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astraxan [27]
2 years ago
10

A rabbit is hopping along at an approximately constant speed of 3.9 m/s. The rabbit passes a crouched cat ready to chase the rab

bit. The cat takes off the instant the rabbit passes the cat, accelerating at 0.5 m/s2. How long does it take the cat to catch the rabbit
Physics
1 answer:
Firlakuza [10]2 years ago
4 0

Answer:

t  = 7,8 s

Explanation:

From the instant, the rabbit passes the cat. The cat star running acceleration of 0,5 m/s² .

When the cat arrives at the speed of 3,9 m/s the cat catches the rabbit

Then for the cat arrives at 3,9 m/s nedds

v = vo + a*t     vo  = 0  then   v = a*t

3,9 ( m/s) = 0,5 ( m/s² ) * t

t  = 7,8 s

v  =  3,9 m/s =

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C. molecules speed up as more thermal energy is added

Explanation:

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If the number of particles is increased in a balloon what happens to the pressure inside of it?
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The air pressure inside the balloon increases as the number of particles increases.
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Multiply the following three numbers and report your answer to the correct number of significant figures: 0.020cm x 50cm x 11.1c
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Answer:11.1

Explanation:

Three significant figures

5 0
3 years ago
Of the following transitions in the Bohr hydrogen atom, the ________ transition results in the absorption of the highest-energy
8_murik_8 [283]

Answer:

The <em><u>n = 2 → n = 3</u></em> transition results in the absorption of the highest-energy photon.

Explanation:

E_n=-13.6\times \frac{Z^2}{n^2}ev

Formula used for the radius of the n^{th} orbit will be,

where,

E_n = energy of n^{th} orbit

n = number of orbit

Z = atomic number

Here: Z = 1 (hydrogen atom)

Energy of the first orbit in H atom .

E_1=-13.6\times \frac{Z^2}{1^2} eV=-13.6 eV

Energy of the second orbit in H atom .

E_2=-13.6\times \frac{Z^2}{(2)^2} eV=-3.40 eV

Energy of the third orbit in H atom .

E_3=-13.6\times \frac{Z^2}{(9)^2} eV=-1.51 eV

Energy of the fifth orbit in H atom .

E_5=-13.6\times \frac{Z^2}{(2)^2} eV=-0.544 eV

Energy of the sixth orbit in H atom .

E_6=-13.6\times \frac{Z^2}{(2)^2} eV=-0.378 eV

Energy of the seventh orbit in H atom .

E_7=-13.6\times \frac{Z^2}{(2)^2} eV=-278 eV

During an absorption of energy electron jumps from lower state to higher state.So,  absorption will take place in :

1) n = 2 → n = 3

2) n=  5 → n = 6

Energy absorbed when: n = 2 → n = 3

E=E_3-E_2

E=(-1.51 eV) -(-3.40 eV)=1.89 eV

Energy absorbed when: n = 5 → n = 6

E'=E_6-E_5

E'=(-0.378 eV)-(-0.544 eV) =0.166 eV

1.89 eV > 0.166 eV

E> E'

So,the n = 2 → n = 3 transition results in the absorption of the highest-energy photon.

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3 years ago
To get maximum current in a circuit, the resistance should be in _____
Tanzania [10]

Answer:

Series is the correct answer

5 0
2 years ago
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