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Ilia_Sergeevich [38]
3 years ago
10

how far the medium (crests and troughs, or compressions and rarefactions) moves from ________ the place the medium is when not m

oving). The ____ energy a wave carries, the _____ is amplitude. amplitude is related to energy by _________________.
Physics
1 answer:
Lyrx [107]3 years ago
8 0

Answer and Explanation:

How far the medium (crests and troughs, or compressions and rarefactions) moves from _*rest position*_ the place the medium is when not moving). The _*more*_ energy a wave carries, the _*larger_ its amplitude. Amplitude is related to energy by _(E = CA²)_.

where C is a constant

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Answer:

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Explanation:

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A 0.15 g honeybee acquires a charge of 22 pC while flying. The electric field near the surface of the earth is typically 100 N/C
Rus_ich [418]

Answer:

1.50\ *10^{-6} }

Explanation:

Given

e=100 N/C

M=0.15 g

q=\ 22\  pC\\=\ 22\ *10^{-2}

The  ratio of the electric force on the bee to the bee's weight can be determined by the following formula

\frac{fe}{M*9.81}

\frac{22*10^{-12\ *\ 100} }{0.15*\ 10^{-3} *\ 9.81}

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A proton moves at a constant velocity of 50m/s along the x axis ,through crossed electric and magnetic fields .The magnetic fiel
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3 years ago
Light of wavelength 580 nm is incident on a slit of width 0.30 mm. An observing screen is placed 2.0 m past the slit. Find the d
uysha [10]

Answer:

Y = 3.87 x 10⁻³ m = 3.87 mm

Explanation:

This problem can be solved by using Young's double-slit experiment formula:

Y = \frac{\lambda L}{d}

where,

Y = fringe spacing = ?

L = slit to screen distance = 2 m

λ = wavelength of light = 580 nm = 5.8 x 10⁻⁷ m

d = slit width = 0.3 mm = 3 x 10⁻⁴ m

Therefore,

Y = \frac{(5.8\ x\ 10^{-7}\ m)(2\ m)}{3\ x\ 10^{-4}\ m}

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3 0
3 years ago
A Bullet Off mass 100 gm is fired From A Gun Off mass 5 Kg. If the backward velocity of the gun's 5 m / s, what is forward veloc
Elena L [17]

Answer:

250 m/s

Explanation:

The mass of the bullet, m₁ = 100 g = 0.1 kg

The mass of the gun, m₂ = 5 kg

The backward velocity of the gun, v₂ = -5 m/s

Given that the momentum is conserved, we have;

The total initial momentum = The total final momentum

The gun and the bullet are at rest, therefore, we have;

The initial momentum = 0

The total final momentum = m₁·v₁ + m₂·v₂

Where;

v₁ = The forward velocity of the bullet

Therefore, we get;

m₁·v₁ + m₂·v₂ = 0

0.1 kg × v₁ + 5 kg × (-5 m/s) = 0

0.1 kg × v₁ = 5 kg × 5 m/s

v₁ = (5 kg × 5 m/s)/(0.1 kg) = 250 m/s

The forward velocity of the bullet, v₁ = 250 m/s

6 0
3 years ago
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