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Fudgin [204]
3 years ago
14

An electrician wraps rubber electrical tape around a copper wire. Since rubber is an insulator, which statement best explains th

e point of wrapping rubber around a copper wire?
Physics
1 answer:
yuradex [85]3 years ago
4 0

A) Rubber stops charges from flowing. This protects people by stopping electricity from flowing.

Explanation:

The statement that best describes the point of wrapping rubber around the copper wire is that the rubber stops charges from flowing. This prevents people from getting electrical shocks by stopping the flow  of electricity.

  • A rubber is an insulator.
  • Insulators are substances that prevents the flow of electricity.
  • The lack free mobile electrons or ions that makes them conductors.
  • When they are wrapped round a conductor such as copper wire, they will halt the flow of charges.
  • Copper is a conductor of both heat and electricity. It has free mobile electrons.

learn more:

Metals brainly.com/question/2474874

#learnwithBrainly

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Which of the following occurs with both a cold front and a mountain breeze?
BigorU [14]

Answer:

b warm air rises

Explanation:

5 0
3 years ago
There is a seasaw that's holding two men. The seesaw has a length of 18m that can pivot from a point at its center. Man 1 has a
Alex

Answer:

Distance=  2.3864m

Explanation:

So that the balance is in equilibrium parallel to the floor, we must match the moment each man makes with respect to the pivot point.

In many cases the point of application of force does not coincide with the point of application in the body. In this case the force acts on the object and its structure at a certain distance, by means of an element that transfers that action of this force to the object.

This combination of force applied by the distance to the point of the structure where it is applied is called the moment of force F with respect to the point. The moment will attempt a rotation shift or rotation of the object. The distance from the force to the point of application is called the arm.

Mathematically it is calculated by expression:

M= F×d

The moment caused by the first man is:

M1= 75kg × (9.81m/s²) × 1.75m= 1287.5625 N×m

The moment caused by the second man must be equal to that caused by the first by which:

M2= 1287.5625 N×m= 55kg × (9.81m/s²) × distance ⇒

⇒distance= (1287.5625 N×m)/( (55kg × (9.81m/s²) )= 2.3864m

At this distance from the pivot point, the second should sit down so that the balance is balanced parallel to the ground.

3 0
3 years ago
How large must the coefficient of static friction be between the tires and road, if a car rounds a level curve of radius 85 m at
tatuchka [14]

Answer:

0.66

Explanation:

By using the formula

u = v^2 / r g

Where u is coefficent of friction

u = 23.5 × 23.5 / (85 × 9.8)

u = 0.66

4 0
3 years ago
Sphere A of mass 0.600 kg is initially moving to the right at 4.00 m/s. sphere B, of mass 1.80 kg is initially to the right of s
anzhelika [568]

A) The velocity of sphere A after the collision is 1.00 m/s to the right

B) The collision is elastic

C) The velocity of sphere C is 2.68 m/s at a direction of -5.2^{\circ}

D) The impulse exerted on C is 4.29 kg m/s at a direction of -5.2^{\circ}

E) The collision is inelastic

F) The velocity of the center of mass of the system is 4.00 m/s to the right

Explanation:

A)

We can solve this part by using the principle of conservation of momentum. The total momentum of the system must be conserved before and after the collision:

p_i = p_f\\m_A u_A + m_B u_B = m_A v_A + m_B v_B

m_A = 0.600 kg is the mass of sphere A

u_A = 4.00 m/s is the initial velocity of the sphere A (taking the right as positive direction)

v_A is the final velocity of sphere A

m_B = 1.80 kg is the mass of sphere B

u_B = 2.00 m/s is the initial velocity of the sphere B

v_B = 3.00 m/s is the final velocity of the sphere B

Solving for vA:

v_A = \frac{m_A u_A + m_B u_B - m_B v_B}{m_A}=\frac{(0.600)(4.00)+(1.80)(2.00)-(1.80)(3.00)}{0.600}=1.00 m/s

The sign is positive, so the direction is to the right.

B)

To verify if the collision is elastic, we have to check if the total kinetic energy is conserved or not.

