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3 years ago

14

A photon with a frequency of 5.48 × 1014 hertz is emitted when an electron in a mercury atom falls to a lower energy level. Iden

tify the color of light associated with this photon.

1 answer:

ludmilkaskok [199]3 years ago

3 0

The color should be green.

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The International Space Station (ISS) orbits Earth at an altitude of 4.08 × 105 m above the surface of the planet. At what veloc

alukav5142 [94]

This question involves the concepts of orbital velocity and orbital radius.

The orbital velocity of ISS must be "7660.25 m/s".

The orbital velocity of the ISS can be given by the following formula:

$v=\sqrt{\frac{GM}{R}}$

where,

v = orbital velocity = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

M = Mass of Earth = 5.97 x 10²⁴ kg

R = orbital radius = radius of earth + altitude = 63.78 x 10⁵ m + 4.08 x 10⁵ m

R = 67.86 x 10⁵ m

Therefore,

$v=\sqrt{\frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(5.97\ x\ 10^{24}\ kg)}{67.86\ x\ 10^5\ m}}$

<u>v = 7660.25 m/s</u>

Learn more about orbital velocity here:

brainly.com/question/541239

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2 years ago

If pressure is increased from 200 kPa to 300 kPa, and the original volume of gas was 1.5 L, what is the new volume? Assume the t

Oxana [17]

Answer:

The answer to your question is: V2 = 1 l

Explanation:

Data

P1 = 200 kPa

P2 = 300 kPa

V1 = 1.5 l

V2 = ?

Formula

P1V1 = P2V2

V2 = (P1V1) / P2

V2 = (200 x 1.5) / 300

V2 = 1 l

6 0

3 years ago

Which statement best describes the law of conservation of energy?

Phantasy [73]

<span>The correct option is C. Energy cannot be created or destroyed. This statement is known as law of conservation of energy, and it implies that whenever a certain form of energy does change, the loss of this form of energy must have converted into an another type of energy. A typical example is an object falling to the ground: initially, the object has gravitational potential energy. As the object falls down, it loses potential energy (since its altitude from the grounf decreases), but it acquires kinetic energy (because its velocity increases). In this example, potential energy has converted into kinetic energy, but the total energy of the object has remained constant.</span>

8 0

3 years ago

Read 2 more answers

A machine part has the shape of a solid uniform sphere of mass 205 g and diameter 4.10 cm. It is spinning about a frictionless a

kondor19780726 [428]

Answer:

Answer is in the following attachment.

Explanation:

3 0

3 years ago

The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me

Sav [38]

Answer:

(a): $\rm meter/ second^2.$

(b): $\rm meter/ second^3.$

(c): $\rm 2ct-3bt^2.$

(d): $\rm 2c-6bt.$

(e): $\rm t=\dfrac{2c}{3b}.$

Explanation:

Given, the position of the particle along the x axis is

$\rm x=ct^2-bt^3.$

The units of terms $\rm ct^2$ and $\rm bt^3$ should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of $\rm ct^2=meter$

Therefore, unit of $\rm c= meter/ second^2.$

(b):

Unit of $\rm bt^3=meter$

Therefore, unit of $\rm b= meter/ second^3.$

(c):

The velocity v and the position x of a particle are related as

$\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.$

(d):

The acceleration a and the velocity v of the particle is related as

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.$

(e):

The particle attains maximum x at, let's say, $\rm t_o$ , when the following two conditions are fulfilled:

$\rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.$
$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Applying both these conditions,

$\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.$

For $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c$

Since, c is a positive constant therefore, for $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0$

Thus, particle does not reach its maximum value at $\rm t = 0\ s.$

For $\rm t_o = \dfrac{2c}{3b}$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.$

Here,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Thus, the particle reach its maximum x value at time $\rm t_o = \dfrac{2c}{3b}.$

7 0

3 years ago