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3 years ago

The handle of a pressure cooker is covered with plastic.why?

1 answer:

8 0

Answer: handle of pressure cooker is covered with thick plastic. ... Plastic is a bad conductor of heat. The pressure cooker is made of metal which can conduct heat. Thus, to protect us from burns, the handles of a pressure cooker is made of plastic

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Explanation:

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What is the net force of a falling bowling ball that weighs 55 N and has 15 N of air resistance?

serg [7]

its 55 :) have a great day queen/king/whatever gender you are

3 0

3 years ago

Read 2 more answers

when a temparature of a coin is 75°C, the coin's diameter increases. if the original diameter of a coin is 1.8*10^-2 m and its c

andrey2020 [161]

Answer:

ΔD = 2.29 10⁻⁵ m

Explanation:

This is a problem of thermal expansion, if the temperature changes are not very large we can use the relation

ΔA = 2α A ΔT

the area is

A = π r² = π D² / 4

we substitute

ΔA = 2α π D² ΔT/4

as they do not indicate the initial temperature, we assume that ΔT = 75ºC

α = 1.7 10⁻⁵ ºC⁻¹

we calculate

ΔA = 2 1.7 10⁻⁵ pi (1.8 10⁻²) ² 75/4

ΔA = 6.49 10⁻⁷ m²

by definition

ΔA = A_f- A₀

A_f = ΔA + A₀

A_f = 6.49 10⁻⁷ + π (1.8 10⁻²)² / 4

A_f = 6.49 10⁻⁷ + 2.544 10⁻⁴

A_f = 2,551 10⁻⁴ m²

the area is

A_f = π D_f² / 4

A_f = $\sqrt{4 A_f /\pi }$

D_f = $\sqrt{4 \ 2.551 10^{-4} /\pi }$

D_f = 1.80229 10⁻² m

the change in diameter is

ΔD = D_f - D₀

ΔD = (1.80229 - 1.8) 10⁻² m

ΔD = 0.00229 10⁻² m

ΔD = 2.29 10⁻⁵ m

5 0

3 years ago

Which of the following in the list is the best conductor: plastic, metal, styrofoam, or glass?

tensa zangetsu [6.8K]

Answer:

Metal

Explanation:

3 0

3 years ago

Read 2 more answers

Helium gas contained in a piston cylinder assembly undergoes an isentropic polytropic process from the given initial state with

jeyben [28]

Answer:

Heat transfer during the process = 0

Work done during the process = - 371.87 KJ

Explanation:

Initial pressure $P_{1}$ = 0.02 bar

Initial temperature $T_{1}$ = 200 K

Final pressure $P_{2}$ = 0.14 bar

Gas constant for helium R = 2.077 $\frac{KJ}{kg k}$

This is an isentropic polytropic process so temperature - pressure relationship is given by the following formula,

$\frac{T_{2} }{T_{1} }$ = $[\frac{P_{2} }{P_{1} } ]^{\frac{\gamma - 1}{\gamma} }$

Put all the values in above formula we get,

⇒ $\frac{T_{2} }{200}$ = $[\frac{0.14 }{0.02 } ]^{\frac{1.4 - 1}{1.4} }$

⇒ $\frac{T_{2} }{200}$ = 1.74

⇒ $T_{2}$ = 348.72 K

This is the final temperature of helium.

For isentropic polytropic process heat transfer to the system is zero.

⇒ ΔQ = 0

Work done W = m × ( $T_{1}$ - $T_{2}$ ) × $\frac{R}{\gamma - 1}$

⇒ W = 1 × ( 200 - 348.72 ) × $\frac{2.077}{1.4 - 1}$

⇒ W = 371.87 KJ

This is the work done in this process. here negative sign shows that work is done on the gas in the compression of gas.

3 0

3 years ago

Consider a semi-infinite (hollow) cylinder of radius R with uniform surface charge density. Find the electric field at a point o

VikaD [51]

Answer:

For the point inside the cylinder: $E = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + 4x_0^2}}$

For the point outside the cylinder: $E = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + x_0^2}}$

where x0 is the position of the point on the x-axis and σ is the surface charge density.

Explanation:

Let us assume that the finite end of the cylinder is positioned at the origin. And the rest of the cylinder lies on the (-x) axis, which is the vertical axis in this question. In the first case (inside the cylinder) we will calculate the electric field at an arbitrary point -x0. In the second case (outside), the point will be +x0.

The cylinder is consist of the sum of the rings with the same radius.

First we will calculate the electric field at point -x0 created by the ring at an arbitrary point x.

We will also separate the ring into infinitesimal portions of length 'ds' where ds = Rdθ.

The charge of the portion 'ds' is 'dq' where dq = σds = σRdθ. σ is the surface charge density.

Now, the electric field created by the small portion is 'dE'.

$dE = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rd\theta}{R^2+x^2}$

The electric field is a vector, and it needs to be separated into its components in order us to integrate it. But, the sum of horizontal components is zero due to symmetry. Every dE has an equal but opposite counterpart which cancels it out. So, we only need to take the component with the sine term.

$dE = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rd\theta}{R^2+x^2} \frac{x}{\sqrt{x^2+R^2}} = dE = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rxd\theta}{(R^2+x^2)^{3/2}}$

We have to integrate it over the ring, which is an angular integration.

$E_{ring} = \int{dE} = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rx}{(R^2+x^2)^{3/2}}\int\limits^{2\pi}_0 {} \, d\theta = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rx}{(R^2+x^2)^{3/2}}2\pi = \frac{1}{2\epsilon_0}\frac{\sigma Rx}{(R^2+x^2)^{3/2}}$

This is the electric field created by a ring a distance x away from the point -x0. Now we can integrate this electric field over the semi-infinite cylinder to find the total E-field:

$E_{cylinder} = \int{E_{ring}} = \frac{\sigma R}{2\epsilon_0}\int\limits^{-\inf}_{-2x_0} \frac{x}{(R^2+x^2)^{3/2}}dx = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + 4x_0^2}}$

The reason we integrate over -2x0 to -inf is that the rings above -x0 and below to-2x0 cancel out each other. Electric field is created by the rings below -2x0 to -inf.

We will only change the boundaries of the last integration.

$E_{cylinder} = \int{E_{ring}} = \frac{\sigma R}{2\epsilon_0}\int\limits^{-\inf}_{x_0} \frac{x}{(R^2+x^2)^{3/2}}dx = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + x_0^2}}$

6 0

3 years ago