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Anna71 [15]
3 years ago
7

Negatively charged particle that orbits the nucleus

Physics
1 answer:
Ede4ka [16]3 years ago
6 0

Answer: Electrons.

Explanation:

Electrons are the negatively charged particles that orbit the nucleus of an atom. Protons, on the other hand, are the positively charged particles that orbit the nucleus of an atom.

in the picture, the protons and neutrons are “in” the nucleus and you’ll also see the the electrons are orbiting all around the nucleus.

If this helped please mark me brainliest.

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If the mass of the sun is 1x, at least one planet will fall into the habitable zone. if I place a planet in orbits 2, 6, and 75, and all planets will orbit the sun successfully.

If the mass of the sun is 2x, at least one planet will fall into the habitable zone. if I place a planet in orbits 84, 1, and 5, and all planets will orbit the sun successfully.

If the mass of the sun is 3x, at least one planet will fall into the habitable zone if I place a planet in orbits 672, and 7 and all planets will orbit the sun successfully.

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2 years ago
What is the anomalous expansivity of water
lora16 [44]

Answer:

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Explanation:do you want me to explain it more??

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3 years ago
A student is performing a double-slit experiment to determine the wavelength of a light source. She has measured the distance be
Harman [31]

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3 years ago
On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini
irinina [24]

Answer:

v_a=0.8176 m/s

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being m_a and m_b the masses of pucks a and b respectively, the initial momentum of the system is

M_1=m_av_a+m_bv_b

Since b is initially at rest

M_1=m_av_a

After the collision and being v'_a and v'_b the respective velocities, the total momentum is

M_2=m_av'_a+m_bv'_b

Both momentums are equal, thus

m_av_a=m_av'_a+m_bv'_b

Solving for v_a

v_a=\frac{m_av'_a+m_bv'_b}{m_a}

v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}

v_a=0.8176 m/s

The initial kinetic energy can be found as (provided puck b is at rest)

K_1=\frac{1}{2}m_av_a^2

K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J

The final kinetic energy is

K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2

K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J

The change of kinetic energy is

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

3 0
3 years ago
A circular coil of wire of radius 5.0 cm has 20 turns and carries a current of 2.0 A. The coil lies in a magnetic field of magni
Korvikt [17]

Answer:a. Magnetic dipole moment is 0.3412Am²

b. Torque is zero(0)N.m

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That is,

U = n*I*A

But Area A is given as pi*radius² since it is a circular coil

Radius given is 5cm converting to meter we divide by 100 so we have our radius to be 0.05m. So area A is

A = 3.142*(0.05)² =7.86*EXP {-3} m²

Current I is 2 A

Number of turns is 20

So magnetic dipole moment U is

U = 20*2*7.86*EXP {-3}=0.3142A.m²

b. Torque is given as the cross product of the magnetic field B and magnetic dipole moment U

Torque = B x U =B*U*Sine(theta)

But since the magnetic field is directed parallel to the plane of the coil from the question, it means that the angle between them is zero and sine zero is equals 0(zero) if you substitute that into the formula for torque you will find out that your torque would equals zero(0)N.m

7 0
3 years ago
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