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Why does self inductance acts as electrical inertia?

Taya2010 [7]

Answer:

self-indulgence of coil is the property by virtue of wich is tends to maintain magnatic flux link with it and opposed any any change in the flux inducing current in it

8 0

3 years ago

What is the final velocity of the ball that is dropped from a height of 200m?

Hoochie [10]

The final velocity of the ball that is dropped from a height of 200m is v = 44.73 m/s .

<h3>What is velocity with example?</h3>

The rate at which an object is travelling in one direction is referred to as its velocity. an automobile traveling north on a highway, or a rocket taking off. Its velocity vector's absolute value always is equal to the motion's speed because it is a scalar.

<h3>Briefing:</h3>

Given the initial velocity of the ball (u) = 0

Distance travelled by the ball (s) = 200m

Acceleration (a) = 10 m/s²

As we know:

v² = u² + 2as

Putting values:

v² = 0+2 × (10 m/s²) × (200 m)

v = 44.73 m/s.

To know more about Velocity visit:

brainly.com/question/18084516

#SPJ9

8 0

1 year ago

A small glass ball rubbed with silk gains a charge of +2.0 uc. the glass ball is placed 12 cm from a small charged rubber ball t

Gre4nikov [31]

The magnitude of the electric force between two obejcts with charge $q_1$ and $q_2$ is given by Coulomb's law:
$F= k_e \frac{q_1 q_2}{r^2}$
where
$k_e = 8.99 \cdot 10^9 N m^2 C^{-2}$ is the Coulomb's constant
and r is the distance between the two objects.

In our problem, the distance is $r=12 cm=0.12 m$ , while the magnitudes of the two charges are
$q_1 = 2.0 \mu C=2.0 \cdot 10^{-6}C$
$q_2 = 3.5 \mu C = 3.5 \cdot 10^{-6} C$
(we can neglect the sign of the second charge, since we are interested only in the magnitude of the force).

So, using the formula and the data of the problem, we find
$F=(8.99 \cdot 10^9 N m^2 C^{-2} ) \frac{(2.0 \cdot 10^{-6} C)(3.5 \cdot 10^{-6} C)}{(0.12 m)^2}=4.37 N$

4 0

3 years ago

A metal block of mass 3 kg is falling downward and has velocity of 0.44 m/s when it is 0.8 m above the floor. It strikes the top

Anton [14]

Answer:

$y_{max} = 0.829\,m$

Explanation:

Let assume that one end of the spring is attached to the ground. The speed of the metal block when hits the relaxed vertical spring is:

$v = \sqrt{(0.8\,\frac{m}{s})^{2} + 2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.4\,m)}$

$v = 2.913\,\frac{m}{s}$

The maximum compression of the spring is calculated by using the Principle of Energy Conservation:

$(3\,kg)\cdot (9.807\,\frac{m}{s^{2}})\cdot (0.4\,m) + \frac{1}{2}\cdot (3\,kg)\cdot (2.913\,\frac{m}{s} )^{2} = (3\,kg) \cdot (9.807\,\frac{m}{s^{2}})\cdot (0.4\,m-\Delta s) + \frac{1}{2}\cdot (2000\,\frac{N}{m})\cdot (\Delta s) ^{2}$

After some algebraic handling, a second-order polynomial is formed:

$12.728\,J = \frac{1}{2}\cdot (2000\,\frac{N}{m} )\cdot (\Delta s)^{2} - (3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot \Delta s$

$1000\cdot (\Delta s)^{2}-29.421\cdot \Delta s - 12.728 = 0$

The roots of the polynomial are, respectively:

$\Delta s_{1} \approx 0.128\,m$

$\Delta s_{2} \approx -0.099\,m$

The first root is the only solution that is physically reasonable. Then, the elongation of the spring is:

$\Delta s \approx 0.128\,m$

The maximum height that the block reaches after rebound is:

$(3\,kg) \cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.4\,m-\Delta s) + \frac{1}{2}\cdot (2000\,\frac{N}{m})\cdot (\Delta s)^{2} = (3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot y_{max}$

$y_{max} = 0.829\,m$

4 0

3 years ago

Read 2 more answers

Do 3-7 for me? It is science and i hate doing science hw.

Serggg [28]

The answer to your question is -4

6 0

3 years ago