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s344n2d4d5 [400]
3 years ago
12

How do you test hydrogen gas ​

Physics
1 answer:
neonofarm [45]3 years ago
4 0

Explanation:

The test for hydrogen consists of holding a burning splint held at the open end of a test tube of gas. If the gas is hydrogen it burns with a loud “squeaky pop” which is the result of the rapid combustion of hydrogen with oxygen to produce water.

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What type of mage can never be formed by a converging lens?
Irina18 [472]

Answer: Real image

Explanation:

converging lens will only produce a real image if the object is located beyond the focal point (i.e., more than one focal length away).

4 0
4 years ago
Which layer lies between the stratosphere and the
Pepsi [2]
The answer is Mesosphere
8 0
3 years ago
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In the periodic table, the most reactive metals are found a. in Group 1, the first column on the left. b. in Period 1, the first
Gnesinka [82]

Answer:

Bottom left corner for whatever group that is

Lithium, sodium, and potassium all react with water

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3 years ago
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A uniform solid sphere has a moment of inertia I about an axis tangent to its surface. What is the moment of inertia of this sph
arsen [322]

Answer:

option E

Explanation:

given,

I is moment of inertia about an axis tangent to its surface.

moment of inertia about the center of mass

I_{CM} = \dfrac{2}{5}mR^2.....(1)

now, moment of inertia about tangent

I= \dfrac{2}{5}mR^2 + mR^2

I= \dfrac{7}{5}mR^2...........(2)

dividing equation (1)/(2)

\dfrac{I_{CM}}{I}= \dfrac{\dfrac{2}{5}mR^2}{\dfrac{7}{5}mR^2}

\dfrac{I_{CM}}{I}=\dfrac{2}{7}

I_{CM}=\dfrac{2}{7}I

the correct answer is option E

4 0
3 years ago
A single slit 1.4 mmmm wide is illuminated by 460-nmnm light. Part A What is the width of the central maximum (in cmcm ) in the
mamaluj [8]

Answer:

<h2> 1.643*10⁻⁴cm</h2>

Explanation:

In a single slit experiment, the distance on a screen from the centre point is expressed as y = \frac{\delta m \lambda d}{a} where;

\delta m is the first two diffraction minima = 1

\lambda is light wavelength

d is the distance of diffraction pattern from the screen

a is the width of the slit

Given \lambda = 460-nm = 460*10⁻⁹m

d = 5.0mm = 5*10⁻³m

a = 1.4mm = 1.4*10⁻³m

Substituting this values into the formula above to get width of the central maximum y;

y = 1*460*10⁻⁹ * 5*10⁻³/1.4*10⁻³

y = 2300*10⁻¹²/1.4*10⁻³

y = 1642.86*10⁻⁹

y = 1.643*10⁻⁶m

Converting the final value to cm,

since 100cm = 1m

x = 1.643*10⁻⁶m

x = 1.643*10⁻⁶ * 100

x = 1.643*10⁻⁴cm

Hence, the width of the central maximum in the diffraction pattern on a screen 5.0 mm away is  1.643*10⁻⁴cm

3 0
3 years ago
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