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likoan [24]
3 years ago
9

If element X has 14 protons, how many electrons does it have?

Physics
1 answer:
Svetlanka [38]3 years ago
5 0
The number of protons and the number of electrons are always the same in an element that is neutral and has no charge. Therefore, this element has 14 electrons.
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Find the angle of refraction of a ray of light that enters a diamond (n=2.419) from air at an angle of 18.0° to the normal. ONLY
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n

Explanation:

hb

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4 years ago
A dog whistle is designed to produce a sound with a frequency beyond that which can be heard by humans (between 20 000 Hz and 27
krek1111 [17]

Answer:

Explanation:

331 m/s / 2.5e4 cyc/s = 0.01324 m ≈ 1.3 cm

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3 years ago
A 16 g piece of Styrofoam carries a net charge of -8.6 µC and floats above the center of a large horizontal sheet of plastic tha
PilotLPTM [1.2K]

Answer:

the charge per unit area on the plastic sheet is - 3.23 x 10⁻⁷ C/m²

Explanation:

given information:

styrofoam mass, m = 16 g = 0.016 kg

net charge, q = - 8.6 μC

to calculate the charge per unit area on the plastic sheet, we can use the following equation:

F_{e} = mg

where

F_{e} = the force between the electric field

m = mass

g = gravitational force

F_{e} =qE

where

q = charge

E = electric field

and

E = σ/2ε₀

where

ε₀ = permitivity

thus

F_{e} =qE

mg = qσ/2ε₀

σ = (2mg ε₀)/q

  = 2 (0.016) (9.8)  (8.85 x 10⁻¹²)/( - 8.6 x 10⁻⁶)

  = - 3.23 x 10⁻⁷ C/m²

8 0
4 years ago
Which occurs when two musical instruments use resonance?
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D, they are both loud
8 0
4 years ago
Read 2 more answers
In 2000, NASA placed a satellite in orbit around an asteroid. Consider a spherical asteroid with a mass of 1.40×1016 kg and a ra
Arturiano [62]

A) 8.11 m/s

For a satellite orbiting around an asteroid, the centripetal force is provided by the gravitational attraction between the satellite and the asteroid:

m\frac{v^2}{(R+h)}=\frac{GMm}{(R+h)^2}

where

m is the satellite's mass

v is the speed

R is the radius of the asteroide

h is the altitude of the satellite

G is the gravitational constant

M is the mass of the asteroid

Solving the equation for v, we find

v=\sqrt{\frac{GM}{R+h}}

where:

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}

M=1.40\cdot 10^{16}kg

R=8.20 km=8200 m

h=6.00 km = 6000 m

Substituting into the formula,

v=\sqrt{\frac{(6.67\cdot 10^{-11})(1.40\cdot 10^{16}kg)}{8200 m+6000 m}}=8.11 m/s

B) 11.47 m/s

The escape speed of an object from the surface of a planet/asteroid is given by

v=\sqrt{\frac{2GM}{R+h}}

where:

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}

M=1.40\cdot 10^{16}kg

R=8.20 km=8200 m

h=6.00 km = 6000 m

Substituting into the formula, we find:

v=\sqrt{\frac{2(6.67\cdot 10^{-11})(1.40\cdot 10^{16}kg)}{8200 m+6000 m}}=11.47 m/s

5 0
3 years ago
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