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cestrela7 [59]
2 years ago
5

What is the energy of the electron in a helium ion with a charge of +1 in an orbit with a value of n=4?

Physics
1 answer:
goldenfox [79]2 years ago
8 0

Answer:

The energy of the electron in a helium ion is -5.44\times10^{-19}\ J

Explanation:

Given that,

Charge = +1

Value of n = 4

Suppose k=2.179\times10^{-18}\ J

We need to calculate the energy of the electron in a helium ion

Using formula of energy

E_{n}=-k\dfrac{z^2}{n^2}

Put the value into the formula

E_{n}=-2.179\times10^{-18}\times\dfrac{2^2}{4^2}

E_{n}=-5.44\times10^{-19}\ J

Hence, The energy of the electron in a helium ion is -5.44\times10^{-19}\ J

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Complete the passage to summarize factors affecting the speed of a wave. The material or substance that a wave moves through is
pantera1 [17]

the first one is medium, the second one is type, and the third one is temperature . if i gave the correct answer, please give best answer x

6 0
2 years ago
Read 2 more answers
Consider two less-than-desirable options. In the first you are driving 30 mph and crash head-on into an identical car also going
DerKrebs [107]

Answer:

The impact force will be same for both the cases.

Explanation:

The rate of change of momentum is known as the Impulse and is given by:

I = \frac{\Delta p}{\Delta t}

where

I = Impulse

\Delta p = change in momentum

\Delta t = time interval

Now,

In first case both the cars are identical and have same velocity and in the second case, the wall is stationary.

Also, in both the cases the car does not bounces off the things it hit.

Thus

\Delta p = 0 - m\times v = - mv

Thus

Impact force, F = \frac{\Delta p}{\Delta t} = \frac{m\Delta v}{\Delta t}

Therefore, impact force is same for both the cases.

5 0
2 years ago
Read 2 more answers
A capacitor with C = 6.00 μF is fully charged by connecting it to a battery that has emf 50.0 V. The capacitor is disconnected f
Arte-miy333 [17]

Answer:

1.99×10^-4coulombs

Explanation:

The charge (Q) across the resistor the directly proportional to the voltage (V) where capacitance of the capacitor(C) is the proportionality constant. Mathematically, Q = CV

If V is the voltage across the resistor, V = IR (according to ohm's law) where I is the current in the resistor and R is the resistance.

We need to calculate the voltage on the resistor first when 0.18A current is passed through it.

V = 0.18 × 185

V = 33.3Volts

The charge Q on the resistor will be;

Q = CV

Were C = 6.00 μF, V = 33.3

Q= 6×10^-6 ×33.3

Q = 0.0001998

Q= 1.99×10^-4Coulombs

4 0
2 years ago
Keisha looks out the window from a tall building at her friend Monique standing on the ground, 8.3 m away from the side of the b
Salsk061 [2.6K]

Answer:

Explanation:

GIVEN DATA:

Distance between keisha and her friend 8.3 m

angle made by keisha toside building 30 degree

height of her friend monique is 1.5 m

from the figure

\Delta ACB

tan 30 = \frac{8.3}{h}

h= \frac{8.3}{tan 30} = 14.376 m

therefore

height of keisha is = h  + 1.5 m

                               = 14.376 + 1.5

= 15.876 \simeq 16 m

therefore option c is correct

5 0
2 years ago
Two 8.0 Ω lightbulbs are connected in a 12 V series circuit. What is the power of both glowing bulbs?
V125BC [204]

Answer:

18 W

Explanation:

Applying,

P = V²/R.................. Equation 1

Where P = Power of both glowing bulbs, V = Voltage, R = Combined Resistance of both bulbs

Since: It is a series circuit,

Then,

R = R1+R2............. Equation 2

Where R1= Resistance of the first bulb, R2 = Resistance of the second bulb

Given: R1 = R2 = 8 Ω

Substitute into equation 1

R = 8+8

R = 16 Ω

Also Given: V = 12 V

Substitute into equation 1

P = 12²/8

P = 144/8

P = 18 W

7 0
2 years ago
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