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MakcuM [25]
3 years ago
9

If you disconnect the wires from the battery and then reconnect them at the opposite ends of the battery, how does that change t

he direction of the charge flow?
The direction of the charge flow is the same as the question
The direction of the charge flow still varies in different part
The direction of the charge flow is reversed

I've heard it was the first option, but my dad is adamant it is the third. How is it the first one (or third)?? Please help, been stuck on this for hours
Physics
1 answer:
IrinaVladis [17]3 years ago
3 0
Correct answer is first option: <span>The direction of the charge flow is the same as the question

Explanation:
When we have battery (or any other source of electricity) we have flow of a charge. Charge flows from positive part of battery over a conductor (most often it is a wire) towards negative part of battery. This is what always happens.

Now lets observe what we have done in this question. We had wires connected to a battery and we had a flow of charge from positive to negative pole. Then we disconnect wires and </span><span>reconnect them at the opposite ends of the battery. The question is does the direction of flow of charge change? Answer is NO. Charge still flows from positive towards negative pole. The direction of flow of charge does not depend on how the wires are connected.

Imagine this:
there is line of people stretching from one end of room to opposite end. You need to take something from front part to back part of room. Imagine people pass that object one to another. Does the direction in which this object moves depend on how this line of people is arranged? Same is for wire and charge.</span>
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What is the resultant of the two vectors shown?
Ksju [112]

Answer

correct answer is C

Head-to-tail method

Here head-to-tail method is employed to determine resultant of vectors.

Let horizontal component be 1 and let vertical component be 2

To add these two vectors. Move vector 2 until its tail touches the head of 1. The tail of resultant of these two vectors touch the tail of vector 1 and head of resultant vector will touch the head of vector 2.

6 0
3 years ago
Read 2 more answers
A potential difference of 3.27 nV is set up across a 2.16 cm length of copper wire that has a radius of 2.33 mm. How much charge
miskamm [114]

Answer:

Charge = 4.9096 x 10⁻⁷ C

Explanation:

First, we find the resistance of the copper wire.

R = ρL/A

where,

R = resistance = ?

ρ = resistivity of copper = 1.69 x 10⁻⁸ Ω.m

L = Length of wire = 2.16 cm = 0.0216 m

A = Cross-sectional area of wire = πr² = π(0.00233 m)² = 1.7 x 10⁻⁵ m²

Therefore,

R = (1.69 x 10⁻⁸ Ω.m)(0.0216 m)/(1.7 x 10⁻⁵ m²)

R = 2.14 x 10⁻⁵ Ω

Now, we find the current from Ohm's Law:

V =IR

I = V/R

I = 3.27 x 10⁻⁹ V/2.14 x 10⁻⁵ Ω

I = 1.52 x 10⁻⁴ A

Now, for the charge:

I = Charge/Time

Charge = (I)(Time)

Charge = (1.52 x 10⁻⁴ A)(3.23 x 10⁻³ s)

<u>Charge = 4.9096 x 10⁻⁷ C</u>

8 0
4 years ago
An astronaut and his space suit have a combined mass of 157 kg. The
alexgriva [62]

Answer:

v₃ = 9.62[m/s]

Explanation:

To solve this type of problem we must use the principle of conservation of linear momentum, which tells us that the momentum is equal to the product of mass by velocity.

We must analyze the moment when the astronaut launches the toolkit, the before and after. In order to return to the ship, the astronaut must launch the toolkit in the opposite direction to the movement.

Let's take the leftward movement as negative, which is when the astronaut moves away from the ship, and rightward as positive, which is when he approaches the ship.

In this way, we can construct the following equation.

-(m_{1}+m_{2})*v_{1}=(m_{1}*v_{2})-(m_{2}*v_{3})

where:

m₁ = mass of the astronaut = 157 [kg]

m₂ = mass of the toolkit = 5 [kg]

v₁ = velocity combined of the astronaut and the toolkit before throwing the toolkit = 0.2 [m/s]

v₂ = velocity for returning back to the ship after throwing the toolkit [m/s]

v₃ = velocity at which the toolkit should be thrown [m/s]

Now replacing:

-(157+5)*0.2=(157*0.1)-(5*v_{3})\\(5*v_{3})= 15.7+32.4\\v_{3}=9.62[m/s]

6 0
3 years ago
Explain why an egg thrown at a concrete wall will break, but an egg thrown at a sheet hanging from the ceiling will not. (Be sur
Firlakuza [10]

An egg thrown at a concrete wall will break, but an egg thrown at a sheet hanging from the ceiling will not due to high momentum and acceleration.

<h3>Why an egg thrown at a concrete wall will break?</h3>

An egg thrown at a concrete wall will break, but an egg thrown at a sheet hanging from the ceiling will not because the momentum and acceleration increases when the egg is thrown downward due to gravity but when we throw an egg in the vertical direction, they move against gravity so the momentum and acceleration decreases.

So we can conclude that an egg thrown at a concrete wall will break, but an egg thrown at a sheet hanging from the ceiling will not due to high momentum and acceleration.

Learn more about momentum here: brainly.com/question/7538238

#SPJ1

4 0
2 years ago
A batter hits a ball at 35 degrees above the horizontal and it is caught 4 seconds later 100 meters from home plate. What is the
Mamont248 [21]

Answer:

u=(26.5 i + 18.5j) m/s

Explanation:

The range of a projectile is given by the formula

d=\frac{u^2}{g} sin 2\theta

where in this case, we have

d = 100 m is the range

u is the initial speed (the magnitude of the initial velocity)

g = 9.8 m/s^2 is the acceleration of gravity

\theta = 35^{\circ} is the angle of projection

Solving for u, we find:

u=\sqrt{\frac{dg}{sin 2\theta}}=\sqrt{\frac{(100)(9.8)}{sin(2\cdot 35^{\circ})}}=32.3 m/s

Now we can easily find the components of the initial velocity:

u_x = u cos \theta = (32.3)(cos 35^{\circ})=26.5 m/s\\u_y = u sin \theta = (32.3)(sin 35^{\circ})=18.5 m/s

So, the initial velocity of the ball is

u=(26.5 i + 18.5  j) m/s

where i and j are the unit vector indicating the horizontal and vertical direction.

7 0
3 years ago
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