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2 years ago

How many electrons does a neutral atom of the element iron (fe) have?

Physics

1 answer:

borishaifa [10]2 years ago

3 0

Answer:

26 electrons

A neutral iron atom contains 26 protons and 30 neutrons plus 26 electrons in four different shells around the nucleus. As with other transition metals, a variable number of electrons from iron's two outermost shells are available to combine with other elements

Explanation:

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Mr Jones launches an arrow horizontally at a rate of 40m/s off of a 78.4 m cliff towards the south, how far south does the arrow

DanielleElmas [232]

Answer:c

Explanation:its the answer because its the answer

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2 years ago

3. A simple way to state Newton’s first law is:

MrRissso [65]

Explanation:

A simple way to state Newton's first law is:

For every action force, there is a reaction force which is equal in magnitude and opposite in direction.

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2 years ago

Physics laws of motion

expeople1 [14]

The acceleration is $0.095 m/s^2$

Explanation:

We can solve the problem by applying Newton's second law of motion: in fact, the net force acting on an object is equal to the product between the mass of the object and its acceleration. Therefore we can write:

$\sum F = ma$

where:

$\sum F$ is the resultant force acting on the object

m is its mass

a is its acceleration

In this problem, we have the following forces acting on the system:

$F_1 = 2500 N$ (forward)

$F_2 = -2400 N$ (backward)

So, Newton's second law can be rewritten as:

$F_1 -F_2 = ma$

where:

m = 1050 kg is the mass of all the students

Solving the formula for a, we find the acceleration of the system:

$a=\frac{F_1-F_2}{m}=\frac{2500+(-2400)}{1050}=0.095 m/s^2$

Learn more about Newton's second law:

brainly.com/question/3820012

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2 years ago

How much force is needed to accelerate a 1000-kg car at a rate of 3 m/s squared

Burka [1]

Answer:

3000 newton force is required

Explanation:

F = ma

F= 1000 kgs x 3 m/s^ 2

F=3000(kgs x m/s^2)

F=3000 newton

8 0

2 years ago

Show that the electric potential along the axis of a uniformly charged disk of radius R and charge density sigma is given by by

OlgaM077 [116]

Explanation:

Area of ring $\ 2{\pi} a d a$

Charge of on ring $d q=-(\ 2{\pi} a d a)$

Charge on disk

$Q=-\left(\pi R^{2}\right)$

$\begin{aligned}d v &=\frac{k d q}{\sqrt{x^{2}+a^{2}}} \\&=2 \pi-k \frac{a d a}{\sqrt{x^{2}+a^{2}}} \\v(1) &=2 \pi c k \int_{0}^{R} \frac{a d a}{\sqrt{x^{2}+a^{2}}} \cdot_{2 \varepsilon_{0}}^{2} R \\&=2 \pi \sigma k[\sqrt{x^{2}+a^{2}}]_{0}^{2} \\&=\frac{2 \pi \sigma}{4 \pi \varepsilon_{0}}[\sqrt{z^{2}+R^{2}}-(21)] \\&=\frac{\sigma}{2}(\sqrt{2^{2}+R^{2}}-2)\end{aligned}$

Note: Refer the image attached

8 0

3 years ago