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2 years ago

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 linear impulse delivered by the hit of a boxer is 202 N · s during the 0.244 s of contact. What is the magnitude of the aver

zlopas [31]

Answer: Magnitude of the average force exerted on the glove by the other boxer is 827.86 N (approximately 828 N).

Explanation: Impulse is defined as the force acting on an object for a short period or interval of time.

Mathematically it is given by the relation:

Impulse = Force $\times$ Time

According to the numerical values given in the question, I = 202 Ns and T = 0.244 s

So, Force F = $\frac{Impulse}{Time}$ = $\frac{202}{0.244}$ = 827.86 N

Magnitude of the average force exerted on the glove by the other boxer is 827.86 N (approximately 828 N).

7 0

2 years ago

Pls help, me, i nees to submit now

Pavlova-9 [17]

Answer:

12m

Explanation:

To obtain the answer to the question given, we must observe the characteristics of image formed by a plane mirror.

The image formed by a plane mirror have the following characteristics:

1. Laterally inverted.

2. Same distance as the object from the mirror.

3. Same height as the object.

4. Virtual.

With the above information, we can calculate the distance between the boy and his image as follow:

Initially:

Object distance (u) = 4m

Image distance (v) = 4m

The boy moved 2m away, therefore:

Object distance (u) = 2 + 4 = 6m

Image distanc(v) = 2 + 4 = 6m

The distance between the boy and his image will be the sum of his distance (u) and image distance (v) i.e (u + v)

The distance between the boy and his image = 6 + 6 = 12m

Therefore, the distance between the boy and his image is 12m.

6 0

2 years ago

The electric field between square the plates of a parallel-plate capacitor has magnitude E. The potential across the plates is m

Helga [31]

Answer:

The magnitude of the electric field between the plates is half its initial value.

Explanation:

We know the electric field E = V/d where V = voltage applied and d = separation between plates.

Since V is constant and V = Ed,

So, E₁d₁ = E₂d₂ where E₁ = initial electric field at separation d₁, d₁ = initial separation of plates, E₂ = final electric field at separation d₂ and d₂ = final separation of plates.

So, E₂ = E₁d₁/d₂

Now, the distance between the plates is twice their original separation. Thus, d₂ = 2d₁

So, E₂ = E₁d₁/2d₁ = E₁/2

So, E₂ = E₁/2

Thus, the magnitude of the electric field between the plates is half its initial value.

5 0

3 years ago

The free-body diagram of a crate is shown. A free body diagram with 4 force vectors. The first vector is pointing downward, labe

satela [25.4K]

The net force acting on the crate is determined as 176 N to the left.

<h3>Net force acting on the crate</h3>

The net force acting on the crate is calculated as follows;

∑F = F1 + F2 + F3 + F4

F(net) = -440y + 176x + 440y - 352x

F(net) = -176 x

The resultant force is pointing in negative x direction.

Thus, the net force acting on the crate is determined as 176 N to the left.

Learn more about net force here: brainly.com/question/14361879

#SPJ1

8 0

1 year ago

A Van de Graaff generator is one of the original particle accelerators and can be used to accelerate charged particles like prot

Rzqust [24]

Answer:

595391.482946 m/s

$3.21875\times 10^{6}$

Explanation:

E = Energy = 1.85 keV

I = Current = 5.15 mA

e = Charge of electron = $1.6\times 10^{-19}\ C$

t = Time taken = 1 second

m = Mass of proton = $1.67\times 10^{-27}\ kg$

Velocity of proton is given by

$v=\sqrt{\dfrac{2E}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 1.85\times 10^3\times 1.6\times 10^{-19}}{1.67\times 10^{-27}}}\\\Rightarrow v=595391.482946\ m/s$

The speed of the proton is 595391.482946 m/s

Current is given by

$I=\dfrac{\Delta Q}{t}\\\Rightarrow \Delta Q=It\\\Rightarrow \Delta Q=5.15\times 10^{-3}\times (1\ sec)\\\Rightarrow Q=5.15\times 10^{-3}\ C$

Number of protons is

$n=\dfrac{Q}{e}\\\Rightarrow n=\dfrac{5.15\times 10^{-3}}{1.6\times 10^{-19}}\\\Rightarrow n=3.21875\times 10^{6}\ protons$

The number of protons is $3.21875\times 10^{6}$

3 0

3 years ago