What are the similarities between strong nuclear force and weak nuclear force

1 answer:

7 0

Alike because they both act on the quarks making up the nucleons and they have very short ranges. The Strong Nuclear Force is an attractive force between protons and neutrons that keep the nucleus together and the Weak Nuclear Force is responsible for the radioactive decay of certain nuclei. Which also makes them very different

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Free brainlyest. if get thes quston right Devlin recorded the number of days of rain for his town over three months. The bar gra

m_a_m_a [10]

Answer:

Weather is the conditions (temperature, wind etc.) at a given time, like on that day. Climate, which is what his data would show, is the conditions over an extended period of time like the 3 months he collected data

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2 years ago

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During the fission reaction shown, how did the target nucleus change ?

Zarrin [17]

Answer:

A. The target nucleus split into two nuclei, each with fewer nucleons than the original.

Explanation:

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2 years ago

If you know the amount of air pressure exerted on a tabletop, how can you calculate the force exerted on the tabletop? Multiply

babymother [125]

Answer:

Multiply the air pressure by the area of the tabletop.

Explanation:

The relationship between pressure, force and area is given by:

$p=\frac{F}{A}$

where in this case, p is the air pressure, F is the force exerted and A the area of the tabletop. By re-arranging the equation, we can solve for F, the force exerted:

$p=\frac{F}{A}\\p\cdot A=\frac{F}{A}\cdot A\\pA=F$

So, the correct answer is:

The force exerted on the tabletop can be found by multiplying the air pressure by the area of the tabletop.

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3 years ago

The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece

Brilliant_brown [7]

Answer:

10.01 cm

Explanation:

Given that,

The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece of fat tissue was 0.13 ms.

The average propagation speed for sound in body tissue is 1540 m/s.

We need to find the depth when the reflection occur. We know that, the distance is double when transmitting and arriving. So,

$v=\dfrac{2d}{t}\\\\d=\dfrac{vt}{2}\\\\d=\dfrac{1540\times 0.13\times 10^{-3}}{2}\\\\d= $$0.1001\ m$