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3 years ago

When Elements neutral atom contains 5 neutrons 4 electrons and 4 protons

Physics

2 answers:

Amanda [17]3 years ago

8 0

Answer:

What is the atomic number of an atom that has 5 neutrons and 4 electrons? A neutral atom will have the same number of electrons is does protons. Since it has 4 protons, it must have an atomic number of 4. (That makes it beryllium.)

Explanation:

mark me brainliest please and thank you Ma'aM/Sir

Marina86 [1]3 years ago

4 0

Answer:

4 protons, 5 neutrons, and 4 electrons are there in an atom of beryllium.

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MariettaO [177]

Answer:

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Explanation:

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3 years ago

Read 2 more answers

In a microwave oven, electrons describe circular motion in a magnetic field within a special tube called amagnetron; as you'll l

QveST [7]

Answer:

The magnetic field strength and the electrons' energy are 0.077 T and 0.8906 eV.

Explanation:

Given that,

Diameter = 2.62 mm

Frequency = 2.15 GHz

(A). We need to calculate the magnetic field strength

Using formula of the magnetic field strength

$B=\dfrac{2\pi mf}{e}$

Where, f = frequency

e = charge of electron

Put the value into the formula

$B=\dfrac{2\times3.14\times9.1\times10^{-31}\times2.15\times10^{9}}{1.6\times10^{-19}}$

$B=0.077\ T$

(B). We need to calculate the energy of electron

Using formula of energy

$E=\dfrac{1}{2}m(r\omega)^2$

$E=\dfrac{1}{2}\times9.1\times10^{-31}\times(1.31\times10^{-3}\times2\pi\times2.15\times10^{9})^2$

$E=1.4249\times10^{-16}\ J$

The energy in eV

$1 eV=1.6\times10^{-16}\ J$

$E=\dfrac{1.4249\times10^{-16}}{1.6\times10^{-16}}$

$E=0.8906\ eV$

Hence, The magnetic field strength and the electrons' energy are 0.077 T and 0.8906 eV.

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3 years ago

the curiosity rover sent to explore the surface of mars has an electric generator powered by heat from the radioactive decay of

pav-90 [236]

An element can be identified by its unique atomic number. When we look in the periodic table, we find that the element with an atomic number of 9292 is uranium. There is only option containing uranium which also confirms the mass number we found. So, the daughter nucleus of the decay is 234^U.

In an alpha decay, a positively charged particle similar to a helium-4 nucleus gets released from the parent nucleus spontaneously. As the composition suggests, an alpha particle consists of two protons and 2 neutrons. The particle does not travel much, but in short range, it carries the most energy.

It's smart to use the thermal energy provided by the radioactive decay to generate electricity. This allows for a stable supply of power without consuming much space which means the saved space can be used for more scientific equipment. The alpha particle, structurally equivalent to the nucleus of a helium atom.

Learn more about nucleus here:

brainly.com/question/23366064

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2 years ago

3. Do Newton's Laws of Motion apply to a Water Spout? If so, how?

PSYCHO15rus [73]

Yes it’s spills out becasue bucket

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3 years ago

9. Consider the elbow to be flexed at 90 degrees with the forearm parallel to the ground and the upper arm perpendicular to the

mojhsa [17]

Answer:

Moment about SHOULDER ∑ τ = 3.17 N / m,

Moment respect to ELBOW Στ= 2.80 N m

Explanation:

For this exercise we can use Newton's second law relationships for rotational motion

∑ τ = I α

The moment is requested on the elbow and shoulder at the initial instant, just when the movement begins.

They indicate the angular acceleration, for which we must look for the moments of inertia of the elements involved

The mass of the forearm with the included weight is approximately 2.3 kg, with a length of about 50cm

Moment about SHOULDER

∑ τ = I α

I = I_forearm + I_sphere

the forearm can be approximated as a fixed bar at one end

I_forearm = ⅓ m L²

the moment of inertia of the mass in the hand, let's approach as punctual

I_mass = m L²

we substitute

∑ τ = (⅓ m L² + M L²) α

let's calculate

∑ τ = (⅓ 2.3 0.5² + 0.5 0.5²) 10

∑ τ = 3.17 N / m

Moment with respect to ELBOW

In this case, the arm exerts an upward force (muscle) that is about 3 cm from the elbow

Στ = I α

I = I_ forearm + I_mass

I = ⅓ m (L-0.03)² + M (L-0.03)²

let's calculate

i = ⅓ 2.3 0.47² + 0.5 0.47²

I = 0.2798 Kg m²

Στ = 0.2798 10

Στ= 2.80 N m

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3 years ago