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777dan777 [17]
3 years ago
12

Given the resistivities below, which material is best described as an insulator?

Physics
1 answer:
koban [17]3 years ago
5 0

Answer: D) 3.2 × 10^8

Explanation: Insulators are best described as materials which are poor conductors of electricity and hence do not allow passage of electricity through them. Resistivity are hence used to describe property or characteristic of a material which does not allow the flow or passage of electric current.

In summary, materials which high resistivity values are referred to as Insulators. Conversely, those with low resistivity or high conductivity values are called conductors while those with intermediate values are named semiconductors.

In the question given above, the material with a resistivity value of 3.2 × 10^8 - - - has a very high resistivity value (320000000Ω), the other options given have very low resistivity values ; 0.0000000017, etc and are most likely to be conductors due to their excessively low resistivity values.

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If a force of 1.0 N pulled up and parallel to the surface of the incline is required to raise the mass back to the top of the in
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The question is incomplete! The complete question along with answer and explanation is provided below.

Question:

A 0.5 kg mass moves 40 centimeters up the incline shown in the figure below. The vertical height of the incline is 7 centimeters.

What is the change in the potential energy (in Joules) of the mass as it goes up the incline?  

If a force of 1.0 N pulled up and parallel to the surface of the incline is required to raise the mass back to the top of the incline, how much work is done by that force?

Given Information:  

Mass = m = 0.5 kg

Horizontal distance = d = 40 cm = 0.4 m

Vertical distance = h = 7 cm = 0.07 m

Normal force = Fn = 1 N

Required Information:  

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Work done = W = ?

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Work done = 0.39 N.m

Explanation:

The potential energy is given by

PE = mgh

where m is the mass of the object, h is the vertical distance and g is the gravitational acceleration.

PE = 0.5*9.8*0.07

PE = 0.343 Joules

As you can see in the attached image

sinθ = opposite/hypotenuse

sinθ = 0.07/0.4

θ = sin⁻¹(0.07/0.4)

θ = 10.078°

The horizontal component of the normal force is given by

Fx = Fncos(θ)

Fx = 1*cos(10.078)

Fx = 0.984 N

Work done is given by

W = Fxd

where d is the horizontal distance

W = 0.984*0.4

W = 0.39 N.m

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