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3 years ago

Once an object enters orbit, what keeps the object moving sideways?

Physics

1 answer:

zubka84 [21]3 years ago

7 0

Answer: Inertia!!

Explanation: I just completed the edg quiz and got that answer correct! Hope its not too late for you!

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How would the model change as the atom forms bonds? The third shell would have eight electrons after the atom gains seven electr

atroni [7]

Answer:

The third shell would be empty, so the eight electrons on the second level would be the outermost after the atom lost one electron

Explanation:

When an atom is bonded with other atoms, a more stable configuration must be reached, which is why the energy of the molecule is less than the energy of the individual atoms, for this to happen in general, electrons are shared or lost and gained in each atom, depending on the electronegative of the same.

If we analyze an atom within the molecule, its last shell is full, in the case of atoms with few electrons in this shell, they are lost and in the case of many electors in this shell, it gains electrons to have eight (8) in total.

When reviewing the different answers, the correct one is:

* The third shell would be empty, so the eight electrons on the second level would be the outermost after the atom lost one electron

4 0

2 years ago

Read 2 more answers

Object a and object b are both in motion when they collide with each other. They then continue in a new direction unaffected by

Harrizon [31]

This is an elastic collision

bcuz i think they move apart after the collision

sorry if im wrong

8 0

3 years ago

A 5.93 kg ball is attached to the top of a vertical pole with a 2.35 m length of massless string. The ball is struck, causing it

Delvig [45]

Answer

given,

mass of ball = 5.93 kg

length of the string = 2.35 m

revolve with velocity of 4.75 m/s

acceleration due to gravity = 9.81 m/s²

T cos θ = mg

T cos θ = $5.93\times 9.81$

T cos θ = 58.17

$T sin \theta =\dfrac{mv^2}{r}$

$T sin \theta =\dfrac{5.93\times 4.75^2}{2.35 sin \theta}$

$T sin^2 \theta =56.93$

$sin^2 \theta = 1 - cos^2 \theta$

$T (1 - cos^2 \theta) =56.93$

$T (1 - (\dfrac{58.17}{T})^2) =56.93$

T² - 56.93T - 3383.75 = 0

T = 93.22 N

$cos \theta = \dfrac{58.17}{93.22}$

θ = 51.39°

6 0

3 years ago

A 5.00-A current runs through a 12-gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has 8.5 * 1028 free ele

evablogger [386]

Answer:

a)n= 3.125 x $10^{19$ electrons.

b)J= 1.515 x $10^{6$ A/m²

c) $V_{d$ =1.114 x $10^{4$ m/s

d) see explanation

Explanation:

Current 'I' = 5A =>5C/s

diameter 'd'= 2.05 x $10^{-3$ m

radius 'r' = d/2 => 1.025 x $10^{-3$ m

no. of electrons 'n'= 8.5 x $10^{28}$

a) the amount of electrons pass through the light bulb each second can be determined by:

I= Q/t

Q= I x t => 5 x 1

Q= 5C

As we know that: Q= ne

where e is the charge of electron i.e 1.6 x $10^{-19$ C

n= Q/e => 5/ 1.6 x $10^{-19$

n= 3.125 x $10^{19$ electrons.

b) the current density 'J' in the wire is given by

J= I/A => I/πr²

J= 5 / (3.14 x (1.025x $10^{-3$ )²)

J= 1.515 x $10^{6$ A/m²

c) The typical speed' $V_{d$ ' of an electron is given by:

$V_{d$ = $\frac{J}{n|q|}$

=1.515 x $10^{6$ / 8.5 x $10^{28}$ x |-1.6 x $10^{-19$ |

$V_{d$ =1.114 x $10^{4$ m/s

d) According to these equations,

J= I/A

$V_{d$ = $\frac{J}{n|q|}$ = $\frac{I}{nA|q|}$

If you were to use wire of twice the diameter, the current density and drift speed will change

Increase in the diameter increase the cross sectional area and decreases the current density as it has inverse relation.

Also drift velocity will decrease as it is inversely proportional to the area

5 0

3 years ago

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What did scientists predict the big bang should have left ??

ELEN [110]

<span>radiation, hydrogen, and helium </span>

4 0

3 years ago

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