Before the collision:

K_i = \frac{1}{2}m_A u_A^2 + \frac{1}{2}m_B u_B^2 =\frac{1}{2}(0.600)(4.00)^2 + \frac{1}{2}(1.80)(2.00)^2=8.4 J

After the collision:

K_f = \frac{1}{2}m_A v_A^2 + \frac{1}{2}m_B v_B^2 = \frac{1}{2}(0.600)(1.00)^2 + \frac{1}{2}(1.80)(3.00)^2=8.4 J

The total kinetic energy is conserved: therefore, the collision is elastic.

C)

Now we analyze the collision between sphere B and C. Again, we apply the law of conservation of momentum, but in two dimensions: so, the total momentum must be conserved both on the x- and on the y- direction.

Taking the initial direction of sphere B as positive x-direction, the total momentum before the collision along the x-axis is:

p_x = m_B v_B = (1.80)(3.00)=5.40 kg m/s

While the total momentum along the y-axis is zero:

p_y = 0

We can now write the equations of conservation of momentum along the two directions as follows:

p_x = p'_{Bx} + p'_{Cx}\\0 = p'_{By} + p'_{Cy} (1)

We also know the components of the momentum of B after the collision:

p'_{Bx}=(1.20)(cos 19)=1.13 kg m/s\\p'_{By}=(1.20)(sin 19)=0.39 kg m/s

So substituting into (1), we find the components of the momentum of C after the collision:

p'_{Cx}=p_B - p'_{Bx}=5.40 - 1.13=4.27 kg m/s\\p'_{Cy}=p_C - p'_{Cy}=0-0.39 = -0.39 kg m/s

So the magnitude of the momentum of C is

p'_C = \sqrt{p_{Cx}^2+p_{Cy}^2}=\sqrt{4.27^2+(-0.39)^2}=4.29 kg m/s

Dividing by the mass of C (1.60 kg), we find the magnitude of the velocity:

v_c = \frac{p_C}{m_C}=\frac{4.29}{1.60}=2.68 m/s

And the direction is

\theta=tan^{-1}(\frac{p_y}{p_x})=tan^{-1}(\frac{-0.39}{4.27})=-5.2^{\circ}

D)

The impulse imparted by B to C is equal to the change in momentum of C.

The initial momentum of C is zero, since it was at rest:

p_C = 0

While the final momentum is:

p'_C = 4.29 kg m/s

So the magnitude of the impulse exerted on C is

I=p'_C - p_C = 4.29 - 0 = 4.29 kg m/s

And the direction is the angle between the direction of the final momentum and the direction of the initial momentum: since the initial momentum is zero, the angle is simply equal to the angle of the final momentum, therefore -5.2^{\circ}.

E)

To check if the collision is elastic, we have to check if the total kinetic energy is conserved or not.

The total kinetic energy before the collision is just the kinetic energy of B, since C was at rest:

K_i = \frac{1}{2}m_B u_B^2 = \frac{1}{2}(1.80)(3.00)^2=8.1 J

The total kinetic energy after the collision is the sum of the kinetic energies of B and C:

K_f = \frac{1}{2}m_B v_B^2 + \frac{1}{2}m_C v_C^2 = \frac{1}{2}(1.80)(1.20)^2 + \frac{1}{2}(1.60)(2.68)^2=7.0 J

Since the total kinetic energy is not conserved, the collision is inelastic.

F)

Here we notice that the system is isolated: so there are no external forces acting on the system, and this means the system has no acceleration, according to Newton's second law:

F=Ma

Since F = 0, then a = 0, and so the center of mass of the system moves at constant velocity.

Therefore, the centre of mass after the 2nd collision must be equal to the velocity of the centre of mass before the 1st collision: which is the velocity of the sphere A before the 1st collision (because the other 2 spheres were at rest), so it is simply 4.00 m/s to the right.

Learn more about momentum and collisions:

brainly.com/question/6439920

brainly.com/question/2990238

brainly.com/question/7973509

brainly.com/question/6573742

#LearnwithBrainly

8 0
3 years ago
Why do planets rings surround them and not the sun
Shalnov [3]

Well, the rings surrounding a planet are made out of rock. A ring surrounding the sun would be impossible since the sun can reach more than 27 million degrees Fahrenheit (15 million degrees Celsius.)

Hope this helped.

7 0
3 years ago
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