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3 years ago

At what speed does a typical electron pass by any given point in the wire?

1 answer:

7 0

A 5.00 A current runs through a 12 gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has 8.5*10^28 free electrons per cubic metre.

a) How many electrons pass through the light bulb each second?

b) What is the current density in the wire? (answer in A/m^2)

c) At what speed does a typical electron pass by any given point in the wire? (answer in m/s)

a) 5.0 A = 5.0 C/s
. Number of electrons in 5.0C = 5.0 / 1.60^-19 = 3.125^19
. 5.0 A ►= 3.125^19 electrons/s

b) A/m² = 5.0 / π(1.025^-3 m)² .. .. ►= 1.52^6 A/m²

c) Charge density (q/m³) = 8.50^28 e/m³ x 1.60^-19 = 1.36^10 C/m³

(q/m³)(m²)(m/s) = q/s (current i in C/s [A])
(m²) = Area
(m/s) = mean drift speed

(q/m³)(A)(v) = i

v = i.[(q/m³)A]ˉ¹

v = 5.0 [1.36^10 * π(1.025^-3 m)²]ˉ¹.. .. ►v = 1.10^-4 m/s

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A trick shot archer shoots an arrow with a velocity of 30.0 m/s at an angle of 20.0 degrees with respect to the horizontal. An a

svlad2 [7]

Answer:

$u'=10.259\ m.s^{-1}$ is the initial velocity of tossing the apple.

the apple should be tossed after $\Delta t=0.0173\ s$

Explanation:

Given:

velocity of arrow in projectile, $v=30\ m.s^{-1}$

angle of projectile from the horizontal, $\theta=20^{\circ}$

distance of the point of tossing up of an apple, $d=30\ m$

Now the horizontal component of velocity:

$v_x=v\ cos\ \theta$

$v_x=30\times cos\ 20^{\circ}$

$v_x=28.191\ m.s^{-1}$

The vertical component of the velocity:

$v_y=v.sin\ \theta$

$v_y=30\times sin\ 20^{\circ}$

$v_y=10.261\ m.s^{-1}$

Time taken by the projectile to travel the distance of 30 m:

$t=\frac{d}{v_x}$

$t=\frac{30}{28.191}$

$t=1.0642\ s$

Vertical position of the projectile at this time:

$h=v_y.t-\frac{1}{2}g.t^2$

$h=10.261\times 1.0642-\frac{1}{2} \times 9.8\times 1.0642^2$

$h=5.3701\ m$

Now this height should be the maximum height of the tossed apple where its velocity becomes zero.

$v'^2=u'^2-2g.h$

$0^2=u'^2-2\times 9.8\times 5.3701$

$u'=10.259\ m.s^{-1}$ is the initial velocity of tossing the apple.

Time taken to reach this height:

$v'=u'-g.t'$

$0=10.259-9.8\times t'$

$t'=1.0469\ s$

We observe that $t>t'$ hence the time after the launch of the projectile after which the apple should be tossed is:

$\Delta t=t-t'$

$\Delta t=1.0642-1.0469$

$\Delta t=0.0173\ s$

8 0

3 years ago

Explain what would happen to an animal if the vegetation around them disappeared

lara [203]

Answer:

he type of vegetation growing in the buffer affects its usefulness to wildlife through the availability of food, foraging and nesting sites, and other habitat needs (Johnson and Beck 1988). The more diverse the habitat, the greater its utility to many species of animals.

Explanation:

3 0

2 years ago

Melissa and Rory live near a bike trail, 21 km apart. They plan to meet at a park mid-way between their homes. They start from t

DiKsa [7]

Answer:

27km.

Explanation:

Jessie will run for a duration of time, which is the time it takes for Melissa and Rory to meet each other.

Together combined, Melissa and Rory runs at 14km/h. The bike trail is 21km, so it will take 1.5 hours.

Jessie runs at 18km/h, so Jessie will run 18 * 1.5 = 27km.

7 0

3 years ago

Maximum range of a projectile is 1.6 m. Then the velocity of projection will be..... (g=10m/s)

qaws [65]

Answer:

4 m/s

Explanation:

From the question given above, the following data were obtained:

Maximum range (Rₘₐₓ) = 1.6 m

Acceleration due to gravity (g) = 10 m/s²

Initial velocity (u) =?

The initial velocity of the projectile can be obtained as follow:

Rₘₐₓ = u² / g

1.6 = u² / 10

Cross multiply

u² = 1.6 × 10

u² = 16

Take the square root of both side

u = √16

u = 4 m/s

Therefore, the velocity of the projectile is 4 m/s

6 0

3 years ago

Use word in a sentence (Related to eclipse) - PROMINENCE

motikmotik

"The prominence of the light expelled by the sun was breathtaking during the eclipse."

Brainliest answer please

4 0

3 years ago