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DanielleElmas [232]
3 years ago
15

A child bounces a 48 g superball on the sidewalk. the velocity change of the superball is from 26 m/s downward to 17 m/s upward.

if the contact time with the sidewalk is 1 800 s, what is the magnitude of the average force exerted on the superball by the sidewalk? answer in units of n.
Physics
1 answer:
Nataly_w [17]3 years ago
3 0
By definition we have the momentum is:
 P = m * v
 Where,
 m = mass
 v = speed
 Before the impact:
 P1 = (0.048) * (26) = 1.248 kg * m / s
 After the impact:
 P2 = (0.048) * (- 17) = -0.816 Kg * m / s.
 Then we have that deltaP is:
 deltaP = P2-P1
 deltaP = (- 0.816) - (1,248)
 deltaP = -2,064 kg * m / s.
 Then, by definition:
 deltaP = F * delta t
 Clearing F:
 F = (deltaP) / (delta t)
 Substituting the values
 F = (- 2.064) / (1/800) = - 1651.2N
 answer:
 the magnitude of the average force exerted on the superball by the sidewalk is 1651.2N
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Answer:

f_{e} = 1.7 cm

Explanation:

The magnification of the compound microscope is given by the product of the magnification of each lens

        M = M₀ m_{e}

        M = - L/f₀  25/f_{e}

Where f₀ and f_{e} are the focal lengths of the lens and eyepiece, respectively, all values ​​in centimeters

In this exercise they give us the magnification (M = 400X), the focal length of the lens (f₀ = 0.6 cm), the distance of the tube (L = 16 cm), let's look for the focal length of the eyepiece (f_{e})

         f_{e} = - L / f₀ 25 / M

Let's calculate

        f_{e} = - 16 / 0.6 25 / (-400)

        f_{e} = 1.67 cm

The minus sign in the magnification is because the image is inverted.

          f_{e} = 1.7 cm

6 0
3 years ago
A 300 MHz electromagnetic wave in air (medium 1) is normally incident on the planar boundary of a lossless dielectric medium wit
Masja [62]

Answer:

Wavelength of the incident wave in air = 1 m

Wavelength of the incident wave in medium 2 = 0.33 m

Intrinsic impedance of media 1 = 377 ohms

Intrinsic impedance of media 2 = 125.68 ohms

Check the explanation section for a better understanding

Explanation:

a) Wavelength of the incident wave in air

The frequency of the electromagnetic wave in air, f = 300 MHz = 3 * 10⁸ Hz

Speed of light in air, c =  3 * 10⁸ Hz

Wavelength of the incident wave in air:

\lambda_{air} = \frac{c}{f} \\\lambda_{air} = \frac{3 * 10^{8} }{3 * 10^{8}} \\\lambda_{air} = 1 m

Wavelength of the incident wave in medium 2

The refractive index of air in the lossless dielectric medium:

n = \sqrt{\epsilon_{r} } \\n = \sqrt{9 }\\n =3

\lambda_{2} = \frac{c}{nf}\\\lambda_{2} = \frac{3 * 10^{6} }{3 * 3 * 10^{6}}\\\lambda_{2} = 1/3\\\lambda_{2} = 0.33 m

b) Intrinsic impedances of media 1 and media 2

The intrinsic impedance of media 1 is given as:

n_1 = \sqrt{\frac{\mu_0}{\epsilon_{0} } }

Permeability of free space, \mu_{0} = 4 \pi * 10^{-7} H/m

Permittivity for air, \epsilon_{0} = 8.84 * 10^{-12} F/m

n_1 = \sqrt{\frac{4\pi * 10^{-7}  }{8.84 * 10^{-12}  } }

n_1 = 377 \Omega

The intrinsic impedance of media 2 is given as:

n_2 = \sqrt{\frac{\mu_r \mu_0}{\epsilon_r \epsilon_{0} } }

Permeability of free space, \mu_{0} = 4 \pi * 10^{-7} H/m

Permittivity for air, \epsilon_{0} = 8.84 * 10^{-12} F/m

ϵr = 9

n_2 = \sqrt{\frac{4\pi * 10^{-7} *1 }{8.84 * 10^{-12} *9 } }

n_2 = 125.68 \Omega

c) The reflection coefficient,r  and the transmission coefficient,t at the boundary.

Reflection coefficient, r = \frac{n - n_{0} }{n + n_{0} }

You didn't put the refractive index at the boundary in the question, you can substitute it into the formula above to find it.

r = \frac{3 - n_{0} }{3 + n_{0} }

Transmission coefficient at the boundary, t = r -1

d) The amplitude of the incident electric field is E_{0} = 10 V/m

Maximum amplitudes in the total field is given by:

E = tE_{0} and E = r E_{0}

E = 10r, E = 10t

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assume the suns total energy output is 4.0 * 10^26 watts, and 1 watt is 1 joule/second. assume 4.3 * 10^-12 J is released from e
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Answer:

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Explanation:

Power is defined as the energy produced (E) per unit of time (t):

P= \frac{E}{t}

This means that the energy produced in the Sun each second (1 s), given the power P=4.0\cdot 10^{26}W, is

E=Pt=(4.0\cdot 10^{26}W)(1s )=4.0\cdot 10^{26} J

Each p-p chain reaction produces an amount of energy of

E_1 = 4.3\cdot 10^{-12} J

in order to get the total number of p-p chain reactions per second, we need to divide the total energy produced per second by the energy produced by each reaction:

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