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3 years ago

Explain why materials such as plastic foam, feathers, and fur are poor conductors of heat.

1 answer:

7 0

The atoms of some materials have no free electrons in their outer orbits. These electrons are busy doing other jobs, like being shared in the orbits of two adjacent atoms. They are so closely held that it is very difficult to pull them away. Most compounds of carbon and hydrogen are like this.

<span>Plastics, whose molecules are made from long combinations of carbon and hydrogen atoms, have few or no free electrons. This means that plastics are poor conductors of electricity (and they are also poor conductors of heat). hope that helped.</span>

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A person pulls a crate of mass M = 63 kg a distance 40.0 m along a horizontal floor by a constant force FP = 130 N, which acts a

GrogVix [38]

Answer:

Check attachment for solution and diagrams

Explanation:

Given that,

Mass of crate m=63kg

Distance travelled d=40m

Horizontal force Fx=130N

Angle the force applied on cord makes with horizontal is θ=23°.

The weight of the crate is given by

W=mg

W=63×9.81

W=618.03N

Horizontal force Fx=130N

Resolving the applied Force F to the horizontal will give

Fx=FCos θ

F=Fx/Cos θ

F=130/Cos23

F=141.2N

a. Check attachment for model diagram

b. Check attachment for free body diagram

c. Check attachment for pictorial representation

d. Work done by gravitational force.

We, know that the body did not move upward, then the distance d=0

Work done is given as

W=F×d

So, d=0

W=F×0

W=0J

So, no work is done by gravity

e. Normal force?

Using newton law of motion

ΣFy = may

Since the body is not moving upward, then ay=0m/s²

N+141.2Sin23-618.03=0

N=618.03-141.2Sin23

N=562.86N

f. Work done by normal force.

The body is not moving upward, then the distance is zero

d=0

Work done by normal=normal force × distance

Wn=562.86×0

Wn=0J

No work is done by the normal force

g. Frictional force?

Since the coefficient of kinetic friction is zero, then the surface is frictionless

So, no frictional force is acting on the body

Fictional force is given as

Fr=μk•N

Given that, μk=0

Fr=0×562.86

Fr=0N

d. Work done by frictional force?

Since the frictional force Is zero, then, no work is done by friction

W(friction ) = frictional force × d

Here, the body moved a distance of 40m

W(fr)=0×40

W(fr)=0J

No work is done by friction

I. Work done by exerted force

The horizontal component of the exerted force is 130N and the body traveled a distance of 40m

Then, work done is given as

Workdone=force ×distance

Work done=130×40

W=5200J

W=5.2KJ

h. Net workdone?

Since no work is lost by friction, then, the net work done is equal to the work done by the exerted force.

Went, = work done by force exerted - work done by friction

Wnet=5200-0

Wnet, =5200

Wnet=5.2KJ

7 0

3 years ago

A boat travels at 15 m/s in a direction 45° east of north for an hour. The boat then turns and travels at 18 m/s in a direction

Nuetrik [128]

Answer:

brainly.com/question/5055678

Explanation:

4 0

3 years ago

30) A force produces power P by doing work W in a time T. What power will be produced by a force that does six times as much wor

schepotkina [342]

Answer:

A) 12P

Explanation:

The power produced by a force is given by the equation

$P=\frac{W}{T}$

where

W is the work done by the force

T is the time in which the work is done

At the beginning in this problem, we have:

W = work done by the force

T = time taken

So the power produced is

$P=\frac{W}{T}$

Later, the force does six times more work, so the work done now is

$W'=6W$

And this work is done in half the time, so the new time is

$T'=\frac{T}{2}$

Substituting into the equation of the power, we find the new power produced:

$P'=\frac{W'}{T'}=\frac{6W}{T/2}=12\frac{W}{T}=12P$

So, 12 times more power.

4 0

3 years ago

Can someone help me answer this??

Bingel [31]

Answer:

7 0

3 years ago

A 22.0 kg child slides down a slide that makes a 37.0° angle with the horizontal. (a) What is the magnitude of the normal force

patriot [66]

Answer:

(a) 172.185 N

(b) $53^{\circ}$

Solution:

As per the question:

Mass of the child, m = 22.0 kg

Angle, $\theta = 37.0^{\circ}$

Now,

(a) The magnitude of the normal force exerted by the slide on the child:

$F_{N} = mgcos\theta$

$F_{N} = 22\times 9.8cos37^{\circ} = 172.185\ N$

Now,

(b) The angle from the horizontal at which the force is directed is:

$90^{\circ} - 37^{\circ} = 53^{\circ}$

6 0

3 years ago