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Zanzabum
3 years ago
5

Plzzzz HELLPPPPPP!!!!

Physics
1 answer:
Rainbow [258]3 years ago
7 0
?? what’s the topic
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A current of 3.60A flows for 15.3s through a conductor. Calculate the number of electrons that pass through a point in the condu
Artyom0805 [142]

3.60 A = 3.60 coulombs of charge per second

(3.60 Coul/sec) x (15.3 sec) = 55.08 coulombs of charge

1 coulomb of charge is carried by 6.25 x 10^18 electrons

Number of electrons =

               (55.08 Coul) x (6.25 x 10^18 e/coul) = <em>3.4425 x 10^20 electrons</em>


7 0
4 years ago
Why does it seem harder to start sliding a couch across a carpeted floor than it oes to keep it moving?
svetlana [45]
When you start sliding the couch, you have to overcome the static friction of the carpeted floor, which is usually very big, especially due to the texture of the carpet. However, as long as you get it moving, you only have to apply the force to overcome the kinetic friction, which is smaller than the static friction. 
6 0
4 years ago
Turning a corner at a typical large intersection in a city means driving your car through a circular arc with a radius of about
Naddik [55]

Answer:

9.89 m/s.

Explanation:

Given that,

The radius of the circular arc, r = 25 m

The acceleration of the vehicle is 0.40 times the free-fall acceleration i.e.,a = 0.4(9.8) = 3.92 m/s²

Let v is the maximum speed at which you should drive through this turn. It can be solved as follows :

a=\dfrac{v^2}{r}\\\\v=\sqrt{ar} \\\\v=\sqrt{3.92\times 25} \\\\=9.89 m/s

So, the maximum speed of the car should be 9.89 m/s.

8 0
3 years ago
An α-particle has a charge of +2e and a mass of 6.64 × 10-27 kg. It is accelerated from rest through a potential difference that
zzz [600]

Answer with Explanation:

We are given that

Charge on alpha particle=q=2 e=2\times 1.6\times 10^{-19} C

1 e=1.6\times 10^{-19} C

Mass of alpha particle=m=6.64\times 10^{-27} kg

Potential difference,V=1.97\times 10^6 V

Magnetic field,B=3.49 T

a.Speed of alpha particle=v=\sqrt{\frac{2 qV}{m}}

By using the formula

v=\sqrt{\frac{2\times 2\times 1.6\times 10^{-19}\times 1.97\times 10^6}{6.64\times 10^{-27}}

v=1.38\times 10^7 m/s

b.Magnetic force,F=qvB=2\times 1.6\times 10^{-19}\times 1.38\times 10^7\times 3.49=1.5\times 10^{-11} N

F=1.5\times 10^{-11} N

c.Radius of circular path, r=\frac{mv^2}{F}

r= \frac{6.64\times 10^{-27}\times (1.38\times 10^7)^2}{1.5\times 10^{-11}}

r=0.084 m

5 0
3 years ago
A 12 N force is used to pull a wagon 10 m to the left in 15 seconds. Which of the following represents the magnitude of the forc
Lana71 [14]

1. - 12N

2.- 2 hours

3. - 0.5 m/s/s

7 0
3 years ago
Read 2 more answers
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