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3 years ago

What is the acceleration of an object with a mass of 15 kg and a coefficient of friction of 0.18

Physics

1 answer:

11111nata11111 [884]3 years ago

7 0

Answer:

a = 1.764m/s^2

Explanation:

By Newton's second law, the net force is F = ma.

The equation for friction is F(k) = F(n) * μ.

In this case, the normal force is simply F(n) = mg due to no other external forces being specified

F(n) = mg = 15kg * 9.8 m/s^2 = 147N.

F(k) = F(n) * μ = 147N * 0.18 = 26.46N.

Assuming the object is on a horizontal surface, the force due to gravity and the normal force will cancel each other out, leaving our net force as only the frictional one.

Thus, F(net) = F(k) = ma

26.46N = 15kg * a

a = 1.764m/s^2

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If you increase the size of the resistor and keep the voltage the same, what will happen to the current

Grace [21]

Answer:

If voltage is kept constant across the resistor itself, it' current will reduce. If the resistance is part of oscillator circuit, frequency response will change. If it is in series with capacitor or inductor, it will change the damping effect.

Explanation:

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3 years ago

What is the frequency and wavelength, in nanometers, of photons capable of just ionizing nitrogen atoms?

nika2105 [10]

Answer:

The frecuency and wavelength of a photon capable to ionize the nitrogen atom are ν = 3.394×10¹⁵ s⁻¹ and λ = 88.31 nm.

Explanation:It is possible to know what are the frequency and wavelength of a photon capable to ionize the nitrogen atom using the equation of the energy of a photon described below.

E = hc/λ (1)

Where h is the Planck constant, c is the speed of light and λ is the wavelength of the photon.

But first, it is neccesary to know the ionization energy of the nitrogen atom. The ionization energy is the energy needed to remove an electron from an atom, for the Nitrogen atom it will lose an electron of its outer orbit from the nucleus, farther snuff, so the electric force is weaker. Experimentally, it is known that it has a value of 14.04 eV. This value is easy to found in a periodic table.

So the nitrogen atom will need a photon with the energy of 14.04 eV to remove the electron from its outer orbit.

Replacing the Planck constant, the speed of light and the energy of the photon in the equation 1, the wavelength can be calculated:

λ = hc/E (2)

Where h = 6.626×10⁻³⁴ J.s and c = 3.00×10⁸ m/s

But the Planck constant can be expressed in electron volts:

1 eV = 1.602 x 10⁻¹⁹ J

h = 6.626x10⁻³⁴ J/1.602x10⁻¹⁹ J . eV .s

h= 4.136x10⁻¹⁵ eV.s

Now, it is convenient to express the speed of light in nanometers:

1nm = 1x10⁻⁹ m

c = 3.00x10⁸ m/ 1x10⁻⁹ m

c = 3x10¹⁷ nm/s

Substituting in equation 2:

λ = (4.136x10⁻¹⁵ eV.s)(3x10¹⁷ nm/s)/14.04 eV

λ = 1240 eV. nm/ 14.04 eV

λ = 88.31 nm

The frenquency is calculated using the equation 2 in the following way:

E = hν (3)

Where ν is the frecuency

ν = E/h

ν = 14.04 eV/4.136×10⁻¹⁵ eV.s

ν = 3.394×10¹⁵ s-1

So the frecuency of a photon, capable to ionize the nitrogen atom, will be 3.394×10¹⁵ s⁻¹ and its wavelength 88.31 nm.

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3 years ago

Potassium is a crucial element for the healthy operation of the human body. Potassium occurs naturally in our environment and th

gregori [183]

Complete Question

Potassium is a crucial element for the healthy operation of the human body. Potassium occurs naturally in our environment and thus our bodies) as three isotopes: Potassium-39, Potassium-40, and Potassium-41. Their current abundances are 93.26%, 0.012% and 6.728%. A typical human body contains about 3.0 grams of Potassium per kilogram of body mass. 1. How much Potassium-40 is present in a person with a mass of 80 kg? 2. If, on average, the decay of Potassium-40 results in 1.10 MeV of energy absorbed, determine the effective dose (in Sieverts) per year due to Potassium-40 in an 80- kg body. Assume an RBE of 1.2. The half-life of Potassium-40 is $1.28 * 10^9$ years.

Answer:

The potassium-40 present in 80 kg is $Z = 0.0288 *10^{-3}\ kg$

The effective dose absorbed per year is $x = 2.06 *10^{-24}$ per year

Explanation:

From the question we are told that

The mass of potassium in 1 kg of human body is $m = 3g= \frac{3}{1000} = 3*10^{-3} \ kg$

The mass of the person is $M = 80 \ kg$

The abundance of Potassium-39 is 93.26%

The abundance of Potassium-40 is 0.012%

The abundance of Potassium-41 is 6.78 %

The energy absorbed is $E = 1.10MeV = 1.10 *10^{6} * 1.602 *10^{-19} = 1.7622*10^{-13} J$

Now 1 kg of human body contains $3.0*10^{-3}\ kg$ of Potassium

So 80 kg of human body contains k kg of Potassium

=> $k = \frac{ 80 * 3*10^{-3}}{1}$

$k = 0.240\ kg$

Now from the question potassium-40 is 0.012% of the total potassium so

Amount of potassium-40 present is mathematically represented as

$Z = \frac{0.012}{100} * 0.240$

$Z = 0.0288 *10^{-3}\ kg$

The effective dose (in Sieverts) per year due to Potassium-40 in an 80- kg body is mathematically evaluated as

$D = \frac{E}{M}$

Substituting values

$D = \frac{1.7622*10^{-13}}{80}$

$D = 2.2*10^{-15} J/kg$

Converting to Sieverts

We have

$D_s = REB * D$

$D_s = 1.2 * 2.2 *10^{-15}$

$D_s = 2.64 *10^{-15}$

for half-life ( $1.28 *10^9 \ years$ ) the dose is $2.64 *10^{-15}$

Then for 1 year the dose would be x

=> $x = \frac{2.64 *10^{-15}}{1.28 * 10^9}$

$x = 2.06 *10^{-24}$ per year

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4 years ago

LiRa [457]

Answer:

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3 years ago