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3 years ago

Define electric current and drift velocity.

Physics

2 answers:

Luba_88 [7]3 years ago

8 0

Answer:

An electric current is the stream of changed particals, such as electrons & ions, moving through an electrical conductor.

The average velocity attained bycharged partical ,such as electrons,in a material due to electric fields

Blizzard [7]3 years ago

3 0

Answer:

Current- the flow of free charges, such as electrons and ions

Drift velocity- the average speed at which these charges move

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3 years ago

Two particles are attracted to each other by the gravitational force between them. Particle 1 has a mass of 12kg, Particle 2 has

svlad2 [7]

Answer: 1.4 x 10^-8N

Explanation:

Given that,

Mass of Particle 1 (m1) = 12kg

Mass of Particle 2 (m2) = 25kg

distance between particles (r) = 1.2m. Gravitational force (F) =

Apply the formula for gravitational force:

F = Gm1m2/r²

where G is the gravitational constant with a value of 6.7 x 10^-11 Nm2/kg2

Then, F = (6.7 x 10^-11 Nm²/kg² x 12kg x 25kg) / (1.2m)²

F = (2.01 x 10^-8Nm²) / (1.44m²)

F = 1.396 x 10^-8N (Rounded to the nearest tenth as 1.4 x 10^-8N)

Thus, the magnitude of the gravitational force acting on the particles is 1.4 x 10^-8 Newton

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4 years ago

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emmainna [20.7K]

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3 years ago

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3 years ago

A 1.40-kg block is on a frictionless, 30 ∘ inclined plane. The block is attached to a spring (k = 40.0 N/m ) that is fixed to a

Oliga [24]

Answer:

the mass drop by 6.5cm before coming to rest.

Explanation:

Given that:

the mass of the block M= 1.40 kg

angle of inclination θ = 30°

spring constant K = 40.0 N/m

mass of the suspended block m = 60.0 g = 0.06 kg

initial downward speed = 1.40 m/s

The objective is to determine how far does it drop before coming to rest?

Let assume it drops at y from a certain point in the vertical direction;

Then :

Workdone by gravity on the mass of the block is:

$w_1g = 1.4*9.81*y*sin30$ = - 6.867y

Workdone by gravity on the mass of the suspended block is:

$w_2g = 0.06*9.81$ = 0.5886

The workdone by the spring = -1/2ky²

= -0.5 × 40y²

= -20 y²

The net workdone is = -20 y² - 6.867y + 0.5886y

According to the work energy theorem

Net work done = Δ K.E

-20 y² - 6.867y + 0.5886 = 1/2 × 0.06 × 1.4²

-20 y² - 6.867y + 0.5886 = 0.5 × 0.1176

-20 y² - 6.867y + 0.5886 = 0.0588

-20 y² - 6.867y + 0.5886 - 0.0588

-20 y² - 6.867y + 0.5298 = 0

multiplying through by (-)

20 y² + 6.867y - 0.5298 = 0

Using the quadratic formula:

$\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$

where;

a = 20 ; b = 6.867 c= - 0.5298

$\dfrac{-(6.867) \pm \sqrt{(6.867)^2-(4*20*-0.5298)}}{2*(20)}$

$= \dfrac{-(6.867) + \sqrt{(6.867)^2-(4*20*-0.5298)}}{2*(20)} \ \ \ OR \ \ \ \dfrac{-(6.867) - \sqrt{(6.867)^2-(4*20*-0.5298)}}{2*(20)}$ = 0.0649 OR −0.408

We go by the positive integer

y = 0.0649 m

y = 6.5 cm

Therefore; the mass drop by 6.5cm before coming to rest.

7 0

4 years ago