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svetlana [45]
2 years ago
9

If an atom has 15 protons, 12 neutrons, and 17 electrons, what is the atom's electrical charge? A. -5 B. -2 C. +3 D. +5

Physics
1 answer:
Nimfa-mama [501]2 years ago
5 0

Answer:

B. -2

Explanation:

The total charge on an atom is the sum of all individual charges present in it. Therefore, the total charge on this atom is given by the following formula:

q = (n_p)(q_p)+(n_e)(q_e)+(n_n)(q_n)

where,

q = total charge on atom = ?

n_p = no. of protons in the atom = 15

n_e = no. of electrons in the atom = 17

n_n = no. of neutrons in the atom = 12

q_p = charge on proton = +1

q_e = charge on electron = -1

q_n = charge on neutron = 0

Therefore,

q = (15)(1)+(17)(-1)+(12)(0)\\q=-2

Hence the correct option is:

<u>B. -2</u>

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Answer:

The Sun looks bigger than other stars because it is so much closer to the Earth. The further away an object is, the smaller it appears, even if it is very big.

Explanation:

However, compared to other stars, our Sun is only a medium-sized star, meaning that some stars are much larger than the Sun and some are much smaller.

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Drivers a and b travel west from boston. driver a leaves three hours earlier traveling at a constant speed of 68mph. driver b fo
Elza [17]

We can answer this using one of the equations of linear motion:

v = d / t

where:

v = velocity

d = distance

t = time

<span>In the problem, we are asked to find for the time in which Driver B will catch up to Driver A. Therefore,  find the time when dA = dB. Rearranging the equation and equation dA and dB will result in:</span>

<span>vA * tA = vB * tB  ---> 1</span>

It was given that:

vA = 68 mph

tA = tB + 3 (since person A was travelling 3 hours earlier)

vB = 85 mph

tB = unknown

Substituting into equation 1:

68 * (tB + 3) = 85 * tB

68 tB + 204 = 85 tB

tB = 12 hrs

Therefore driver B would catch up to driver A after 12 hrs.

 

 

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2 years ago
a thunderclap sends a sound wave through the air and the ocean below The thunderclap sound wave has a constant frequency of 50 H
Montano1993 [528]

Answer:

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Explanation:

amonbgus

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3 years ago
Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field st
Black_prince [1.1K]

Answer:

Part A:

E_{midpoint}=0

Part B:

E_{center}=2711.7558 N/C

Explanation:

Part A:

Formula of Electric Field Strength:

E=\frac{1}{4\pi\epsilon}\frac{xQ}{(x^2+R^2)^{3/2}}

Where:

x is the distance from the ring

R is the radius of the ring

\epsilon is constant permittivity of free space=8.854*10^-12 farads/meter

Q is the charge

For right Ring E at the midpoint can be calculated as:

x for right plate=25/2=12.5 cm=0.125 m

Radius=R=10/2=5 cm=0.05 m

E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{right}=9208.1758 N/C

For Left Ring E at the midpoint can be calculated as:

Since charge on both plates is +ve and same in magnitude, the electric field will be same for both plates.

E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{left}=9208.1758 N/C

Electric Field at midpoint:

Both rings have same magnitude but the direction of fields will be opposite as they have same charge on them.

E_{midpoint}=E_{left}-E_{right}\\E_{midpoint}=9208.1758-9208.1758\\E_{midpoint}=0

Part B:

At center of left ring:

Due to left ring Electric field at center is zero because x=0.

E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0)*(20*10^{-19})}{((0)^2+(0.05)^2)^{3/2}}\\E_{left}=0 N/C

Due to right ring Electric field at center of left ring:

Now: x=25 cm= o.25 m (To the center of left ring)

E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.25)*(20*10^{-19})}{((0.25)^2+(0.05)^2)^{3/2}}\\E_{right}=2711.7558 N/C

Electric Field Strength at center of left ring is same as that of right ring.

E_{center}=2711.7558 N/C

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