The atomic mass of an atom is determined by which of the following?

Reducing, reusing, and recycling in your office is likely to _______.

iris [78.8K]

Conserve natural resources, energy and landfill space.

4 0

3 years ago

How would the composition of an atom change if both the atomic number and mass number each increase by one? A) The atom would ha

weqwewe [10]

Choice-B is the correct one.

-- The atomic number is the number of protons in the nucleus.
-- Each proton in the nucleus is usually matched by one electron in the 'cloud'.
-- The addition of a proton OR a neutron increases the mass number by 1 .
-- Electrons have such small mass that they don't figure into the atomic mass at all.

8 0

3 years ago

A man runs at an average speed of 5.0m/s how long will it take him to run 5.2km on a perfectly straight line

Anni [7]

The time taken is 1040 s.

<h3>What is speed?</h3>

The term speed refers to the rate at which the distance changes per unit time. This is why we define speed as the ratio of the distance to time for a body that is moving along a straight line.

Now;

We must first convert the distance to meters;

distance = 5.2km or 5200m

Speed = distance/time

time = distance/speed

time = 5200m/5 m/s

time = 1040 s

Learn more about speed:brainly.com/question/28224010

#SPJ1

8 0

1 year ago

Describe three pieces of evidence that support the Big Bang theory

lisov135 [29]

Explanation :

According to astronomers, the whole universe is started with a giant explosion called as Big Bang. Big Bang theory shows that the universe is extended from high density state.

There are some evidence for big bang as :

(1) There are some red shifts of different galaxies which means that the universe is expanding.

(2) Due to the expanding of universe, some of the new elements are created like hydrogen, deuterium etc.

(3) Microwaves are detected by orbiting detectors.

All this parameters shows that big bang theory was correct.

6 0

3 years ago

Read 2 more answers

Planet 1 orbits Star 1 and Planet 2 orbits Star 2 in circular orbits of the same radius. However, the orbital period of Planet 1

hichkok12 [17]

Answer:

The mass of Star 2 is Greater than the mass of Start 1. (This, if we suppose the masses of the planets are much smaller than the masses of the stars)

Explanation:

First of all, let's draw a free body diagram of a planet orbiting a star. (See attached picture).

From the free body diagram we can build an equation with the sum of forces between the start and the planet.

$\sum F=ma$

We know that the force between two bodies due to gravity is given by the following equation:

$F_{g} = G\frac{m_{1}m_{2}}{r^{2}}$

in this case we will call:

M= mass of the star

m= mass of the planet

r = distance between the star and the planet

G= constant of gravitation.

so:

$F_{g} =G\frac{Mm}{r^{2}}$

Also, if the planet describes a circular orbit, the centripetal force is given by the following equation:

$F_{c}=ma_{c}$

where the centripetal acceleration is given by:

$a_{c}=\omega ^{2}r$

where

$\omega = \frac{2\pi}{T}$

Where T is the period, and $\omega$ is the angular speed of the planet, so:

$a_{c} = ( \frac{2\pi}{T})^{2}r$

or:

$a_{c}=\frac{4\pi^{2}r}{T^{2}}$

so:

$F_{c}=m(\frac{4\pi^{2}r}{T^{2}})$

so now we can do the sum of forces:

$\sum F=ma$

$F_{g}=ma_{c}$

$G\frac{Mm}{r^{2}}=m(\frac{4\pi^{2}r}{T^{2}})$

in this case we can get rid of the mass of the planet, so we get:

$G\frac{M}{r^{2}}=(\frac{4\pi^{2}r}{T^{2}})$

we can now solve this for $T^{2}$ so we get:

$T^{2} = \frac{4\pi ^{2}r^{3}}{GM}$

We could take the square root to both sides of the equation but that would not be necessary. Now, the problem tells us that the period of planet 1 is longer than the period of planet 2, so we can build the following inequality:

$T_{1}^{2}>T_{2}^{2}$

So let's see what's going on there, we'll call:

$M_{1}$ = mass of Star 1

$M_{2}$ = mass of Star 2

So:

$\frac{4\pi^{2}r^{3}}{GM_{1}}>\frac{4\pi^{2}r^{3}}{GM_{2}}$

we can get rid of all the constants so we end up with:

$\frac{1}{M_{1}}>\frac{1}{M_{2}}$

and let's flip the inequality, so we get:

$M_{2}>M_{1}$

This means that for the period of planet 1 to be longer than the period of planet 2, we need the mass of star 2 to be greater than the mass of star 1. This makes sense because the greater the mass of the star is, the greater the force it applies on the planet is. The greater the force, the faster the planet should go so it stays in orbit. The faster the planet moves, the smaller the period is. In this case, planet 2 is moving faster, therefore it's period is shorter.

6 0

3 years ago