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2 years ago

15

Vera and her three sisters received the same amount of money to go to the school festival. Each girl spent $12. Afterward, the g

irls had $24 altogether. determine the amount of money each girl recived

1 answer:

rodikova [14]2 years ago

7 0

Answer:

$18

Explanation:

Each of the was qiven $18

All together =24

24 divide it by 4 (because there were 4 people there)

= 6 (6+6+6+6)

= 24

=6+12=18

.Hence each of them was given $18

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Mark all of the antimatter particlesa) Proton b) Electron c)Anti-top d) Gluon e) Tau Neutrino

Bond [772]

Answer:

c) Anti-top

Explanation:

Protons are particles which belong to the hadron group of particles. The antimatter equivalent particle is antiproton

Electrons are particles that belong to the fermion group of particles. The antimatter equivalent is antielectron

Antitop quarks are particles that belong to the group family of particles. They are the antimatter equivalent of top quarks.

Gluons are particles that belong to the group of particles called bosons.

Tau neutrinos are particles that belong to the fermion group of particles. The antimatter equivalent is Tau antineutrino.

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3 years ago

The current supplied by a battery as a function of time is I(t) = (0.88 A) e^(-t*6 hr). What is the total number of electrons tr

Nimfa-mama [501]

Answer:

$N = 1.2 \times 10^{23}$

Explanation:

As we know that the current through the battery is given as

$i = (0.88 A) e^{-t/6}$

here from above equation we know that current will become zero when time elapsed is very large

so here we can say that charge will flow through the battery from t = 0 to t = infinite

now we have

$q = \int i dt$

$q = \int (0.88 A) e^{-t/6} dt$

$q = 0.88\times -6(3600)\times (e^{-t/6})$

$q = -1.90 \times 10^{4} (0 - 1)$

$q = 1.90 \times 10^{4} C$

As we know that

$q = Ne$

$N = \frac{q}{e}$

$N = \frac{1.90 \times 10^4}{1.6 \times 10^{-19}}$

$N = 1.2 \times 10^{23}$

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3 years ago

A 0.0400-g positive charged ball with charge q = 6.40 μC is resting on a flat, frictionless horizontal surface. For a time of t

Leto [7]

Answer:

The height is 0.1014 m

Explanation:

Given that,

Mass = 0.0400 g

Charge $q= 6.40\ \mu C$

Time t = 0.0420 s

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We need to calculate the electric force on the particle

Using formula of electric force

$F=qE$

Put the value into the formula

$F_{e}=6.40\times10^{-6}\times7.80\times10^{2}$

$F_{e}=0.004992= 0.499\times10^{-2}\ N$

We need to calculate the gravitational force

Using formula of force

$F=mg$

Put the value into the formula

$F_{g}=0.0400\times10^{-3}\times9.8$

$F_{g}=0.000392 = 0.392\times10^{-3}\ N$

We need to calculate the net force

$F_{net}=F_{e}-F_{g}$

$F_{net}=0.499\times10^{-2}-0.392\times10^{-3}$

$F_{net}=0.004598= 0.4598\times10^{-2}\ N$

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Using newton's law

$F = ma$

$a = \dfrac{F}{m}$

$a=\dfrac{0.4598\times10^{-2}}{0.0400\times10^{-3}}$

$a =114.95\ m/s^2$

We need to calculate the height

Using equation of motion

$s = ut+\dfrac{1}{2}at^2$

$s=0+\dfrac{1}{2}\times114.95\times(0.0420)^2$

$s=0.1014\ m$

Hence, The height is 0.1014 m

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