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3 years ago

8

If a window is double glazed rather than single glazed there is less condensation on the inside of the glass. Explain why.

1 answer:

sashaice [31]3 years ago

7 0

Because it tends to be hotter inside therefore when the cold air hits the inside of the glass, the cold glass makes it condense however outside its cold therefore it does condense on the outside of the window

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In the attachment there is a density column where there is colour

Anettt [7]

Answer:

For your kind information, read the question thoroughly!

The bottom is the container part and not the liquid.

I hope it will be useful.

4 0

3 years ago

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Classify Use periodic table to classify the elements potassium (K) ,Bromine (Br) and argon, (Ar) according to know likely their

kodGreya [7K]

Potassium-A
Bromine-B
Argon-C

8 0

3 years ago

What is the maximum value of the magnetic field at a<br> distance2.5m from a 100-W light bulb?

MA_775_DIABLO [31]

To solve this problem we will apply the concepts related to the intensity included as the power transferred per unit area, where the area is the perpendicular plane in the direction of energy propagation.

Since the propagation occurs in an area of spherical figure we will have to

$I = \frac{P}{A}$

$I = \frac{P}{4\pi r^2}$

Replacing with the given power of the Bulb of 100W and the radius of 2.5m we have that

$I = \frac{100}{4\pi (2.5)^2}$

$I = 1.2738W/m^2$

The relation between intensity I and $E_{max}$

$I = \frac{E_max^2}{2\mu_0 c}$

Here,

$\mu_0$ = Permeability constant

c = Speed of light

Rearranging for the Maximum Energy and substituting we have then,

$E_{max}^2 = 2I\mu_0 c$

$E_{max}=\sqrt{2I\mu_0 c }$

$E_{max} = 2(1.2738)(4\pi*10^{-7})(3*10^8)$

$E_{max} = 30.982 V/m$

Finally the maximum magnetic field is given as the change in the Energy per light speed, that is,

$B_{max} = \frac{E_{max}}{c}$

$B_{max} = \frac{30.982 V /m}{3*10^8}$

$B_{max} = 1.03275 *10{-7} T$

Therefore the maximum value of the magnetic field is $B_{max} = 1.03275 *10{-7} T$

3 0

3 years ago

A circuit has a resistance of 10ohmsband a current of 42 amps.caculate the voltage

Anon25 [30]

R=U/I so
U=RxI
U= 10 x 42
U= 420 volts

3 0

3 years ago

Determine the thrust produced if 1.5 x 10^3 kg of gas exits the combustion chamber each second, with a speed of 4.00 x 10^3 m/s.

ozzi

Answer:

The thrust is $6\times 10^6\ N$

Explanation:

Given that,

Mass of gas, $m=1.5\times 10^3\ kg$

The rate at which the gas is expelling, $\dfrac{dv}{dt}=4\times 10^{3}\ m/s$

We need to find the thrust produced by the gas.

We know that force is equal to the rate of change of momentum. So,

$F=\dfrac{p}{t}$

Also, p = mv

$F=\dfrac{mv}{t}$

So,

$F=1.5\times 10^3\times 4\times 10^3\\\\F=6\times 10^6\ N$

So, the thrust is $6\times 10^6\ N$

3 0

3 years ago