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3 years ago

What happens when an atom is exposed to an energy source such as electricity?

2 answers:

5 0

I think the correct answer from the choices listed above is the first option. When an atom is exposed to an energy source such as electricity, then the electrons may jump to a higher energy level. This happens because it needs to be in a stable configuration. Hope this answers the question.

Alex73 [517]3 years ago

4 0

A) The electrons may jump to a higher energy level.

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A pogo stick has a spring with a force constant of 2.50 × 104 N/m , which can be compressed 11.2 cm. what maximum height, in met

AleksandrR [38]

Answer:

h = 36.4 cm

Explanation:

given,

spring constant = 2.5 x 10⁴ N/m

compressed distance = 11.2 cm = 0.112 m

mass of the child = 44 kg

maximum height = ?

by energy of conservation

$K.E_i +P.E_i= K.E_f + P.E_f$

$\dfrac{1}{2}kx^2 = mgh$

$\dfrac{1}{2}\times 2.5 \times 10^4 \times 0.112^2 = 44\times 9.8 \times h$

$156.8 = 44 \times 9.8 \times h$

$h = \dfrac{156.8}{44 \times 9.8}$

h = 0.364 m

h = 36.4 cm

7 0

3 years ago

It turns out that Mercury and Mars have the same gravity as one another – that is, you would weigh the same on the surface of Me

Tamiku [17]

Answer:

Explanation:

The value of acceleration due to gravity of mars is same as that the value of acceleration due to gravity of Mercury.

The value of acceleration due to gravity of a planet depends on its mass and the radius.

The formula for the acceleration due to gravity is given by

$g=\frac{GM}{R^{2}}$

Where, M be the mass of the planet, R be the radius of the planet.

As according to the question, the value of acceleration due to gravity for Mercury is same as that of Mars.

$\frac{GM_{mercury}}{R_{mercury}^{2}}=\frac{GM_{mars}}{R_{mars}^{2}}$

$\frac{M_{mercury}}{R_{mercury}^{2}}=\frac{M_{mars}}{R_{mars}^{2}}$

As teh radius of Mercury is small, and acceleration due to gravity is same, it is possible because the mass of Mercury is more than the mass of Mars.

4 0

3 years ago

After the change of state occurs, the temperature _____________.

Artyom0805 [142]

Confused about how much fun you have done to the other day and then I will be sure you guys are gonna right I know you have your own words but you know that I don’t think you know what I mean but you don’t

4 0

4 years ago

A closed container initially holds 50 monatomic Aparticles that have a combined energy of 480 units. After 100 monatomic B parti

Molodets [167]

Answer:

"8 units" is the appropriate answer.

Explanation:

According to the question,

Throughout equilibrium all particles are of equivalent intensity, and as such the integrated platform's total energy has been uniformly divided across all individuals.

Now,

The total energy will be:

= $480+720$

= $1200 \ units$

The total number of particles will be:

= $50+100$

= $150$

hence,

Energy of each A particle or each B particle will be:

= $\frac{1200}{150}$

= $8 \ units$

5 0

3 years ago

Three guns are aimed at the center of a circle, and each fires a bullet simultaneously. The directions in which they fire are 12

ikadub [295]

Answer:

The unknown mass of the bullet is $m1=3.751x10^{-3} kg$

Explanation:

According to Newton's laws of motion, when a net external force acts on a body of mass m , it results in change in momentum of the body and is given by:

$F=\frac{P}{dt}$

Where:

is the linear momentum of the body

As a consequence, when there are no external forces acting on the body the total momentum remains conserved i.e.

Given:

$m_{2}=5.79x10^{-3}kg \\m_{3}=5.79x10^{-3}kg\\v_{2}=v_{3}=392 \frac{m}{s}\\$

For momentum along the y-direction to be zero, it is achieved when the equal masses are moving at angles of

θ1=180°, θ2=60°, θ3=-60°

Therefore, from conservation of momentum along x - direction:

$m_{1}*v_{1}*cos(180)+m_{2}*v_{2}*cos(60)+m_{3}*v_{3}*cos(-60)=0\\$ $m_{1}*605\frac{m}{s}*cos(180)+5.79x10^{-3}kg *392\frac{m}{s}*cos(60)+5.79x10^{-3}kg*392\frac{m}{s}*cos(-60)=0\\$

$-m_{1}*605+5.79x10^{-3}kg*196\frac{m}{s}+5.79x10^{-3}kg*196\frac{m}{s}=0\\m_{1}*605kg= 2.26968\frac{kg*m}{s}\\m_{1}=3.75 x10^{-3} kg$

3 0

3 years ago