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Lady bird [3.3K]
3 years ago
8

A motorcycle weighs 2,009 newtons. The mass of the motorcycle is kilograms. Use g = 9.8 N/kg for gravity.

Physics
2 answers:
natta225 [31]3 years ago
4 0

Answer:

Mass of motorcycle = 205 kg

Explanation:

Weight is the product of mass and acceleration due to gravity,

That is

         Weight = Mass x Acceleration due to gravity

Weight is represented by W, mass is represented by m and acceleration due to gravity is represented by g,

         W = mg

Here weight of motorcycle is 2,009 newtons and acceleration due to gravity is 9.8 N/kg,

Substituting

                 W = mg

               2009 = m x 9.8

                   m =205 kg

Mass of motorcycle = 205 kg

kramer3 years ago
3 0

We Know, W=mg, m=W/g

2009 N / 9.8 N/Kg = 205 Kg.

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A beam of white light passes through a uniform thickness of air. If the intensity of the scattered light in the middle of the gr
k0ka [10]

This question is incomplete, the complete question is;

A beam of white light passes through a uniform thickness of air. If the intensity of the scattered light in the middle of the green part of the visible spectrum (λ=520nm) is I, find the intensity (in terms of I) of scattered light

a) In the middle of the red part of the spectrum (λ= 665 nm)

b) In the middle of the violet part of the spectrum (λ = 420 nm)

Answer:

a) the intensity of scattered light ( in terms of I ) In the middle of the red part of the spectrum is 0.3739I

b) the intensity of scattered light ( in terms of I ) In the middle of the violet part of the spectrum is 2.3497I  

Explanation:

Given the data in the question,

the visible spectrum (λ=520nm) = I

we know that; intensity of scattered light is proportional to 1 / λ⁴  

I ∝ ( 1 / λ⁴ )

so

a)

I_R / I = ( λ / λ_R )⁴

As the middle of the green part of the visible spectrum λ is 520nm and middle of the red part of the spectrum λ_R is 665 nm

we substitute

I_R / I = ( 520 / 665 )⁴

I_R / I = ( 0.781954887 )⁴

I_R / I = 0.3739

I_R  =  0.3739I  { in terms of I  }

Therefore, the intensity of scattered light ( in terms of I ) In the middle of the red part of the spectrum is 0.3739I

b)

I_V / I = ( λ / λ_V )⁴

As the middle of the green part of the visible spectrum λ is 520nm and middle of the red part of the spectrum λ_R is 420 nm

we substitute

I_V / I = ( 520 / 420 )⁴

I_V / I = ( 1.238095 )⁴

I_V / I = 2.3497

I_V = 2.3497I  { in terms of I  }

Therefore, the intensity of scattered light ( in terms of I ) In the middle of the violet part of the spectrum is 2.3497I  

5 0
3 years ago
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Which layer in the atmosphere contains the most water vapor and gases? mesosphere thermosphere troposphere stratosphere
gregori [183]
The troposphere contains 80% of the the atmosphere, including the majority of gases and water vapor.
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4 0
3 years ago
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When the cylinder is displaced slightly along its vertical axis it will oscillate about its equilibrium position with a frequenc
Nesterboy [21]

Answer:

w = √[g /L (½ r²/L2 + 2/3 ) ]

When the mass of the cylinder changes if its external dimensions do not change the angular velocity DOES NOT CHANGE

Explanation:

We can simulate this system as a physical pendulum, which is a pendulum with a distributed mass, in this case the angular velocity is

          w² = mg d / I

In this case, the distance d to the pivot point of half the length (L) of the cylinder, which we consider long and narrow

         d = L / 2

The moment of inertia of a cylinder with respect to an axis at the end we can use the parallel axes theorem, it is approximately equal to that of a long bar plus the moment of inertia of the center of mass of the cylinder, this is tabulated

        I = ¼ m r2 + ⅓ m L2

        I = m (¼ r2 + ⅓ L2)

now let's use the concept of density to calculate the mass of the system

        ρ = m / V

        m = ρ V

the volume of a cylinder is

         V = π r² L

          m =  ρ π r² L

let's substitute

        w² = m g (L / 2) / m (¼ r² + ⅓ L²)

        w² = g L / (½ r² + 2/3 L²)

        L >> r

         w = √[g /L (½ r²/L2 + 2/3 ) ]

When the mass of the cylinder changes if its external dimensions do not change the angular velocity DOES NOT CHANGE

4 0
4 years ago
An electron is projected with an initial speed vi = 4.60 × 105 m/s directly toward a very distant proton that is at rest. Becaus
frutty [35]

Answer:

2.99\times 10^{-19}\ m

Explanation:

<u>Given:</u>

  • u = initial velocity of the electron = 4.60\times 10^5\ m/s
  • v = final velocity of the electron = 3u
  • x = initial position of the electron from the proton = very distant =  \infty

<u>Assume:</u>

  • m = mass of an electron = 9.1\times10^{-31}\ kg
  • e = magnitude of charge on an electron = 1.6\times10^{-19}\ C
  • p = magnitude of charge on an proton = 1.6\times10^{-19}\ C
  • k = Boltzmann constant = 9\times 10^9\ Nm^2/C^2
  • y = final position of the electron from the proton
  • \Delta K = change in kinetic energy of the electron
  • W = work done by the electrostatic force
  • F = electrostatic force
  • r = instantaneous distance of the electron from the proton

Let us first calculate the work done by the electrostatic force.

W=\int Fdr\\\Rightarrow W = \int \dfrac{kep}{r^2}dr\\\Rightarrow W = kep\int \dfrac{1}{r^2}dr\\\Rightarrow W = kep\left | \dfrac{1}{r} \right |_{y}^{x}\\\Rightarrow W = kep\left ( \dfrac{1}{x}-\dfrac{1}{y} \right )\\\Rightarrow W = kep\left ( \dfrac{1}{x}-\dfrac{1}{\infty} \right )\\\Rightarrow W =\dfrac{kep}{x}

Using the principle of the work-energy theorem,

As only the electrostatic force is assumed to act between the two charges, the kinetic energy change of the electron will be equal to the work done by the electrostatic force on the electron due to proton.

\therefore \Delta K = W\\\Rightarrow \dfrac{1}{2}m(v^2-u^2)= \dfrac{kep}{x}\\\Rightarrow \dfrac{1}{2}m((3u)^2-u^2)= \dfrac{kep}{x}\\\Rightarrow \dfrac{1}{2}m(8u^2)= \dfrac{kep}{x}\\\Rightarrow x= \dfrac{2kep}{8mu^2}\\\Rightarrow x= \dfrac{2\times 9\times 10^9\times 1.6\times10^{-19}\times 1.6\times10^{-19}}{8\times 9.1\times10^{-31}\times (4.60\times 10^5)^2}\\\Rightarrow x=2.99\times 10^{-10}\ m\leq

Hence, the electron is at a distance of 2.99\times 10^{-10}\ m when the electron instantaneously has speed of three times the initial speed.

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