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3 years ago

Every atom contains a .............. which is positively charged

Physics

1 answer:

Arisa [49]3 years ago

5 0

Answer:

Proton

Explanation:

The nucleus has an overall positive charge as it contains the protons. Every atom has no overall charge (neutral). This is because they contain equal numbers of positive protons and negative electrons.

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The age of the universe is around 100,000,000,000,000,000s. A top quark has a lifetime of roughly 0.000000000000000000000001s. W

Maslowich

Answer:

10⁴¹ s quark top lives have been in the history of the universe.

Explanation:

You need to determine how many quark top lives there have been in the history of the universe, that is, what is the age of the universe divided by the lifetime of a top quark. Expressed in a formula, this is:

$t\frac{Age of the universe}{Lifetime of a top quark}$

Yo know that the "Age of the universe" is 100,000,000,000,000,000 which can also be expressed as 10¹⁷ s .

You also know that the "Lifetime of a top quark" is 0.000000000000000000000001 which can also be expressed as 10⁻²⁴ s.

Then $t=\frac{10^{17} }{10^{-24} }$

Recalling that the result of dividing two powers of the same base is another power with the same base where the exponent is the subtraction of the initial exponents, it is possible to calculate this division as follows:

$t=10^{17-(-24)}$

$t=10^{17+24}$

t=10⁴¹ s

So 10⁴¹ s quark top lives have been in the history of the universe.

5 0

3 years ago

Which of the following is not an example of kinetic energy being converted to potential energy?

KengaRu [80]

The list of choices you provided with your question
is utterly devoid of any such examples.

6 0

4 years ago

Read 2 more answers

Aluminum oxide can be produced during rocket launches. Show that the sum of positive and negative charges in a unit of Al2O3 equ

Triss [41]

Answer:

The sum of positive and negative charges in a unit of Al2O3 equals zero.

Aluminium has a charge of +3 while Oxygen has a charge of -2 on each ion.

Al203 has 2 Al atoms and 3 O atoms.

Charge on Al2O3 = 2(charge on Al ion) + 3(charge on O ion)

= 2(3) + 3(-2)

= 6 - 6

= 0

Explanation:

Aluminium has 3 electrons in the outermost shell and has the tendency to lose those 3 electrons to form a positive ion and have a complete outermost shell.

Whereas, Oxygen has 6 electrons in the outermost and has the tendency to accept two more electrons to form a negative ion and have a complete outermost shell.

4 0

3 years ago

Can someone solve this problem and explain to me how you got it

evablogger [386]

Answer:

question5: F=74312.5N

question6: charge at the end of antenna=0.37N

Explanation:

Coulomb's law: the magnitude of the force of attraction or repulsion due to two charges is proportional to the product of the magnitude of the charges and inversely proportional to the square of distance between the charges.

⇒ $F\alpha\frac{q1*q2}{r^{2}}$

∴ $F=k\frac{q1*q2}{r^{2}}$

where $F$ is the force of attraction or repulsion

$k$ is Coulumb's constant= $9*10^{9}Nm^{2}C^{-2}$

$q1$ and $q2$ are the magnitude of the charges

$r$ is the distance between two charges

The force between the two charges is attractive if they are of different polarity

The force between the two charges is repulsive if they are of same polarity

Question5:

Given: q1=0.041 C, q2=0.029 C, r=12 m

therefore by Coulumb's law,

$F=9*10^{9}*\frac{0.041*0.029}{12^{2}}$

$F=74312.5N$

Question6:

Given: q1= $3*10^{-18}C$ , r=5 m, F= $4*10^{-11}N$

therefore by Coulumb's law,

$4*10^{-11}=9*10^{9}*\frac{3*10^{-18}*q2}{5^{2}}$

⇒ $q2=\frac{4*10^{-11}*25}{9*10^{9}*3*10^{-18}} \\=0.37C$

4 0

4 years ago

block of mass 5kgriding on a horizontal frictionlessxy-plane surface is subjected tothree applied forces:→F1= 12√2N[ 45◦]→F2= (8

dsp73

Answer:

(i) See attached image for the drawing

(ii) net force given in component form: (20, 20)N with magnitude: $\sqrt{800} \,\,\,N$

Explanation:

First try to write all forces in vector component form:

The force F1 acting at 45 degrees would have multiplication factors of $\frac{\sqrt{2} }{2}$ on both axes, to take care of the sine and cosine projections. Therefore, the:

x-component of F1 is $F1_x=12\,\sqrt{2} \frac{\sqrt{2} }{2} =12\,\,N$

y-component of F1 is $F1_y=12\,\sqrt{2} \frac{\sqrt{2} }{2} =12\,\,N$

As far as force F2, it is given already in x and y components, then:

x-component of F2 = 8 N

y-component of F2 = -6 N (negative meaning pointing down the y-axis)

Force F3 has only component (upwards) in the y-direction

x-component of F3 = 0 N

y-component of F3 =14 N

The additions of all these component by component, gives the resultant force (R) acting on the 5 kg mass:

x-component of R = 12 + 8 = 20 N

y-component of R = 12 + 14 - 6 = 20 N

Therefore, the acceleration that the mass receives due to this force is given in component form as:

x-component of acceleration: 20 N / 5 kg = $4\,\,\,m/s^2$

y-component of acceleration: 20 N / 5 kg = $4\,\,\,m/s^2$

Now we can calculate the components of the velocity of this mass after 2 seconds of being accelerated by this force, using the formula of acceleration times time:

x-component of the velocity is: $v_x=4\,*\,2=8\,m/s$

y-component of the velocity is: $v_y=4\,*\,2=8\,m/s$

7 0

3 years ago