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Ilia_Sergeevich [38]
3 years ago
7

When the motion of one or both of the particles is at an angle to the line of impact, the impact is said to be ________

Physics
1 answer:
Nana76 [90]3 years ago
7 0

Answer: Oblique impact

Explanation:

When the motion of one or both of the particles is at an angle to the line of impact, the impact is said to be oblique impact.

On the other hand, when the directions of motion of the two colliding particles are moving along a line of impact, then it's refered to as central impact.

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Stars of spectral type A and F are considered ________.
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<u>B. the stars of spectral type A and F are considered reasonably to have habitable planets but much less likely to have planets with complex plant - or animal - like life.</u>

Explanation:

The appropriate spectral range for habitable stars is considered to be "late F" or "G", to "mid-K" or even late "A". <em>This corresponds to temperatures of a little more than 7,000 K down to a little less than 4,000 K</em> (6,700 °C to 3,700 °C); the Sun, a G2 star at 5,777 K, is well within these bounds. "Middle-class" stars (late A, late F, G , mid K )of this sort have a number of characteristics considered important to planetary habitability:

• They live at least a few billion years, allowing life a chance to evolve. <em>More luminous main-sequence stars of the "O", "B", and "A" classes usually live less than a billion years and in exceptional cases less than 10 million.</em>

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• They emit sufficient radiation at wavelengths conducive to photosynthesis.

• Liquid water may exist on the surface of planets orbiting them at a distance that does not induce tidal locking.

<u><em>Thus , the stars of spectral type A and F are considered reasonably to have habitable planets but much less likely to have planets with complex plant - or animak - like life.</em></u>

4 0
3 years ago
How many words can be made with five letters
dangina [55]
The last word is gear
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What happens when magnesium reacts with fluorine
slamgirl [31]
It creates a white crystal salt  MgF2
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3 years ago
The Ha line of the Balmer series is emitted in the transition from n-3 to n 2. Compute the wavelength of this line for H and 2H.
Citrus2011 [14]

Explanation:

According to Rydberg's formula, the wavelength of the balmer series is given by:

\frac{1}{\lambda}=R(\frac{1}{2^2}-\frac{1}{3^2})

R is Rydberg constant for an especific hydrogen-like atom, we may calculate R for hydrogen and deuterium atoms from:

R=\frac{R_{\infty}}{(1+\frac{m_e}{M})}

Here, R_{\infty} is the "general" Rydberg constant, m_e is electron's mass and M is the mass of the atom nucleus

For hydrogen, we have, M=1.67*10^{-27}kg:

R_H=\frac{1.09737*10^7m^{-1}}{(1+\frac{9.11*10^{-31}kg}{1.67*10^{-27}kg})}\\R_H=1.09677*10^7m^{-1}

Now, we calculate the wavelength for hydrogen:

\frac{1}{\lambda}=R_H(\frac{1}{2^2}-\frac{1}{3^2})\\\lambda=[R_H(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=[1.0967*10^7m^{-1}(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=6.5646*10^{-7}m=656.46nm

For deuterium, we have M=2(1.67*10^{-27}kg):

R_D=\frac{1.09737*10^7m^{-1}}{(1+\frac{9.11*10^{-31}kg}{2*1.67*10^{-27}kg})}\\R_D=1.09707*10^7m^{-1}\\\\\lambda=[R_D(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=[1.09707*10^7m^{-1}(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=6.5629*10^{-7}=656.29nm

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3 years ago
A square sheet of rubber has sides that are 20 cm long. If you stretch it so that the area of the rubber is four times larger, h
oee [108]

Answer:

400

Explanation:

Trust me its correct

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3 years ago
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