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slega [8]
2 years ago
11

Removing an electron from a neutral atom will result in an atom that?

Physics
2 answers:
Alex73 [517]2 years ago
8 0

Removing an electron from a neutral atom will result in an atom that is positive.

son4ous [18]2 years ago
7 0

Answer:

give me brainllest if this is right.

Explanation:

No, the atom will get a positive charge.

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The flywheel of an engine has moment of inertia 2.50 kg m2 about its rotation axis. What constant torque is required to bring it
MrRissso [65]

Answer:

Explanation:

From the question we are told that

   The moment of inertia is  I = 2.50 \ kg \cdot m^2

    The final  angular speed is w_f =  400 rev/min  =  \frac{400 * 2\pi}{60}  = 41.89 \ rad/s

     The time taken is  t =  8.0 s

      The initial angular speed is  w_i  =  0\ rad/s

Generally the average angular acceleration is mathematically represented as

        \alpha  =  \frac{w_f - w_i }{t}

=>     \alpha  =  \frac{41.89}{8}

=>      \alpha  = 5.24 \ rad/s^2

Generally the torque is mathematically represented as

   \tau  =  I  *  \alpha

=>    \tau   =  5.24 *  2.50

=>     \tau   =  13.09 \  N \cdot m

5 0
3 years ago
What would the speed wave be ?
Mkey [24]
It would be 12hz because it
6 0
3 years ago
Read 2 more answers
A 5kg ball is on top of the school building at a height of 40m above the ground.
mojhsa [17]

Answer:

A-Caclcuate the potential energy of the ball at that height

Explanation:

(a). Mass of the Body = 10 kg.

Height = 10 m.

Acceleration due to gravity = 9.8 m/s².

Using the Formula,Potential Energy = mgh

= 10 × 9.8 × 10 = 980 J.

(b). Now, By the law of the conservation of the Energy, Total amount of the energy of the system remains constant.

∴ Kinetic Energy before the body reaches the ground is equal to the Potential Energy at the height of 10 m.

∴ Kinetic Energy = 980 J.

(c). Kinetic Energy = 980 J.

Mass of the ball = 10 kg.

∵ K.E. = 1/2 × mv²

∴ 980 = 1/2 × 10 × v²

∴ v² = 980/5

⇒ v² = 196

∴ v = 14 m/s.

3 0
2 years ago
The spring is released and a 0.10-kilogram plastic sphere is fired from the launcher. Calculate the maximum speed with which the
tigry1 [53]

Answer:

A) The elastic potential energy stored in the spring when it is compressed 0.10 m is 0.25 J.

B) The maximum speed of the plastic sphere will be 2.2 m/s

Explanation:

Hi there!

I´ve found the complete problem on the web:

<em>A toy launcher that is used to launch small plastic spheres horizontally contains a spring with a spring constant of 50. newtons per meter. The spring is compressed a distance of 0.10 meter when the launcher is ready to launch a plastic sphere.</em>

<em>A) Determine the elastic potential energy stored in the spring when the launcher is ready to launch a plastic sphere.</em>

<em>B) The spring is released and a 0.10-kilogram plastic sphere is fired from the launcher. Calculate the maximum speed with which the plastic sphere will be launched. [Neglect friction.] [Show all work, including the equation and substitution with units.]</em>

<em />

A) The elastic potential energy (EPE) is calculated as follows:

EPE = 1/2 · k · x²

Where:

k = spring constant.

x = compressing distance

EPE = 1/2 · 50 N/m · (0.10 m)²

EPE = 0.25 J

The elastic potential energy stored in the spring when it is compressed 0.10 m is 0.25 J.

B) Since there is no friction, all the stored potential energy will be converted into kinetic energy when the spring is released. The equation of kinetic energy (KE) is the following:

KE = 1/2 · m · v²

Where:

m = mass of the sphere.

v = velocity

The kinetic energy of the sphere will be equal to the initial elastic potential energy:

KE = EPE = 1/2 · m · v²

0.25 J = 1/2 · 0.10 kg · v²

2 · 0.25 J / 0.10 kg = v²

v = 2.2 m/s

The maximum speed of the plastic sphere will be 2.2 m/s

6 0
3 years ago
O
Sergio [31]

see

below

Explanation:

refractive index = speed of light in vacuum / speed of light in medium

light travels at a speed of 3.0 x 10^8 m/s in vacuum

refractive index = 3.0 x 10^8 / 2.0 x 10^8

refractive index = 1.5

hope this helps, please mark it

4 0
2 years ago
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