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bezimeni [28]
3 years ago
9

A copper wire has a circular cross section with a radius of 1.25 mm. If the wire carries a current of 3.70 A, find the drift spe

ed in mm/s of the electrons in this wire. (Note: the density of charge carriers (electrons) in a copper wire is n = 8.46 × 1028 electrons/m3)
Physics
1 answer:
yulyashka [42]3 years ago
8 0

Answer:

0.0557 mm / s

Explanation:

r = 1.25 mm = 1.25 x 10^-3 m, i = 3.7 A, n = 8.46 x 10^28 per cubic metre,

e = 1.6 x 10^-19 C

Let vd be the drift velocity

use the formula for the current in terms of drift velocity

i = n e A vd

3.7 = 8.46 x 10^28 x 1.6 x 10^-19 x 3.14 x 1.25 x 1.25 x 10^-6 x vd

vd = 5.57 x 10^-5 m/s

vd = 0.0557 mm / s

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The density of air is 1.3 kg/m^3. What mass of air is contained in a room measuring 2.5m x 4m x 10m? Give your answer in kg. Bra
krek1111 [17]
Mass= density x volume
1.3 kg/m^3 x ( 2.5x4x10) m^3
= 130 kg
7 0
3 years ago
Notice that in each conversion factor the numerator equals the denominator when units are taken into account. A common error in
navik [9.2K]

Answer:

he factor for the temporal part 1.296 107 s² = h²

 m / s² = 12960 km / h²

Explanation:

This is a unit conversion exercise.

In the unit conversion, the size of the object is not changed, only the value with respect to which it is measured is changed, for this reason in the conversion the amount that is in parentheses must be worth one.

In this case, it is requested to convert a measure km/h²

Unfortunately, it is not clearly indicated what measure it is, but the most used unit in physics is   m / s² , which is a measure of acceleration. Let's cut this down

the factor for the distance is 1000 m = 1 km

the factor for time is 3600 s = 1 h

let's make the conversion

        m / s² (1km / 1000 m) (3600 s / 1h)²

note that as time is squared the conversion factor is also squared

        m / s² = 12960 km / h²

the factor for the temporal part 1.29 107 s² = h²

6 0
3 years ago
Un satélite geoestacionario se encuentra a una distancia de 120.000 km sobre la superficie de Júpiter. Determine: a. El periodo
Lisa [10]

Answer:

a) a geostationary satellite is that it is always at the same point with respect to the planet,

b) f = 2.7777 10⁻⁵ Hz

c)                           d)   w = 1.745 10⁻⁴ rad / s

Explanation:

a) The definition of a geostationary satellite is that it is always at the same point with respect to the planet, that is, its period of revolutions is the same as the period of the planet

  •                T = 10 h (3600 s / 1h) = 3.6 104 s

b) the period the frequency are related

                T = 1 / f

                 f = 1 / T

                 f = 1 / 3.6 104

                 f = 2.7777 10⁻⁵ Hz

c) the distance traveled by the satellite in 1 day

The distance traveled is equal to the length of the circumference

                 d = 2pi (R + r)

                 d = 2pi (69 911 103 + 120 106)

                 d = 1193.24 m

d) the angular velocity is the angle traveled between the time used.

                 .w = 2pi /t

                  w = 2pi / 3.6 10⁴

                  w = 1.745 10⁻⁴ rad / s

how fast is

                  v = w r

                  v = 1.75 10-4 (69.911 106 + 120 106)

                  v = 190017 m / s

5 0
3 years ago
A hummingbird can a flutter its wings 4800 times per minute what is the frequency of wing flutters per second
Scilla [17]

the answer i got is 80 i hope this helpss!!!!

7 0
3 years ago
Read 2 more answers
Compare the catching of two different water balloons.
Stels [109]

Answer:

a. The volume V₁ and V₂

b. The case that involves the greatest momentum change = Case B

c. The case that involves the greatest impulse = Case B

d. b. The case that involves the greatest force = Case B

Explanation:

Here we have

Case A: V₁ = 150-mL, v₁ = 8 m/s

Case B: V₂ = 600-mL, v₁ = 8 m/s

a. The variable that is different for the two cases is the volume V₁ and V₂

b. The momentum change is by the following relation;

ΔM₁ = Mass, m × Δv₁

The mass of the balloon are;

Δv₁ = Change in velocity = Final velocity - Initial velocity

Mass = Density × Volume

Density of water = 0.997 g/mL

Case A, mass = 150 × 0.997 = 149.55 g

Case B, mass = 600 × 0.997 = 598.2 g

The momentum change is;

Case A: Mass, m × Δv₁ = 149.55 g/1000 × 8 m/s = 1.1964 g·m/s

Case B:  Mass, m × Δv₁ = 598.2/1000 × 8 = 4.7856 g·m/s

Therefore Case B has the greatest momentum change

The case that has the gretest momentum change = Case B

c. The momentum change = impulse therefore Case B involves the greatest impulse

d. Here we have;

Impulse = Momentum change = F_{average} × Δt = mΔV

∴ F_{average} = m·ΔV/Δt

∴ For Case A F_{average} = 149.55×8/Δt =  1196.4/Δt N

For Case B  F_{average} = 598.2×8/Δt =  4785.6/Δt

Where Δt is the same for Case A and Case B,  F_{average}  for Case B >>  F_{average}  for Case B

Therefore, Case B involves the greatest force.

4 0
3 years ago
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