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3 years ago

A bullet of 5 gm is fired from a pistol of 1.5 kg. If the recoil velocity of pistol is 1.5 m/s,

Physics

1 answer:

Andreyy893 years ago

6 0

Answer:

-450 m/s

Explanation:

Momentum is conserved.

p₀ = p

0 = (1.5 kg) (1.5 m/s) + (0.005 kg) v

v = -450 m/s

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What happens to the resistance of a wire as it gets wider

Alchen [17]

Answer:

it snaps

Explanation:

the more force you put on it, the wired out it gets than it snaps. I think

5 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

3 0

3 years ago

What is the electric force acting between two charges of -0.0045 C and -0.0025 C that are 0.0060 m apart? Use Fe=kq1q2/r^2 and k

oksano4ka [1.4K]

Answer:

D. 2.8 × 10⁹ N

Explanation:

The force between two charges is directly proportional to the amount of charges at the two points and inversely proportional to the square of distance between the two points.

Fe= k Q₁Q₂/r²

Q₁= -0.0045 C

Q₂= -0.0025 C

r= 0.0060 m

k= 9.00 × 10 ⁹ Nm²/C²

Fe= (9.00 × 10 ⁹ Nm²/C²×-0.0045 C×-0.0025 C)/0.0060²

=2.8 × 10⁹ N

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3 years ago

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Why the weight of a body decreases with increases in distance from the earth's surface

Tom [10]

because as the distance increases the gravitational force decreases so the weight of a body decreases

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A cart is rolling across the floor at a constant velocity of 2 m/s to the right. The cart is pushed again to the right, and it s

OlgaM077 [116]

The force of the push and acceleration of the cart are related such as The unbalanced push causes the cart to accelerate.

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3 years ago