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2 years ago

How many electrons should be in an atom’s valence shell (with the exception of hydrogen and helium) to become stable? 2 4 6 8.

Physics

1 answer:

podryga [215]2 years ago

4 0

Answer:

Explanation:

octet rule: Atoms lose, gain, or share electrons in order to have a full valence shell of eight electrons. Hydrogen is an exception because it can hold a maximum of two electrons in its valence level.

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Which disadvantage of nuclear power use poses the greatest threat to the environment?

Mekhanik [1.2K]

The correct answer is The storage and management of radioactive wastes

Explanation:

In general, nuclear reactions (changes in the nucleus of an atom such as fission) release a lot of energy including a lot of heat. Moreover, this heat is used by humans to obtain electricity and other types of energy, which is known as a nuclear power. This type of power is considered positive because it does not emit carbon and it is quite efficient.

However, in most cases, it is a threat to the environment and living beings because storing and managing the wastes of this type of power is difficult and expensive. Indeed, dealing with the wastes of nuclear power requires complex infrastructure, and any accident or leaking leads to serious consequences from the death of those exposed to the wastes to permanent loss of diversity or changes in nearby areas.

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3 years ago

Read 2 more answers

Applied force is the force of support exerted by an object that holds up another object.

kenny6666 [7]

False, applied force is when a person or an object pushes on another object

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3 years ago

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Is this true or false a seismograph records the air movements caused by sseismic waves

olasank [31]

The answer to this question is False.

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3 years ago

A leaky 10-kg bucket is lifted from the ground to a height of 11 m at a constant speed with a rope that weighs 0.9 kg/m. Initial

nalin [4]

Answer:

the work done to lift the bucket = 3491 Joules

Explanation:

Given:

Mass of bucket = 10kg

distance the bucket is lifted = height = 11m

Weight of rope= 0.9kg/m

g= 9.8m/s²

initial mass of water = 33kg

x = height in meters above the ground

Let W = work

Using riemann sum:

the work done to lift the bucket =∑(W done by bucket, W done by rope and W done by water)

= $\int\limits^a_b {(Mass of Bucket + Mass of Rope + Mass of water)*g*d} \, dx$

Work done in lifting the bucket (W) = force × distance

Force (F) = mass × acceleration due to gravity

Force = 9.8 * 10 = 98N

W done by bucket = 98×11 = 1078 Joules

Work done to lift the rope:

At Height of x meters (0≤x≤11)

Mass of rope = weight of rope × change in distance

= 0.8kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = $\int\limits^1 _0 {9.8*0.8(11-x)} \, dx$

Note : upper limit is 11 not 1, problem with math editor

W done = 7.84 (11x-x²/2)upper limit 11 to lower limit 0

W done = 7.84 [(11×11-(11²/2)) - (11×0-(0²/2))]

=7.84(60.5 -0) = 474.32 Joules

Work done in lifting the water

At Height of x meters (0≤x≤11)

Rate of water leakage = 36kg ÷ 11m = $\frac{36}{11}$ kg/m

Mass of rope = weight of rope × change in distance

= $\frac{36}{11}$ kg/m × (11-x)m = 3.27kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = $\int\limits^1 _0 {9.8*3.27(11-x)} \, dx$

Note : upper limit is 11 not 1, problem with math editor

W done = 32.046 (11x-x²/2)upper limit 11 to lower limit 0

W done = 32.046 [(11×11-(11²/2)) - (11×0-(0²/2))]

= 32.046(60.5 -0) = 1938.783 Joules

the work done to lift the bucket =W done by bucket+ W done by rope +W done by water)

the work done to lift the bucket = 1078 +474.32+1938.783 = 3491.103

the work done to lift the bucket = 3491 Joules

8 0

3 years ago

A thin-walled cylindrical pressure vessel is subjected to an internal gauge pressure, p=75 psip=75 psi. It had a wall thickness

Mekhanik [1.2K]

To solve this problem we must apply the concept related to the longitudinal effort and the effort of the hoop. The effort of the hoop is given as

$\sigma_h = \frac{Pd}{2t}$

Here,

P = Pressure

d = Diameter

t = Thickness

At the same time the longitudinal stress is given as,

$\sigma_l = \frac{Pd}{4t}$

The letters have the same meaning as before.

Then he hoop stress would be,

$\sigma_h = \frac{Pd}{2t}$

$\sigma_h = \frac{75 \times 8}{2\times 0.25}$

$\sigma_h = 1200psi$

And the longitudinal stress would be

$\sigma_l = \frac{Pd}{4t}$

$\sigma_l = \frac{75\times 8}{4\times 0.25}$

$\sigma_l = 600Psi$

The Mohr's circle is attached in a image to find the maximum shear stress, which is given as

$\tau_{max} = \frac{\sigma_h}{2}$

$\tau_{max} = \frac{1200}{2}$

$\tau_{max} = 600Psi$

Therefore the maximum shear stress in the pressure vessel when it is subjected to this pressure is 600Psi

6 0

3 years ago