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yKpoI14uk [10]
3 years ago
12

A cart traveling at 0.3 m/s collides with stationary object. After the collision, the cart rebounds in the opposite direction. T

he same cart again traveling at 0.3 m/s collides with a different stationary object. This time the cart is at rest after the collision. In which collision is the impulse on the cart greater?
A. The first collision.

B. Cannot be determined without knowing the rebound speed of the first collision.

C. The second collision.

D. Cannot be determined without knowing the mass of the cart.

E. The impulses are the same.
Physics
1 answer:
Nady [450]3 years ago
6 0
The first collision because a greater amount of momentum must be taken and used in order to push the cart back, giving it a greater mass and impulse
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A 12.0 khz, 16.0 v source connected to an inductor produces a 4.00 a current. what is the inductance?
Varvara68 [4.7K]
Inductive reactance (Z) = ω L  =  2Πf L = (2Π) (12,000) (L)

I = V / Z

4 A = 16v / (24,000Π L)

Multiply each side by (24,000 Π L):

96,000 Π L = 16v

Divide each side by  (96,000 Π) :

L = 16 / 96,000Π  =  5.305 x 10⁻⁵ Henry

L = 53.05 microHenry
4 0
3 years ago
The normal eye, myopic eye and old age
yanalaym [24]

Answer:

1)    f’₀ / f = 1.10, the relationship between the focal length (f'₀) and the distance to the retina (image) is given by the constructor's equation

2) the two diameters have the same order of magnitude and are very close to each other

Explanation:

You have some problems in the writing of your exercise, we will try to answer.

1) The equation to be used in geometric optics is the constructor equation

          \frac{1}{f} = \frac{1}{p} + \frac{1}{q}

where p and q are the distance to the object and the image, respectively, f is the focal length

* For the normal eye and with presbyopia

the object is at infinity (p = inf) and the image is on the retina (q = 15 mm = 1.5 cm)

        \frac{1}{f'_o} = 1/ inf + \frac{1}{1.5}

        f'₀ = 1.5 cm

this is the focal length for this type of eye

* Eye with myopia

the distance to the object is p = 15 cm the distance to the image that is on the retina is q = 1.5 cm

           1 / f = 1/15 + 1 / 1.5

           1 / f = 0.733

            f = 1.36 cm

this is the focal length for the myopic eye.

In general, the two focal lengths are related

         f’₀ / f = 1.5 / 1.36

         f’₀ / f = 1.10

The question of the relationship between the focal length (f'₀) and the distance to the retina (image) is given by the constructor's equation

2) For this second part we have a diffraction problem, the point diameter corresponds to the first zero of the diffraction pattern that is given by the expression for a linear slit

          a sin θ= m λ

the first zero occurs for m = 1, as the angles are very small

          tan θ = y / f = sin θ / cos θ

for some very small the cosine is 1

          sin θ = y / f

where f is the distance of the lens (eye)

           y / f = lam / a

in the case of the eye we have a circular slit, therefore the system must be solved in polar coordinates, giving a numerical factor

           y / f = 1.22 λ / D

           y = 1.22 λ f / D

where D is the diameter of the eye

          D = 2R₀

          D = 2 0.1

          D = 0.2 cm

           

the eye has its highest sensitivity for lam = 550 10⁻⁹ m (green light), let's use this wavelength for the calculation

         

* normal eye

the focal length of the normal eye can be accommodated to give a focus on the immobile retian, so let's use the constructor equation

      \frac{1}{f} = \frac{1}{p} + \frac{1}{q}

sustitute

       \frac{1}{f} = \frac{1}{25} + \frac{1}{1.5}

       \frac{1}{f}= 0.7066

        f = 1.415 cm

therefore the diffraction is

        y = 1.22  550 10⁻⁹  1.415  / 0.2

        y = 4.75 10⁻⁶ m

this is the radius, the diffraction diameter is

       d = 2y

       d_normal = 9.49 10⁻⁶ m

* myopic eye

In the statement they indicate that the distance to the object is p = 15 cm, the retina is at the same distance, it does not move, q = 1.5 cm

       \frac{1}{f} = \frac{1}{15} + \frac{1}{ 1.5}

        \frac{1}{f}= 0.733

         f = 1.36 cm

diffraction is

        y = 1.22 550 10-9 1.36 10-2 / 0.2 10--2

        y = 4.56 10-6 m

the diffraction diameter is

        d_myope = 2y

         d_myope = 9.16 10-6 m

         \frac{d_{normal}}{d_{myope}} = 9.49 /9.16

        \frac{d_{normal}}{d_{myope}} =  1.04

we can see that the two diameters have the same order of magnitude and are very close to each other

8 0
3 years ago
A large lightning bolt consists of a 18.2~\text{kA}18.2 kA current that moved 30.0~\text{C}30.0 C of charge. Assuming a constant
marissa [1.9K]

Answer:

1.65\times {-3} \text { s}

Explanation:

Current is the rate of flow of charge.

I=\dfrac{Q}{t}

t = \dfrac{Q}{I}

t = \dfrac{30.0}{18.2\times10^3} =1.65\times {0-3} = 1.65\times {-3} \text { s}

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