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3 years ago

3. What is electric current? The flow of moving electrons electrons that move one time

Physics

2 answers:

3241004551 [841]3 years ago

7 0

Answer:

An electric current is a stream of charged particles, such as electrons or ions, moving through an electrical conductor or space. It is measured as the net rate of flow of electric charge through a surface or into a control volume. ... In electric circuits the charge carriers are often electrons moving through a wire.

Flauer [41]3 years ago

5 0

Answer:

The flow of moving electrons

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Two identical conducting spheres, A and B, carry equal charge. They are separated by a distance much larger than their diameters

Vesnalui [34]

Answer:

C. $\frac{3F}{8}$

Explanation:

Let initial charges on both spheres be, $q$

$F=\frac{Kq^2}{d^2} \ \ \ \ \ \ \ \ \ \ \_i$

When the sphere C is touched by A, the final charges on both will be, $\frac{q}{2}$

#Now, when C is touched by B, the final charges on both of them will be:

$q_c=q_d=\frac{q/2+q}{2}\\\\=\frac{3q}{4}\\$

Now the force between A and B is calculated as:

$F\prime=\frac{k\times\frac{q}{2}\times \frac{3q}{4}}{d^2}\\F\prime=\frac{3F}{8}$

Hence the electrostatic force becomes 3F/8

4 0

2 years ago

An object glides on a horizontal tabletop with a coefficient of kinetic friction of 0.5. If its initial velocity is 4.3 m/s, how

Shkiper50 [21]

Answer:

Time, t = 0.87 seconds

Explanation:

Given that,

Initial velocity of the object, u = 4.3 m/s

The coefficient of kinetic friction between horizontal tabletop and the object is 0.5

We need to find the time taken by the object for the object to come to rest i.e. final velocity will be 0.

Using first equation of motion to find it as :

$v=u+at$

a is the acceleration, here, $a=\mu g$

$0=u+\mu gt$

$t=\dfrac{u}{\mu g}\\\\t=\dfrac{4.3}{0.5\times 9.8}\\\\t=0.87\ s$

So, the time taken by the object to come at rest is 0.87 seconds. Hence, this is the required solution.

8 0

3 years ago

A skateboarder, starting from rest, rolls down a 12.8-m ramp. When she arrives at the bottom of the ramp her speed is 8.89 m/s.

melamori03 [73]

Answer:

a) a = 3.09 m/s²

b) aₓ = 2.60 m/s²

Explanation:

a) The magnitude of her acceleration can be calculated using the following equation:

$V_{f}^{2} = V_{0}^{2} + 2ad$

<u>Where</u>:

$V_{f}$ : is the final speed = 8.89 m/s

$V_{0}$ : is the initial speed = 0 (since she starts from rest)

a: is the acceleration

d: is the distance = 12.8 m

$a = \frac{V_{f}^{2}}{2d} = \frac{(8.89 m/s)^{2}}{2*12.8 m} = 3.09 m/s^{2}$

Therefore, the magnitude of her acceleration is 3.09 m/s².

b) The component of her acceleration that is parallel to the ground is given by:

$a_{x} = a*cos(\theta)$

<u>Where</u>:

θ: is the angle respect to the ground = 32.6 °

$a_{x} = 3.09 m/s^{2}*cos(32.6) = 2.60 m/s^{2}$

Hence, the component of her acceleration that is parallel to the ground is 2.60 m/s².

I hope it helps you!

7 0

3 years ago

A spaceship far from all other objects uses its impulse power system to attain a speed of 104 m/s. The crew then shuts off the p

asambeis [7]

Answer:

Velocity remains the same at 104 m/s

Explanation:

According to Newton's 1st law of motion, an object subjected to no force or net force equal 0 would maintain its velocity. In our case the crew shuts off the power, spaceship is in space and far from all other objects (so no gravity whatsoever) would have no force acting on it. Therefore its velocity would stay the same at 104 m/s

4 0

2 years ago

A concave mirror with a focal length of 32 cm produces an image whose distance from the mirror is one-third the object distance.

iVinArrow [24]

Answer:

part A ⇒ u = 1.28 m

part B ⇒v = 0.43 m

Explanation:

for u is the distance to the object from the mirror and v is the distance from the mirror to the image.

Part A:

the mirror equation is given by:

1/f = 1/v + 1/u

but we told that, v = 1/3u:

1/f = 3/v + 1/u = 4/u

1/f = 4/u

f = u/4

u = 4f

= 4×(32×10^-2)

= 1.28 m

Therefore the distance from the mirror to the object is 1.28 m.

part B:

v = 1/3×u = 1/3×(1.28) = 0.43 m

5 0

3 years ago