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Anna [14]
1 year ago
8

Pot holder should have high insulation and low _____.

Physics
1 answer:
kipiarov [429]1 year ago
3 0

Potholder should have high insulation and low conductivity, therefore the correct answer is the option B

<h3>What is insulation?</h3>

Insulation is a type of material used to create barriers to the transmission of the form of energy which either is in form of heat or electricity.

For outdoor trips in cold weather, several thin layers act as better insulating barriers for heat transfer.

The ability of an electric charge or heat to pass through a material is measured by its conductivity. A material is considered a conductor if it offers very little resistance to the flow of thermal or electric energy.

Thus, Potholders should be highly insulated and have low conductivity, therefore the correct answer is the option B

Learn more about insulation from here

brainly.com/question/14363642

#SPJ1

your question seems incomplete, the complete question is

To be effective, a pot holder should have low _____. viscosity conductivity malleability density

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If the sprinter from the previous problem accelerates at that rate for 20 m, and then maintains that velocity for the remainder
kakasveta [241]

Question:

A 63.0 kg sprinter starts a race with an acceleration of 4.20m/s square. What is the net external force on him? If the sprinter from the previous problem accelerates at that rate for 20m, and then maintains that velocity for the remainder for the 100-m dash, what will be his time for the race?

Answer:

Time for the race will be t = 9.26 s

Explanation:

Given data:

As the sprinter starts the race so initial velocity = v₁ = 0

Distance = s₁ = 20 m

Acceleration = a = 4.20 ms⁻²

Distance = s₂ = 100 m

We first need to find the final velocity (v₂) of sprinter at the end of the first 20 meters.

Using 3rd equation of motion

(v₂)² - (v₁)² = 2as₁ = 2(4.2)(20)

v₂ = 12.96 ms⁻¹

Time for 20 m distance = t₁ = (v₂ - v ₁)/a

t₁ = 12.96/4.2 = 3.09 s

He ran the rest of the race at this velocity (12.96 m/s). Since has had already covered 20 meters, he has to cover 80 meters more to complete the 100 meter dash. So the time required to cover the 80 meters will be

Time for 100 m distance = t₂ = s₂/v₂

t₂ = 80/12.96 = 6.17 s

Total time = T = t₁ + t₂ = 3.09 + 6.17 = 9.26 s

T = 9.26 s

5 0
3 years ago
The vector position of an object is given by r what is the torque acting on the object about the origin when a force f = (−12.5i
attashe74 [19]
Let the vector position of the object in the (x-y) plane be 
\vec{r} = x \hat{i} + y \hat{j}

The applied force is
\vec{f} = -12.5 \hat{i}&#10;

By definition, the applied torque is
\vec{T} = \vec{r} \times \vec{f} = (x\hat{i} + y\hat{j}) \times (-12.5y \hat{i}) = 12.5\hat{k}

Answer: 12.5y \, \hat{k}

7 0
3 years ago
1. An express train, traveling at 36 m/s, is accidentally sidetracked onto a local train track. The express engineer spots a loc
Colt1911 [192]

Answer:

(i) 12 seconds

(ii) 216 meters from the initial position

(iii) 132 meters from the initial position

(iv) No

Explanation:

Speed of express train =36 m/s

Speed of local train =11 m/s

The initial distance between the local train and passenger train =100 m.

Due to the application of breaks, the express train slows at the rare of 3.0 m/s^2.

So, the acceleration of the express train, a=-3 m/s^2.

(i) Let t be the time the express train takes to stop.

From the equation of motion,

v=u+at

where, v: final velocity, u: initial velocity, a: constant acceleration, t: time taken to change the speed from u to v.

In this case, v=0, u=36 m/s, a=-3 m/s^2

So, 0=36+(-3)t

\Rightarrow t= 36/3=12 seconds.

(ii) To compute the distance traveled, s, till the express train stops, using

v^2=u^2+2as

\Rightarrow 0^2=36^2+2(-3)s

\Rightarrow s=\frac{36\times36}{6}

\Rightarrow s=216 meters.

(iii) The local train is moving at a speed of 11 m/s

So, in 12 seconds, the distance, d, traveled by the local train

d= 11x12=132 meters [as distance= speed x time]

(iv) Let 0 be the reference position which is the initial position of the express train.

So, at the initial time, the position of the local train is at 100m.

After 12 seconds:

The position of the express train is at 216 m [using part (ii)]

and the position of the local train is at 100+132=232m  [using part (iii)].

So, the local train is still ahead of the express train, hence the trains didn't collide.

6 0
3 years ago
If you double the mass but don't change the net force what is the result on the acceleration
vazorg [7]
F is force, m is mass and a<span> is acceleration. The math behind this is quite simple. If you double the force, you double the acceleration, but if you double the mass, you cut the acceleration in half.</span>
5 0
3 years ago
The biological roles of complex organic molecules are determined by their shape -- the way atoms and electrons create charge dis
Hatshy [7]

Answer:

a) P_α =  exp (-ΔE / kT),  b)   P_β = 0.145 , d)  ΔE = 309.7 meV

Explanation:

The expression for the number of molecules or particles in a given state in Boltzmann's expression

            n = n₀ exp (-ΔE / kT)

Where k is the Bolztmann constant and T the absolute temperature

The probability is defined as the number of molecules in a given state over the total number of particles

          P = n / n₀ = exp (- ΔE / kT)

Let's apply this expression to our case

a) P_α = n_α / n₀ = exp (-ΔE / kT)

b) the Boltzmann constant

       k = 1,381 10⁻²³ J / K (1 eV / 1.6 10⁻¹⁹ J) = 8.63 10⁻⁵ eV / K

       kT = 8.63 105 300 = 2,589 10⁻² eV

       P_β = exp (- 50 10⁻³ /2.589 10⁻² = exp (-1.931)

       P_β = 0.145

c) If the temperature approaches absolute zero, the so-called is very high, so there is no energy to reach the excited state, therefore or all the molecules go to the alpha state

d) For molecules to spend ¼ of the time in this beta there must be ¼ of molecules in this state since the decay is constant.

        P_β = ¼ = 0.25

     

       P_β = exp (- ΔE / kT)

       ΔE = -kT ln P_β

       ΔE = - 2,589 10⁻² ln 0.25

       ΔE = 0.3097 eV

       ΔE = 309.7 meV

3 0
3 years ago
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