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saul85 [17]
3 years ago
12

The charge per unit length on a long, straight filament is -92.0 μC/m. Find the electric field 10.0 cm above the filament.

Physics
1 answer:
Pepsi [2]3 years ago
4 0

Answer:

E = 1.655 x 10⁷ N/C towards the filament

Explanation:

Electric field due to a line charge is given by the expression

E = [tex]\frac{\lambda}{2\pi\times\epsilon_0\times r}[/tex]

where λ is linear charge density of line charge , r is distance of given point from line charge and ε₀ is a constant called permittivity and whose value is

8.85 x 10⁻¹².

Putting the given values in the equation given above

E = \frac{92\times10^{-6}}{2\times3.14\times8.85\times10^{-12}\times10^{-1}}

E = 1.655 x 10⁷ N/C

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An 8.0-ohm resistor and a 6.0-ohm resistor are connected in series with a battery. The potential difference across the 6.0-ohm r
ipn [44]

Answer:

a) 14 Ω

b) 2.0 A

c) 28 V

Explanation:

a) The total resistance of resistors in series is the sum:

R = R₁ + R₂

R = 8.0 Ω + 6.0 Ω

R = 14 Ω

b) The current in the 6.0 Ω resistor can be found with Ohm's law:

V = IR

12 V = I (6.0 Ω)

I = 2.0 A

c) Since the resistors are in series, they have the same current.  So the total voltage is:

V = IR

V = (2.0 A) (14 Ω)

V = 28 V

6 0
2 years ago
Select the correct answer.
chubhunter [2.5K]
I believe it’s self-referent encoding
3 0
2 years ago
Coach Hogue is riding his motorcycle in a circle on wet pavement. Suddenly the bike slides out from under him. What failed to pr
Arisa [49]

Answer:

The correct option is;

Force of Friction

Explanation:

As coach Hogue rode his motorcycle round in circle on the wet pavement, the motorcycle and the coach system tends to move in a straight path but due to intervention by the coach they maintain the circular path

The motion equation is

v = ωr and we have the centripetal acceleration given by

α = ω²r and therefore centripetal  force is then

m×α = m × ω²r = m × v²/r

The force required to keep the coach and the motorcycle system in their circular path can be obtained by the impressed force of friction acting towards the center of the circular motion.

5 0
3 years ago
Suppose a certain car supplies a constant deceleration of A meter per second per second. If it is traveling at 90km/hr. When. th
aksik [14]

Answer:

i)-6.25m/s

ii)18 metres

iii)26.5 m/s or 95.4 km/hr

Explanation:

Firstly convert 90km/hr to m/s

90 × 1000/3600 = 25m/s

(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)

0 = (25)^2 + 2A(50)

0 = 625 + 100A....then moved the other value to one

-625 = 100A

Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)

(ii) Firstly convert 54km/hr to m/s

In which this is 54 × 1000/3600 = 15m/s

then apply the same formula as that in (i)

0 = (15)^2 + 2(-6.25)s

-225 = -12.5s

Hence the stopping distance = 18metres

(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question

0 = u^2 + 2(-6.25)(56)

u^2 = 700

Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s

In km/hr....26.5 × 3600/1000 = 95.4 km/hr

3 0
3 years ago
Calculate the number of electrons passing a point in the wire in 1 min when the current is 1 A
Vera_Pavlovna [14]

Explanation:

When one coulomb charge passes through any cross section of the wire per second,the current passing is one ampere. Charge of electron ,e=1.6X10^-19C. n=1/(1.6X10^-19)=6.25X10^18.Sep 17, 2017

7 0
3 years ago
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