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Naya [18.7K]
3 years ago
10

A bird is standing on an electric transmission line carrying 3000 A of current. A wire like this has about 3.0 x 10-5 22 of resi

stance per meter. The bird's feet are 6 cm apart. The bird itself has a resistance of about 4 x 105 12. The bird experiences a potential difference of 0.0054 V. What current goes through the bird?
Physics
1 answer:
Darya [45]3 years ago
6 0

Answer:

13.5 x 10^-9 A

Explanation:

Yes

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MIDDLE SCHOOL SCIENCE<br> yee yee please answer 6, 7, and 8
Nikolay [14]

Answer:

what r the questions i can’t see them

Explanation:

5 0
3 years ago
how many kilograms off mercury would fill a 5litre container if the density of mercury is 13.6grams per cm3
Sidana [21]

Answer:

68kg

Explanation:

1 cm^3 is the same as 1 mL and there are 5000mL in 5L

Therefore if the density is 13.6g/mL we multiply 13.6 by 5000 to get the amount of grams required = 68000g which is 68kg

7 0
3 years ago
A steel spur pinion has a module of 2 mm, 17 teeth cut on the 20° full-depth system, and a face width of 20 mm. At a speed of 16
erastovalidia [21]

Answer:

The value of bending stress on the pinion 35.38 M pa

Explanation:

Given data

m = 2 mm

Pressure angle \phi = 20°

No. of teeth T = 17

Face width (b) = 20 mm

Speed N = 1650 rpm

Power = 1200 W

Diameter of the pinion gear

D = m T

D = 2 × 17

D = 34 mm

Velocity of the pinion gear

V =\pi D( \frac{N}{60} )

V = 3.14 (0.034) \frac{(1650)}{60}

V = 2.93 \frac{m}{s}

Form factor for the pinion gear is

Y = 0.303

Now

K_{v} = \frac{6.1 +0.303}{6.1} = 1.049

Force on gear tooth

F = \frac{P}{V}

F = \frac{1200}{2.93}

F = 408.73 N

Now the bending stress is given by the formula

\sigma = \frac{K_{v} F}{m b y}

\sigma = \frac{(1.049)(408.73)}{(0.002)(0.02)(0.303)}

\sigma = 35.38 M pa

This is the value of bending stress on the pinion

8 0
3 years ago
A long straight wire carries a conventional current of 0.7 A. What is the approximate magnitude of the magnetic field at a locat
vfiekz [6]

Answer:

2.64 x 10⁻⁶T

Explanation:

The magnitude of the magnetic field produced by a long straight wire carrying current is given by Biot-Savart law as follows: "The magnetic field strength is directly proportional to the current on the wire and inversely proportional to the distance from the wire". This can be written mathematically as;

B = (μ₀ I) / (2π r)                ----------------(i)

B is magnetic field

I is current through the wire

r is the distance from the wire

μ₀  is the magnetic constant = 4π x 10⁻⁷Hm⁻¹

From the question;

I = 0.7A

r = 0.053m

Substitute these values into equation (i) as follows;

B =  (4π x 10⁻⁷ x 0.7) / (2π x 0.053)

B = 2.64 x 10⁻⁶T

Therefore the approximate magnitude of the magnetic field at that location is 2.64 x 10⁻⁶T

5 0
3 years ago
2.
anzhelika [568]

Answer:

44.72m/s

Explanation:

use th formula:vf²=vi²at

and then substitute the values

remember the units

8 0
3 years ago
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