surface of the Earth dropped by a spaceship hovering from a height of 17m, and the gravity of the Earth $g=9.81\frac{m}{s^{2}}$ .

$t=\sqrt{\frac{2(17m)}{9.81\frac{m}{s^{2} } }} =\sqrt{\frac{34m}{9.81\frac{m}{s^{2} } } }=1.86s$

8 0

2 years ago

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PLEASE HELPPP ASAP, I'LL GIVE BRAINLEST​

PLEASE HELPPP ASAP, I'LL GIVE BRAINLEST