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vovikov84 [41]
4 years ago
7

Does ionic bonds loose or gain electrons ?

Physics
1 answer:
Vlada [557]4 years ago
5 0
I believe they lose and gain they slowly even out use a periodic table for help
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An electric field with a magnitude of 6.0 × 104 N/C is directed parallel to the positive y axis. A particle with a charge q = 4.
jekas [21]

Answer:

Electric force, F=2.88\times 10^{-14}\ N

Explanation:

It is given that,

Magnitude of electric field, E=6\times 10^4\ N/C

Charge, q=4.8\times 10^{-19}\ C

The electric field is directed parallel to the positive y axis. We need to find the  force on the charge particle. It is given by :

F=q\times E

F=4.8\times 10^{-19}\ C\times 6\times 10^4\ N/C

F=2.88\times 10^{-14}\ N

So, the electric force on the charge is 2.88\times 10^{-14}\ N. Hence, this is the required solution.

5 0
4 years ago
A proton in a high-energy accelerator is given a kinetic energy of 50.0 GeV. Determine (a) the momentum and (b) the speed of the
koban [17]

(a) The momentum of the proton is determined as 5.17 x 10⁻¹⁸ kgm/s.

(b) The speed of the proton is determined as  3.1 x 10⁹ m/s.

<h3>Momentum of the proton</h3>

The momentum of the proton is calculated as follows;

K.E = ¹/₂mv²

where;

  • m is mass of proton = 1.67 x 10⁻²⁷ kg
  • v is speed of the proton = ?
<h3>Speed of the proton</h3>

v² = 2K.E/m

v² = (2 x 50 x 10⁹ x 1.602 x 10⁻¹⁹ J)/(1.67 x 10⁻²⁷)

v² = 9.6 x 10¹⁸

v = 3.1 x 10⁹ m/s

<h3>Momentum of the proton</h3>

P = mv = (1.67 x10⁻²⁷ x 3.1 x 10⁹) = 5.17 x 10⁻¹⁸ kgm/s

Learn more about momentum here: brainly.com/question/7538238

#SPJ4

4 0
2 years ago
Felix has a bucket of golf balls. The table shows the number of golf balls of each color in the
Gre4nikov [31]

Answer:c

Explanation:

4 0
3 years ago
Left arm has area 10.0 cm2, right has area 5.00 cm2. 100 g of water isadded to right side. a) Determine the length of the water
Masja [62]

Answer:

0.49  cm

Explanation:

We are given that

A_1=10 cm^2

A_2=5 cm^2

Mass of water,m=100 g

a.\rho_w=1 g/cm^3

Volume,V=\frac{mass}{density}=\frac{100}{1}=100 cm^3

Length of water column in the right arm=L=\frac{V}{A_2}=\frac{100}{5}=20 cm

b.\rho_m=13.6 g/cm^3

In equilibrium condition  

Pressure at point A=Pressure at point B

P+\rho_mg(h+h_2)=P+\rho_wgL

hA_1=h_2A_2

h_2=\frac{hA_1}{A_2}

13.6\times 9.8(h+h(\frac{A_1}{A_2}))=1\times 9.8\times 20

13.6\times h(\frac{A_2+A_1}{A_2})=20

h=\frac{20A_2}{13.6(A_2+A_1)}

h=\frac{20\times 5}{13.6\times (10+5)}=0.49 cm

4 0
4 years ago
What kind of fluctuations in weather patterns do you expect to see in your area? Will the fluctuations or changes vary by month,
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