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3 years ago

6

A 5000kg freight car moving at 2 m/s East collides with a 10,000kg freight car at rest. Upon collision, they got stuck and moved

with same velocity. Find their final velocity after collision

1 answer:

mr_godi [17]3 years ago

6 0

Answer:

A

Explanation:

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Resonance in air column is used in which of these musical instruments

zvonat [6]

Answer: The correct answer is "trumpet".

Explanation:

Resonance occurs when the applied frequency on the object is equal to the natural frequency of the object.

Wind instruments use resonance in the air column to amplify the sound. It makes the sound louder. Sound is produced by using lips or vibrating reeds or buzzing sound into the mouthpiece.

Therefore, resonance in air column is used in trumpet musical instruments.

7 0

3 years ago

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A sample of oxygen gas at 25.0Â°c has its pressure tripled while its volume is halved. What is the final temperature of the gas?

Marat540 [252]

Answer:

447 K

Explanation:

25 C = 25 + 273 = 298 K

Assuming ideal gas, we can apply the ideal gas law

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

$T_2 = T_1\frac{P_2}{P_1}\frac{V_2}{V_1}$

Since pressure is tripled, then $P_2 / P_1 = 3$ . Volume is halved, then $V_2 / V_1 = 0.5$

$T_2 = 298*3*0.5 = 447 K$

4 0

3 years ago

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A wave is moving at 18 m/s. If its wavelength is 3 meters, what is its frequency?

Vlada [557]

here's the solution,

we know that,

=》

$wave \: speed = wavelength \times frequency$

so,

=》

$18 = 3 \times f$

=》

$f = \dfrac{18}{3}$

=》

$f = 6$

frequency = 6 hertz

7 0

3 years ago

Write any two uses of simple machines.

Nana76 [90]

Simple machines could be used to reduce effort or extend the ability of people to perform tasks beyond their normal capabilities.
Examples include pulley, lever, and incline plane

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3 years ago

For crystal diffraction experiments, wavelengths on the order of 0.25 nm are often appropriate.

Kamila [148]

Answer:

A) E = 4.96 x 10³ eV

B) E = 4.19 x 10⁴ eV

C) E = 3.73 x 10⁹ eV

Explanation:

A)

For photon energy is given as:

$E = hv$

$E = \frac{hc}{\lambda}$

where,

E = energy of photon = ?

h = 6.625 x 10⁻³⁴ J.s

λ = wavelength = 0.25 nm = 0.25 x 10⁻⁹ m

Therefore,

$E = \frac{(6.625 x 10^{-34} J.s)(3 x 10^8 m/s)}{0.25 x 10^{-9} m}$

$E = (7.95 x 10^{-16} J)(\frac{1 eV}{1.6 x 10^{-19} J})$

<u>E = 4.96 x 10³ eV</u>

<u></u>

B)

The energy of a particle at rest is given as:

$E = m_{0}c^2$

where,

E = Energy of electron = ?

m₀ = rest mass of electron = 9.1 x 10⁻³¹ kg

c = speed of light = 3 x 10⁸ m/s

Therefore,

$E = (9.1 x 10^{-31} kg)(3 x 10^8 m/s)^2\\$

$E = (8.19 x 10^{-14} J)(\frac{1 eV}{1.6 x 10^{-19} J})\\$

<u>E = 4.19 x 10⁴ eV</u>

<u></u>

C)

The energy of a particle at rest is given as:

$E = m_{0}c^2$

where,

E = Energy of alpha particle = ?

m₀ = rest mass of alpha particle = 6.64 x 10⁻²⁷ kg

c = speed of light = 3 x 10⁸ m/s

Therefore,

$E = (6.64 x 10^{-27} kg)(3 x 10^8 m/s)^2\\$

$E = (5.97 x 10^{-10} J)(\frac{1 eV}{1.6 x 10^{-19} J})\\$

<u>E = 3.73 x 10⁹ eV</u>

8 0

2 years ago