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4 years ago

7

Which of the following is not a means to reduce the rate of heat transfer by conduction through a wall?

1 answer:

DENIUS [597]4 years ago

6 0

The correct answer is 3

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One end of a rope is tied to the handle of a horizontally-oriented and uniform door. a force f is applied to the other end of th

Vlada [557]

For rotational equilibrium of the door we can say that torque due to weight of the door must be counter balanced by the torque of external force

$F\times L = mg \times \frac{L}{2}$

here weight will act at mid point of door so its distance is half of the total distance where force is applied

here we know that

$mg = 145 N$

now we will have

$F = \frac{mg}{2}$

$F = \frac{145}{2} = 72.5 N$

so our applied force is 72.5 N

7 0

4 years ago

The membrane that surrounds a certain type of living cell has a surface area of 5.3 x 10-9 m2 and a thickness of 1.1 x 10-8 m. A

kotykmax [81]

Answer:

$2.1\times 10^{-12} c$

Explanation:

We are given that

Surface area of membrane= $5.3\times 10^{-9} m^2$

Thickness of membrane= $1.1\times 10^{-8} m$

Assume that membrane behave like a parallel plate capacitor.

Dielectric constant=5.9

Potential difference between surfaces=85.9 mV

We have to find the charge resides on the outer surface of membrane.

Capacitance between parallel plate capacitor is given by

$C=\frac{k\epsilon_0 A}{d}$

Substitute the values then we get

Capacitance between parallel plate capacitor= $\frac{5.9\times 8.85\times 10^{-12}\times 5.3\times 10^{-9}}{1.1\times 10^{-8}}$

$C=0.25\times 10^{-12}F$

V= $85.9 mV=85.9\times 10^{-3}$

$Q=CV$

$Q=0.25\times 10^{-12}\times 85.9\times 10^{3}=2.1\times 10^{-12} c$

Hence, the charge resides on the outer surface= $2.1\times 10^{-12} c$

5 0

3 years ago

Read 2 more answers

Please help asap asap<br> wdtwygcbhyeegdlu tcgbihjlnkdm;fnbkhjg hv

kicyunya [14]

Answer:

B

Explanation:

Bbbb

6 0

3 years ago

Two boxes are at rest on a smooth, horizontal surface. The boxes are in contact with one another. If box 1 is pushed with a forc

Maslowich

Answer:

mass of box 1 = 2.20 kg

mass of box 2 = 5.93 kg

Explanation:

Let the mass of box 1 and box 2 is respectively

$m_1$ and $m_2$

so we will have

Force applied on box 1 then acceleration

$a = \frac{F}{m_1 + m_2}$

$1.50 = \frac{12.2}{m_1 + m_2}$

$m_1 + m_2 = 8.13$

Now we know that contact force between them in above case is given as

$F_n = m_2a$

$8.90 = m_2(1.5)$

$m_2 = 5.93 kg$

now we have

$m_1 = 2.20 kg$

8 0

3 years ago

Inside a NASA test vehicle, a 3.50-kg ball is pulled along by a horizontal ideal spring fixed to a friction-free table. The forc

erastova [34]

F(of spring)=230x=ma=3.5(5)=17.5=230x; x=0.07m.

3 0

3 years ago