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3 years ago

11

How many number of electron shells does calcium have ?

2 answers:

Furkat [3]3 years ago

8 0

20 electrons have calcium

Natalka [10]3 years ago

7 0

20 electrons and 2 valence electrons

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A center-seeking force related to acceleration is _______ force.

zvonat [6]

<span>The question is 'a centre seeking force related to acceleration is ............... force. The answer is centripetal force. Motion in a curved path is an accelerated motion and it requires a force that will direct the moving object towards the centre of curvature of the path of motion. This centre seeking force is known as centripetal force.</span>

5 0

3 years ago

Column A is in the x-axis, and column B is on the y-axis. Which titles should replace A and B

pogonyaev

Column A: x-axis, input, domain

Column B: y-axis, output, range

Those are other ways to describe them

hope i helped:)

8 0

2 years ago

When responding to sound, the human eardrum vibrates about its equilibrium position. suppose an eardrum is vibrating with an amp

vodomira [7]

745.92 hz (b)
this is hte answer only because i seen it on the question lol

5 0

2 years ago

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

sladkih [1.3K]

Answer:

The work done shall be 14715 Joules

Explanation:

The work done by a force 'F' in a displacement 'dy' is given by

$W=m(y)g\times dy$

At any position 'y' the weight shall be sum of weft of water and weight of string

$\therefore m(y)=m_{water}(y)+m_{string}(y)\\\\m(y)=30(1-\frac{y}{10})+0.9y$

Thus applying values we get

$W=\int m(y)g\times dy\\\\W=\int_{0}^{10}(30(1-\frac{y}{10}+0.9y)\times g)dy\\\\Solving\\W=294.3\int_{0}^{10}(1-\frac{y}{10}+0.9y)dy\\\\W=14715J$

8 0

2 years ago

One end of a rope is tied to the handle of a horizontally-oriented and uniform door. a force f is applied to the other end of th

Vlada [557]

For rotational equilibrium of the door we can say that torque due to weight of the door must be counter balanced by the torque of external force

$F\times L = mg \times \frac{L}{2}$

here weight will act at mid point of door so its distance is half of the total distance where force is applied

here we know that

$mg = 145 N$

now we will have

$F = \frac{mg}{2}$

$F = \frac{145}{2} = 72.5 N$

so our applied force is 72.5 N

7 0

3 years ago