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3 years ago

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Consider a magnesium wire (σ = 2.2 ✕ 107 Ω−1 · m−1) with a cross-sectional area of 1 mm2 (similar to your connecting wires) and

carrying 0.3 amperes of current, which is about what you get in a circuit with a round bulb and two batteries in series. Calculate the strength of the very small electric field required to drive this current through the wire.

1 answer:

omeli [17]3 years ago

4 0

Answer: 1.36 x 10^-2 N/C

Explanation:

I/A = σ E

E = I / (A σ)

E = 0.3A / (1x10^-6 m^2 2.2 x10^7/ ohm m)

E = 1.36 x^-2 N/C

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The local newspaper has published the passage below. Last week, scientists released incredible images of aurora australis, known

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Answer:

The statement as aurora australis known as northern lights is incorrect. As the designation of aurora australis is for the southern lights i.e. which occur in the southern hemisphere.

Explanation:

Aurora or natural lights is a phenomenon that occurs at the poles of the Earth due to interaction between the Earth's magnetic field and cosmic rays. This interaction results in the beautiful display of colors on both poles. These are named, aurora borealis or aurora australis depending on their geographical location. If they occur on the northern pole they are termed as aurora borealis while those occurring on the southern pole are named aurora australis.

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Under ordinary conditions of temperature and pressure, the particles in a gas are Select one: a. closely packed. b. very far fro

klemol [59]

Answer:

The particle of a gas are very far apart from each other.

Explanation:

The states of matter comprises of solid, liquid and gas. The molecules of gas are free because they are not tightly held together by molecular force. The force between the molecules has been broken, hence the particles possesses the freedom to move about thereby possessing high kinetic energy (energy possessed by a body due to its motion). Since this molecules can easily move freely, they are always far apart from each other under ordinary temperature and pressure.

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3 years ago

Two blocks are connected by a light string that passes over two frictionless pulleys. The block of mass m2 is attached to a spri

irina1246 [14]

(BELOW YOU CAN FIND ATTACHED THE IMAGE OF THE SITUATION)

Answer:

$d=\frac{2g(m1-m2)}{k}$

Explanation:

For this we're going to use conservation of mechanical energy because there are nor dissipative forces as friction. So, the change on mechanical energy (E) should be zero, that means:

$E_{i}=E_{f}$

$K_{i}+U_{i}=K_{f}+U_{f}$ (1)

With $K_{i}$ the initial kinetic energy, $U_{i}$ the initial potential energy, $K_{f}$ the final kinetic energy and $U_{f}$ the final potential energy. Note that initialy the masses are at rest so $K_{i} = 0$ , when they are released the block 2 moves downward because m2>m1 and finally when the mass 2 reaches its maximum displacement the blocks will be instantly at rest so $K_{f} =0$ . So, equation (1) becomes:

$U_{i}=U_{f}$ (2)

At initial moment all the potential energy is gravitational because the spring is not stretched so $U_{i}=U_{gi}$ and at final moment we have potential gravitational energy and potential elastic energy so $U_{f}=U_{gf}+U_{ef}$ , using this on (2)

$U_{gi}=U_{gf}+U_{ef}$ (3)

Additional if we define the cero of potential gravitational energy as sketched on the figure below (See image attached), $U_{gi}=0$ and we have by (3) :

$0= U_{gf}+U_{ef}$ (4)

Now when the block 1 moves a distance d upward the block 2 moves downward a distance d too (to maintain a constant length of the rope) and the spring stretches a distance d, so (4) is:

$0=-m1gd+m2gd+\frac{kd^{2}}{2}$

dividing both sides by d

$0=-m1g+m2g+\frac{kd}{2}$

$g(m1-m2)= \frac{kd}{2}$

$d=\frac{2g(m1-m2)}{k}$ , with k the constant of the spring and g the gravitational acceleration.

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3 years ago

Calculate the drop voltage for a battery of internal resistance 2 ohm and emf of 24 volt

Ivenika [448]

Answer:

<em>The drop voltage is 0.3 V</em>

Explanation:

Electromotive Force EMF

When connecting a battery of internal resistance Ri and EMF ε to an external resistance Re, the current through the circuit is:

$\displaystyle i=\frac{\varepsilon }{R_e+R_i}$

The battery has an internal resistance of Ro=2 Ω, ε=24 V and is connected to an external resistance of Re=158 Ω. Thus, the current is:

$\displaystyle i=\frac{24 }{158+2}$

$\displaystyle i=\frac{24 }{160}$

i = 0.15 A

The drop voltage is the voltage of the internal resistance:

$V_i = i.R_i$

$V_i = 0.15*2$

$\boxed{V_i = 0.3\ V}$

The drop voltage is 0.3 V

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2 years ago

The speed of a car increases uniformly from 6 m/s to 42 m/s in 9 second calculate the acceleration of the car

djyliett [7]

The acceleration of the car which changes the speed uniformly is 4 m/s²

<h3>What is acceleration?</h3>

Acceleration can be defined as the rate change of velocity with time.

acceleration a = (Δv) / (Δt)

Given is the initial velocity 6m/s and final velocity 42m/s, the time take fir this change in speed is 9s.

Substitute the values, we have

a = (42-6)/9

a = 36/9

a =4 m/s²

Thus, the acceleration is 4 m/s²

Learn more about acceleration.

brainly.com/question/12550364

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