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lesantik [10]
3 years ago
9

When a potassium atom forms an ion, it loses one electron. What is the electrical charge of the potassium ion?

Physics
1 answer:
fomenos3 years ago
6 0
The correct answer is answer choice C. +1. Since electrons have negative charges, losing one electron will cause the atom to have a positive charge of 1. This charge comes from the protons, which, until one electron was lost, balanced out the negative charge of the electrons and caused the atom to be neutral.
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a 1000 khz am radio station broadcasts with a power of 20 kw. how many photons does the transmitter antenna emit each second
Gemiola [76]

1000 khz am radio station broadcasts with a power of 20 kw number of photon emitted per second is 30.16 x 10^30 photon/s.

The frequency of the radio station is:
f
=
1000
k
H
z
=
 1
×
10^6Hz
The transmit power is: P = 20kW = 20 X 10^3 W

The transmit power is: h = 6.63 x 10 ^-34 m^2.kg/s

The number of photon emitted per second = N = P/hf = <u>30.16 x 10^30 </u>photon/s.

1000 khz am radio station broadcasts with a photon of 20 kw1000 khz am radio station broadcasts with a power of 20 kw1000 khz am radio station broadcasts with a power of 20 kw1000 khz am radio station broadcasts with a power of 20 kw1000 khz am radio station broadcasts with a power of 20 kw.1000 khz am radio station broadcasts with a power of 20 kw1000 khz am radio station broadcasts with a power of 20 kw1000 khz am radio station broadcasts with a power of 20 kw.


Learn more about photon on:
brainly.com/question/20912241

#SPJ4

6 0
1 year ago
A boat loaded with a barrel of water floats in a swimming pool. When the water in the barrel is poured overboard, the swimming p
Lady bird [3.3K]

Answer: Remain unchanged

Explanation:

The boat with water barrel overboard floats in swimming pool when weight of the water displaced by the boat is equal to the buoyant force acting on the boat.

When the water in the barrel is poured overboard, the level of the swimming pool level would remain unchanged as the weight of the boat  with the water and barrel would remain unchanged ( as the density and volume of the whole system remains same) and hence, the weight of the water (of the swimming pool) displaced by the boat would remain same.

A boat loaded with a barrel of water floats in a swimming pool. When the water in the barrel is poured overboard, the swimming pool level will <u>remain unchanged. </u>

5 0
3 years ago
HELP I WILL GIVE BRAINLIEST!
Ann [662]

Answer:

Answer:B

Explanation:

Because it all stayed consistant

4 0
2 years ago
Two isolated, concentric, conducting spherical shells have radii R1 = 0.500 m and R2 = 1.00 m, uniform charges q1=+2.00 µC and q
scZoUnD [109]

Complete Question

The diagram for this question is shown on the first uploaded image  

Answer:

a E =1.685*10^3 N/C

b E =36.69*10^3 N/C

c E = 0 N/C

d V = 6.7*10^3 V

e   V = 26.79*10^3V

f   V = 34.67 *10^3 V

g   V= 44.95*10^3 V

h    V= 44.95*10^3 V

i    V= 44.95*10^3 V

Explanation:

From the question we are given that

       The first charge q_1 = 2.00 \mu C = 2.00*10^{-6} C

       The second charge q_2 =1.00 \muC = 1.00*10^{-6}

      The first radius R_1 = 0.500m

      The second radius R_2 = 1.00m

 Generally \ Electric \ field = \frac{1}{4\pi\epsilon_0}\frac{q_1+\ q_2}{r^2}

And Potential \ Difference = \frac{1}{4\pi \epsilon_0}   [\frac{q_1 }{r}+\frac{q_2}{R_2} ]

The objective is to obtain the the magnitude of electric for different cases

And the potential difference for other cases

Considering a

                      r  = 4.00 m

           E = \frac{((2+1)*10^{-6})*8.99*10^9}{16}

                = 1.685*10^3 N/C

Considering b

           r = 0.700 m \ , R_2 > r > R_1

This implies that the electric field would be

            E = \frac{1}{4\pi \epsilon_0}\frac{q_1}{r^2}

             This because it the electric filed of the charge which is below it in distance that it would feel

            E = 8*99*10^9  \frac{2*10^{-6}}{0.4900}

               = 36.69*10^3 N/C

   Considering c

                      r  = 0.200 m

=>   r

 The electric field = 0

     This is because the both charge are above it in terms of distance so it wont feel the effect of their electric field

       Considering d

                  r  = 4.00 m

=> r > R_1 >r>R_2

Now the potential difference is

                  V =\frac{1}{4\pi \epsilon_0} \frac{q_1 + \ q_2}{r} = 8.99*10^9 * \frac{3*10^{-6}}{4} = 6.7*10^3 V

This so because the distance between the charge we are considering is further than the two charges given  

          Considering e

                       r = 1.00 m R_2 = r > R_1

                V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2}  ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{1.00} \frac{1.00*10^{-6}}{1.00} ] = 26.79 *10^3 V

          Considering f

              r = 0.700 m \ , R_2 > r > R_1

                      V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2}  ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.700} \frac{1.0*10^{-6}}{1.00} ] = 34.67 *10^3 V

          Considering g

             r =0.500\m , R_1 >r =R_1

   V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2}  ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V

          Considering h

                r =0.200\m , R_1 >R_1>r

  V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2}  ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V

           Considering i    

   r =0\ m \ , R_1 >R_1>r

  V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2}  ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V

8 0
3 years ago
An object is hung from a spring balance attached to the ceiling of an elevator cab. The balance reads 65 N when the cab is stand
lana66690 [7]

Answer:

(A) Reading will be 65 N

(B) Net force on the elevator will be 49.076 N    

Explanation:

We have given the balance force = 65 N

Acceleration due to gravity g=9.8m/sec^2

We know that W=mg

So 65=m\times 9.8

m = 6.632 kg

(a) In first case as the as the speed is constant so the force on the elevator will be 65 N

(B) In second case as the elevator is decelerating at a rate of 2.4m/sec^2

So net acceleration = 9.8-2.4=7.4m/sec^2

So net force on elevator will be = m× net acceleration = 6.632×7.4 = 49.076 N

7 0
3 years ago
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