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3 years ago

13

daca corpuri de masa m1=5kg si m2=3kg , se deplaseaza unul spre altul cu aceeasi viteza. Daca in urma ciocnirii plastice se dega

ja o cantitate de caldura de 10 j ,aflati viteza de deplasare a corpului

1 answer:

sergeinik [125]3 years ago

3 0

Answer:

Speed of two blocks is given as 1.63 m/s

Explanation:

Two blocks of given masses are moving towards each other with same speeds

so here we can use momentum conservation for this collision

$m_1v - m_2v = (m_1 + m_2) v_f$

so final speed after collision is given as

$v_f = \frac{5v - 3v}{5 + 3}$

$v_f = \frac{v}{4}$

now we have loss of energy given as

$\Delta E = \frac{1}{2}m_1v^2 + \frac{1}{2}m_2v^2 - \frac{1}{2}(m_1 + m_2)(\frac{v}{4})^2$

$10 = 4 v^2 - 4\frac{v^2}{16}$

$10 = \frac{15}{4} v^2$

$v = 1.63 m/s$

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Meeta used an elastic tape to measure the length of her window to stitch a curtain. Do you think she will be able to stitch a cu

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Answer:

No

Explanation:

She will not be able to measure the length of her window accurately due to instrumental error from her choice of instrument. The elastic nature of her tape would alter the measurement because it will stretch as she is taking her readings, thus reducing the true measurement of the length of her window.

To measure the length of her window, she could use an inelastic tape rule or a metre rule. These instruments would eliminate instrumental error.

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3 years ago

In nature, where do fusion reactions take place?

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Fusion reactions happen in stars

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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 .

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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

A circular coil of radius r = 5 cm and resistance R = 0.2 is placed in a uniform magnetic field perpendicular to the plane of th

Yuri [45]

Answer:

the question is incomplete, the complete question is

"A circular coil of radius r = 5 cm and resistance R = 0.2 ? is placed in a uniform magnetic field perpendicular to the plane of the coil. The magnitude of the field changes with time according to B = 0.5 e^-t T. What is the magnitude of the current induced in the coil at the time t = 2 s?"

2.6mA

Explanation:

we need to determine the emf induced in the coil and y applying ohm's law we determine the current induced.

using the formula be low,

$E=-\frac{d}{dt}(BACOS\alpha )\\$

where B is the magnitude of the field and A is the area of the circular coil.

First, let determine the area using $\pi r^{2} \\$ where r is the radius of 5cm or 0.05m

$A=\pi *(0.05)^{2}\\ A=0.00785m^{2}\\$

since we no that the angle is at $0^{0}$

we determine the magnitude of the magnetic filed

$B=0.5e^{-t} \\t=2s$

$E=-(0.5e^{-2} * 0.00785)$

$E=-0.000532v\\$

the Magnitude of the voltage is 0.000532V

Next we determine the current using ohm's law

$V=IR\\R=0.2\\I=\frac{0.000532}{0.2} \\I=0.0026A$

$I=2.6mA$

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