Identify the amplitude in the wave image below.* F. G H J

A ball has a momentum of 35 kg·m/s. If the mass of the ball is 5.0 kg, what is the speed of the ball?

lana66690 [7]

Answer:

Simply divide the kgm/s by the 5kg

35kgm/s ÷ 5.0kg = 7m/s

4 0

3 years ago

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The following equation is an example of decay? 232/90 TH---4/2 HE +228/88 RA?

DaniilM [7]

Answer:

Alpha decay

Explanation:

Alpha decay is one of the three major types of decays, others being, beta decay and gamma decay.
When a radioactive isotope undergoes alpha decay it emits alpha particles. An alpha particle is equivalent to the nucleus of Helium atom.
Therefore, an atom undergoing decay, its atomic mass is decreased by 4 and its atomic number is decreased by 2.
Thus, since 232/90 Th, has undergone alpha decay its mass number is reduced by 4 to 228 and its atomic number by 2 to 88, and becomes 228/88 Ra.

5 0

3 years ago

The wavelength of an EM wave is 800 nm. Calculate its frequency, angular frequency, wavenumber, wave vector amplitude, energy, a

irga5000 [103]

Answer:

Given: Wavelength (λ) = 800 nm

Frequency $\nu=\frac{c}{\lambda} \Rightarrow \nu = \frac{3\times10^8}{800\times 10^{-9}} =3.75\times10^{14} Hz$

Angular frequency $\omega=2\pi \nu \Rightarrow \omega = 2\pi 3.75\times10^{14} = 23.55\times10^{14} rad/s$

Wavenumber, $k=\frac{2\pi}{\lambda} \Rightarrow k =\frac{2\pi}{800\times 10^{-9}}=7.85\times 10^6m^{-1}$

wave vector amplitude |k| = k = $7.85\times 10^6m^{-1}$

energy, $E = h \nu \Rightarrow E = 6.63\times 10^{-34}\times 3.75\times10^{14} = 24.86\times 10^{-20}J$

momentum, $p=\frac{h}{\lambda} = \frac{6.63\times 10^{-34}}{800\times 10^{-9}}= 8.28\times 10^{-28}Ns$

3 0

3 years ago

Which of the following is a device

marusya05 [52]

Answer:

spectrograph

hope it is helpful to you

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3 years ago

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If temp. gets cold resistance in thermostat increases so voltage across it increases AND LED lights brighter. I understand every

Valentin [98]

Normally, when something gets colder, its electrical resistance gets smaller. This is true of component-A in the drawing ... a simple resistor.

The component labeled 'B' has a strange and unusual symbol, and it's not a simple resistor. It's a "thermistor". The word "thermal" always has something to do with heat, and "thermistor" comes from "thermal resistor. These things can be manufactured either way ... using different materials, a thermistor can be manufactured so that its resistance goes UP, or goes DOWN, or doesn'tchange when it gets colder. I'm pretty sure that's what's going on here.

When this circuit gets colder, resistance-A gets smaller, but resistance-B either gets bigger OR doesn't change. Either way, the voltage across B increases. Since the LED is connected directly across B, the current through it depends on that voltage, so the LED gets more current, and becomes brighter, when A and B both get colder.

This circuit could actually be a very useful device. If you took out the LED and put a voltmeter in its place, then the reading on the voltmeter would tell you the temperature of wherever you put the two components A and B.

5 0

3 years ago