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valkas [14]
3 years ago
9

Date:

Physics
1 answer:
tia_tia [17]3 years ago
4 0

Answer:

don't know I just want point for answers

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Current Flow and Ohm's Law

Ohm's law is the most important, basic law of electricity. It defines the relationship between the three fundamental electrical quantities: current, voltage, and resistance. When a voltage is applied to a circuit containing only resistive elements (i.e. no coils), current flows according to Ohm's Law, which is shown below.

I = V / R 

Where: 

I =

Electrical Current (Amperes)

V =

Voltage (Voltage)

R =

Resistance (Ohms)

    

Ohm's law states that the electrical current (I) flowing in an circuit is proportional to the voltage (V) and inversely proportional to the resistance (R). Therefore, if the voltage is increased, the current will increase provided the resistance of the circuit does not change. Similarly, increasing the resistance of the circuit will lower the current flow if the voltage is not changed. The formula can be reorganized so that the relationship can easily be seen for all of the three variables.

The Java applet below allows the user to vary each of these three parameters in Ohm's Law and see the effect on the other two parameters. Values may be input into the dialog boxes, or the resistance and voltage may also be varied by moving the arrows in the applet. Current and voltage are shown as they would be displayed on an oscilloscope with the X-axis being time and the Y-axis being the amplitude of the current or voltage. Ohm's Law is valid for both direct current (DC) and alternating current (AC). Note that in AC circuits consisting of purely resistive elements, the current and voltage are always in phase with each other.

Exercise: Use the interactive applet below to investigate the relationship of the variables in Ohm's law. Vary the voltage in the circuit by clicking and dragging the head of the arrow, which is marked with the V. The resistance in the circuit can be increased by dragging the arrow head under the variable resister, which is marked R. Please note that the vertical scale of the oscilloscope screen automatically adjusts to reflect the value of the current.

See what happens to the voltage and current as the resistance in the circuit is increased. What happens if there is not enough resistance in a circuit? If the resistance is increased, what must happen in order to maintain the same level of current flow?


4 0
3 years ago
What the kinetic energy quantities in calculation pls help me​
Rashid [163]

Answer:

KE = 0.5 * m * v², where: m - mass, v - velocity.

Explanation:

In classical mechanics, kinetic energy (KE) is equal to half of an object's mass (1/2*m) multiplied by the velocity squared. For example, if a an object with a mass of 10 kg (m = 10 kg) is moving at a velocity of 5 meters per second (v = 5 m/s), the kinetic energy is equal to 125 Joules, or (1/2 * 10 kg) * 5 m/s 2.

3 0
2 years ago
A floating boat has a weight of 100 N. What is the buoyant force on the boat?
lara31 [8.8K]
If the boat is floating, then it's just sitting there, and not accelerating
up or down.  That means the vertical forces on it must be balanced. 

So if its weight (acting downward) is 100 newtons, then the buoyant
force on it (acting upward) must also be 100 newtons.
4 0
3 years ago
A body travel according to the law x(t)=2t+3t3 calculate the accelaration of body at t=2s
faust18 [17]

Answer:

28m/s

Explanation:

Given law,

⇒ x(t) = 2t + 3t³

Where, t = 2s

Then,

⇒ x(2) = 2(2) + 3(2)³

⇒ 4 + 3(8)

⇒ 4 + 24

⇒ 28 m/s

Acceleration is 28m/s

4 0
3 years ago
A long, thin rod parallel to the y-axis is located at x = - 1 cm and carries a uniform positive charge density λ = 1 nC/m . A se
zheka24 [161]

Answer:

The electric field at origin is 3600 N/C

Solution:

As per the question:

Charge density of rod 1, \lambda = 1\ nC = 1\times 10^{- 9}\ C

Charge density of rod 2, \lambda = - 1\ nC = - 1\times 10^{- 9}\ C

Now,

To calculate the electric field at origin:

We know that the electric field due to a long rod is given by:

\vec{E} = \frac{\lambda }{2\pi \epsilon_{o}{R}

Also,

\vec{E} = \frac{2K\lambda }{R}                  (1)

where

K = electrostatic constant = \frac{1}{4\pi \epsilon_{o} R}

R = Distance

\lambda = linear charge density

Now,

In case, the charge is positive, the electric field is away from the rod and towards it if the charge is negative.

At x = - 1 cm = - 0.01 m:

Using eqn (1):

\vec{E} = \frac{2\times 9\times 10^{9}\times 1\times 10^{- 9}}{0.01} = 1800\ N/C

\vec{E} = 1800\ N/C     (towards)

Now, at x = 1 cm = 0.01 m :

Using eqn (1):

\vec{E'} = \frac{2\times 9\times 10^{9}\times - 1\times 10^{- 9}}{0.01} = - 1800\ N/C

\vec{E'} = 1800\ N/C     (towards)

Now, the total field at the origin is the sum of both the fields:

\vec{E_{net}} = 1800 + 1800 = 3600\ N/C

7 0
3 years ago
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