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geniusboy [140]
3 years ago
7

Object A of mass 0.70 kg travels horizontally on a frictionless surface at 20 m/s. It collides with object B, which is initially

stationary and rebounds with 70% of its initial kinetic energy. What is the magnitude of the change in momentum of object A?
Physics
1 answer:
AnnZ [28]3 years ago
6 0

Answer:

25.71 kgm/s

Explanation:

Let K₁ and K₂ be the initial and final kinetic energies of object A and v₁ and v₂ its initial and final speeds.

Given that K₂ = 0.7K₁

1/2mv₂² = 0.7(1/2mv₁²)

v₂ = √0.7v₁ = √0.7 × 20 m/s = ±16.73 m/s

Since A rebounds, its velocity = -16.73 m/s and its momentum change, p₂ = mΔv = m(v₂ - v₁) = 0.7 kg (-16.73 - 20) m/s = 0.7( -36.73) = -25.71 kgm/s.

Th magnitude of object A's momentum change is thus 25.71 kgm/s

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<h2>Answer:</h2>

1.8 x 10⁻⁵J

<h2>Explanation:</h2>

The energy (E) stored in a capacitor of capacitance, C,  when a voltage, V, is supplied is given by;

E = \frac{1}{2} x C x V²              -------------------(i)

Now, from the question;

C = 2.00μF = 2.00 x 10⁻⁶F

V = 18.0V

Substitute these values into equation (i) as follows;

E = \frac{1}{2} x 2.00 x 10⁻⁶ x 18.0

E = 1.8 x 10⁻⁵J

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Explanation:

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What is the maximum number of lines per centimeter a diffraction grating can have and produce a complete first-order spectrum fo
Pachacha [2.7K]

Answer:

14,300 lines per cm

Explanation:

Answer:

14,300 cm per line

Explanation:

λ400 nm to 400nm

We can find the maximum number of lines per centimeter, which is reciprocal of the least distance separating two adjacent slits, using the following equation.

mλ = dsin (θ)

In this equation,

m is the order of diffraction.

λ is the wavelength of the incident light.

d is the distance separating the centers of the two slits.

θ is the angle at which the mth order would diffract.

To find the least separation that allows the observation of one complete order of spectrum of the visible region, we use the maximum wavelength of the visible region is 700 nm.

d =  mλ / sin (θ)

As we want the distance d to be the smallest then sin (θ) must be the greatest, and the greatest value of the sin (θ) is 1. For that we also use the longest wavelength because using the smallest wavelength, the longest wavelength would not be diffracted.

d =  mλ / sin (θ)

d =  1 x 700nm / 1

  = 700 nm

So, the least separation that would allow for the possibility of observing complete first order of the visible region spectra is 700 nm, and knowing the least separation we can find the maximum number of lines per cm, which is the reciprocal of the number of lines per cm.

n = 1/d

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<u>The maximum number of lines per cm, that would allow for the observation of the complete first order visible spectra.</u>

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