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3 years ago

An electron has _____

Physics

2 answers:

Soloha48 [4]3 years ago

8 0

Answer:

option (D)

Explanation:

The elementary particles of atom are protons, neutrons and electrons. The protons and neutrons are residing in the nucleus and electron revolves around the nucleus in fixed stationary orbits.

The mass of proton and a neutron is almost same and equal to 1.67 x 10^-27 Kg.

The mass of an electron is 9.1 x 10^-31 Kg.

The mass of electron is much less than the mass of proton or a neutron.

Proton is 1837 times heavier than the electron.

Marrrta [24]3 years ago

3 0

a proton and neutron weigh about the same, while an electron is about 1/2000 of the weight of either a proton or neutron

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Answer:

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3 years ago

In a nuclear reactor, the control rods are: A.made of graphite B.used to keep the reactor cool C.used to absorb free neutrons D.

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3 years ago

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A railroad freight car, mass 15,000 kg, is allowed to coast along a level track at a speed of 2.0 m/s. It collides and couples w

gayaneshka [121]

Answer:

The speed of the two cars after coupling is 0.46 m/s.

Explanation:

It is given that,

Mass of car 1, m₁ = 15,000 kg

Mass of car 2, m₂ = 50,000 kg

Speed of car 1, u₁ = 2 m/s

Initial speed of car 2, u₂ = 0

Let V is the speed of the two cars after coupling. It is the case of inelastic collision. Applying the conservation of momentum as :

$m_1u_1+m_2u_2=(m_1+m_2)V$

$V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}$

$V=\dfrac{15000\ kg\times 2\ m/s+0}{(15000\ kg+50000\ kg)}$

V = 0.46 m/s

So, the speed of the two cars after coupling is 0.46 m/s. Hence, this is the required solution.

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3 years ago

Two airplanes leave an airport at the same time.The velocity of the first airplane is 700 m/h at a heading of 31.3 the velocity

MArishka [77]

Here in order to find out the distance between two planes after 3 hours can be calculated by the concept of relative velocity

$v_{12} = v_1 - v_2$

here

speed of first plane is 700 mi/h at 31.3 degree

$v_1 = 700 cos31.3\hat i + 700 sin31.3\hat j$

$v_1 = 598.12\hat i + 363.7\hat j$

speed of second plane is 570 mi/h at 134 degree

$v_2 = 570 cos134 \hat i + 570 sin134 \hat j$

$v_2 = -396\hat i + 410\hat j$

now the relative velocity is given as

$v_{12} = (598.12 + 396)\hat i + (363.7 - 410)\hat j$

$v_{12} =994.12\hat i -46.3 \hat j$

now the distance between them is given as

$d = v* t$

$d = (994.12 \hat i - 46.3 \hat j)* 3$

$d = 2982.36\hat i - 138.9\hat j$

so the magnitude of the distance is given as

$d = \sqrt{2982.36^2 + 138.9^2}$

$d = 2985.6$ miles

so the distance between them is 2985.6 miles

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3 years ago

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