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3 years ago

How do you determine the number of electrons in the shells

Physics

1 answer:

snow_tiger [21]3 years ago

4 0

Periodic table

Find your element

Locate the element atomic number

Determine the number of electrons

Look for the atomic mass of the element

subtract the atomic number from the atomic number

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a 25 kg object is pushed with a horizontal force of 5 N esat across a table. If the force of friction is 2 N, what is the accele

Julli [10]

Answer:

Apply Newton's second law in the moving direction.

Explanation:

$\begin{aligned}F &= ma\\5N-2N &= 25kg\times a\\a & = \frac{3N}{25kg}\\&= 0.12ms^{-2}\end{aligned}$

Friction force applies in the opposite direction of motion; as a restriction.

8 0

3 years ago

A 2.20 kg mass is at the origin. A

Karolina [17]

Answer:

Explanation:

F1 = G•2.2•4.66/3² (pointed right)

F2 = G•2.2•4.66/3² (pointed left)

subtract the two to get zero

3 0

2 years ago

Read 2 more answers

Which statement describes the redox reaction involved in photosynthesis?

NeTakaya

Answer:

D. O2 is removed from the atmosphere, and CO2 is released

Explanation:

Hope it helps you

3 0

3 years ago

Read 2 more answers

An object of mass 6.00 kg falls with an aceleration of 8.00 m/s2. The magnitude of air resitance must be ____ N

allsm [11]

Using Newton's Second Law, we can find the air resistance. We know the net force is equal to mass times acceleration. $F_{net} = m*a = (6.00kg)(8.00 \frac{m}{s^2}) = 48N 

F_{g} - F_{d} = 48N

48N = (6.00kg)(9.81m/s^2) - F_{d} 

F_{d} = 10.86N$

7 0

3 years ago

A bucket of water of mass 10 kg is pulled at constant velocity up to a platform 30 meters above the ground. This takes 10 minute

Rudik [331]

Answer:

W = 2352 J

Explanation:

Given that:

mass of the bucket, M = 10 kg

velocity of pulling the bucket, v = 3 $m.min^{-1}$

height of the platform, h = 30 m

time taken, t = 10 min

rate of loss of water-mass, m = $0.4 kg.min^{-1}$

Here, according to the given situation the bucket moves at the rate,

$v=3 m.min^{-1}$

The mass varies with the time as,

$M=(10-0.4t) kg$

Consider the time interval between t and t + ∆t. During this time the bucket moves a distance

∆x = 3∆t meters

So, during this interval change in work done,

∆W = m.g∆x

<u>For work calculation:</u>

$W=\int_{0}^{10} [(10-0.4t).g\times 3] dt$

$W= 3\times 9.8\times [10t-\frac{0.4t^{2}}{2}]^{10}_{0}$

$W= 2352 J$

5 0

3 years ago