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3 years ago

The lowest energy state of a quantized system is

Physics

1 answer:

son4ous [18]3 years ago

3 0

A state in which an atom has more energy than it does at its ground state.

Hopes this helps!

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5 litres of alcohol have a mass of 4kg. calculate the density of alcohol in g/cm.

Aleks04 [339]

Answer: 0.8 g/cm

Explanation:

p= m/V

= 4 kg/ 5 liter

= 0.8

3 0

3 years ago

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If a car has a velocity of 85 km/hr, how long will it take to accelerate to 45 km/hr if the acceleration is -3 km/hr/sec?

weqwewe [10]

Answer:

Time, t = 13.34 seconds.

Explanation:

Given the following data;

Initial velocity, u = 85km/hr to meters per seconds = 85*1000/3600 = 23.61 m/s

Final velocity, v = 45km/hr to meters per seconds = 45*1000/3600 = 12.5 m/s

Acceleration, a = -3 km/hr/sec to meters per seconds square = -3*1000/3600 = -0.833m/s²

To find the time;

Acceleration = (v - u)/t

-0.833 = (12.5 - 23.61)/t

-0.833t = -11.11

t = 11.11/0.833

Time, t = 13.34 seconds.

6 0

2 years ago

Solve the science problem

scZoUnD [109]

Answer:

What is the problem I cant help unless you have the problem.

Explanation:

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2 years ago

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A 20 g bullet is shot from a 10 kg gun with a velocity of 400 m/s. What

Nady [450]

Answer: -.80m/s

Explanation:

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3 years ago

A point charge with charge q1 = 4.00 μC is held stationary at the origin. A second point charge with charge q2 = -4.40 μC moves

Bezzdna [24]

Answer:

W=0.94J

Explanation:

Electrostatic potential energy is the energy that results from the position of a charge in an electric field. Therefore, the work done to move a charge from point 1 to point 2 will be the change in electrostatic potential energy between point 1 and point 2.

This energy is given by:

$U=\frac{K\left |q_1 \right |\left |q_2 \right |}{r}\\$

So, the work done to move the chargue is:

$W=U_1-U_2\\W=\frac{K\left |q_1 \right |\left |q_2 \right |}{r_1}-\frac{K\left |q_1 \right |\left |q_2 \right |}{r_2}\\r_1=\sqrt{((0.155 m)^2+0 m)^2}=0.115m\\r_2=\sqrt{((0.245 m)^2+(0.270 m)^2}=0.365m\\W=K\left |q_1 \right |\left |q_2 \right |(\frac{1}{r_1}-\frac{1}{r_2})\\W=8.99*10^9\frac{Nm^2}{c^2}(4.00*10^{-6}C)(4.40*10^{-6}C)(\frac{1}{0.115m}-\frac{1}{0.365})\\W=0.94J$

The work is positive since the potential energy in 1 is greater than 2.

5 0

3 years ago