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3 years ago

What is the charge value of 45 electrons

Physics

1 answer:

tamaranim1 [39]3 years ago

8 0

Answer:

$7.2\times 10^{-18}C$

Explanation:

-Let x be the charge of 45 electrons

-An electron has a relatively law mas.

-Using the law of proportions, the charge value of 45 electrons is calculated as:

$1e=1.6 \times 10^{-19} coulomb\\45e=x\\\\x=\frac{45\times1.6 \times 10^{-19} coulomb}{1e}\\\\x=7.2\times 10^{-18}C$

Hence, the electron charge of 45 electrons is $7.2\times 10^{-18}C$

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A 4.79 g bullet moving at 642.3 m/s penetrates a tree trunk to a depth of 4.35 cm. Use work and energy considerations to find th

vladimir1956 [14]

Answer:

Force, $F=2.27\times 10^4\ N$

Explanation:

Given that,

Mass of the bullet, m = 4.79 g = 0.00479 kg

Initial speed of the bullet, u = 642.3 m/s

Distance, d = 4.35 cm = 0.0435 m

To find,

The magnitude of force required to stop the bullet.

Solution,

The work energy theorem states that the work done is equal to the change in its kinetic energy. Its expression is given by :

$F.d=\dfrac{1}{2}m(v^2-u^2)$

Finally, it stops, v = 0

$F.d=-\dfrac{1}{2}m(u^2)$

$F=\dfrac{-mu^2}{2d}$

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F = -22713.92 N

$F=2.27\times 10^4\ N$

So, the magnitude of the force that stops the bullet is $2.27\times 10^4\ N$

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3 years ago

Hey Tori's are the rim of a canyon yells hello toward the opposite side she has an echo hello four seconds later if the speed of

krek1111 [17]

Answer:

The wall is 680 meter away from the person.

Explanation:

Given data

Speed of sound = 340 $\frac{m}{s}$

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Therefore the distance between the person & wall is

D = speed × Time

D = 340 × 2

D = 680 meter

Therefore the wall is 680 meter away from the person.

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A 5-kg projectile is fired over level ground with a velocity of 200 m/s at an angle of 25°above the horizontal. Just before it h

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Answer:

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Explanation:

Given;

mass of the object, m = 5kg

initial velocity of the projectile, v₁ = 200 m/s

final velocity of the projectile, v₂ = 150 m/s

To determine the the change in the thermal energy of the projectile and air, we consider change in potential and kinetic enrgy of the projectile. Since the projectile was fired over level ground, change in potential energy is zero.

Then, change in thermal energy of the projectile, KE = Δ¹/₂mv²

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