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3 years ago

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Two cannonballs are dropped from a second-floor physics lab at height h above the ground. Ball B has four times the mass of ball

A. When the balls pass the bottom of a first-floor window at height above the ground, the relation between their kinetic energies, KA and KB, is:_______.A) KA- 4KB B) KA 2KB C) KA KB D) KB 4KA.

1 answer:

frez [133]3 years ago

6 0

Formula of kinetic energy

$e = \frac{1}{2} m {v}^{2}$

Therefore Kinetic energy of ball B is 4 times more than ball A.

ans is KB=4KA

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For atomic hydrogen, the Paschen series of lines occurs when nf = 3, whereas the Brackett series occurs when nf = 4 in the equat

zvonat [6]

Answer:

$(\lambda_{max} )_{brackett} < (\lambda_{min} )_{paschen}$

So the two wavelength range will over lap

Explanation:

The Rydberg equation is given by

$\frac{1}{\lambda} =\frac{2\pi mk^2e^4}{h^3c} t (\frac{1}{n^2_f}-\frac{1}{n^2_i} )$

m is the mass of electron

k = 1/4π∈₀

∈₀ = is the permitivity of free space

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c is the speed of light in vaccum

z is the atomic number = 1

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where R is the Rydberg constant = 1.097373 × 10⁷m⁻¹

For Paschen series of H spectrum

$n_f = 3$

$n_i = 5,6,7 ...$

in Paschen series of H spectrum

The maximum wavelength occur for $n_i = 4$

$\frac{1}{\lambda_m_a_x } =(1.097373 \times 10^7)(\frac{1}{9} - \frac{1}{16} )\\\\\lambda_m_a_x=1874.6nm$

The minimum wavelength occur for $n_i$ = ∞

$\frac{1}{\lambda_m_i_n } =(1.097373 \times 10^7)(\frac{1}{9} - \frac{1}{_o_o} )\\\\\lambda_m_a_x=820.14nm$

The brackett series of H spectrum

The maximum wavelength occur for $n_i = 4$

$\frac{1}{\lambda_m_a_x } =(1.097373 \times 10^7)(\frac{1}{16} - \frac{1}{25} )\\\\\lambda_m_a_x=4050.05nm$

The minimum wavelength occur for $n_i$ = ∞

$\frac{1}{\lambda_m_i_n } =(1.097373 \times 10^7)(\frac{1}{16} - \frac{1}{_o_o} )\\\\\lambda_m_a_x=1458.03nm$

$(\lambda_{max} )_{brackett} < (\lambda_{min} )_{paschen}$

So the two wavelength range will over lap

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4 years ago

Which equation represents the law of conservation of energy in a closed system?

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Answer:

KE + PE = KE + PE

Explanation:

In a closed system, the mechanical energy of the system is constant.

Mechanical energy is given by the sum of kinetic energy and potential energy; mathematically:

U = KE + PE

where

KE is the kinetic energy

PE is the potential energy

This means that if we consider two situations, one at the beginning and one at the end, the value of U will not change if the system is closed; this means that the sum KE + PE will remain the same, so we can write:

KE + PE = KE + PE

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4 years ago

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BaLLatris [955]

Answer:

Part a)

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Part b)

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Part c)

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Part d)

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Explanation:

Part a)

As we know that the orbital speed of the satellite is given as

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now we will have

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now we have

$M_{mars} = 6.4191 \times 10^{23} kg$

$R_{mars} = 3.397 \times 10^6 m$

now we have

$2480 = \sqrt{\frac{(6.67 \times 10^{-11})(6.4191 \times 10^{23})}{r}}$

$r = 6.96 \times 10^6 m$

Part b)

Here force between mars and satellite is the gravitational attraction force which is given as

$F_g = \frac{GM_{mars} m}{r^2}$

$F_g = \frac{(6.67\times 10^{-11})(6.4191 \times 10^{23})(3400)}{(6.96\times 10^6)^2}$

$F_g = 3.004 \times 10^3 N$

Part c)

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Part d)

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$\frac{r_1}{r_2} = (\frac{1}{2})^2$

so we have

$r_2 = 4r_1$

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4 years ago

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Elanso [62]

Answer:

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