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Delicious77 [7]
3 years ago
15

How do you test hydrogen gas

Physics
2 answers:
Luda [366]3 years ago
7 0

Answer:

You can test for hydrogen gas by putting a burning splinter near the gas. The splinter will gives out a 'pop' sound (mini explosion) if it is hydrogen gas

Vanyuwa [196]3 years ago
3 0

Explanation:

A splint is lit and held near the opening of the tube, then the stopper is removed to expose the splint to the gas. If the gas is flammable, the mixture ignites. This test is most commonly used to identify hydrogen, which extinguishes with a distinctive 'squeaky pop' sound.

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Can someone pls help me with this?
alexira [117]
C because they are both going in a constant speed
6 0
3 years ago
3.A golf ball is hit of it’s tee, 200m down the fairway.
Alinara [238K]

<u>Answers:</u>


3. The diagrams showing the forces acting on the golf ball are in the figure attached. Let’s have a detailed look:


a) Here the ball is under 2 forces:


F1 which is called The Normal force and is perpendicular to the surface of the tee where the ball rests


-F1, related to the gravity force, to the weight of the ball, and has the same magnitude, but the opposite direction (That’s why it has a negative sign).



In this case the sum of the forces is 0



b) Here we have again forces F1 and –F1, but in this very moment the club strikes the ball and we have:


F2, the force of the strike





c) While the ball is in its flight, it is under the following forces:


F1, the force of the lift through the air


-F2, the gravity force

-F3, the force of air resistance, also called drag


F4, the tangencial force of the ball flight




4. Here are the sizes and directions of the resultant forces:


i)Two forces of the same magnitude or size are applied to this block, but in opposite directions (in the x-axis). This is expressed as:



F=-10N+10N=0 The resulting force is zero



ii) Two forces of different size and opposite directions in the y-axis are applied to this block. The sum of the forces is:



F=30N-40N=-10N This means the resulting force is 10N applied downwards



iii) In this case the only force applied to the block is -5N applied downwards



iv) Here there are four forces applied to the block.  

In the y-axis we have to forces of the same size but opposite directions:



F1=10N-10N=0 This means the applied force in the y-axis is zero



In the x-axis we have two forces of different size and opposite directions:



F2=-15N+10N=-5N This means the resulting force is applied to the –x-side



4 0
3 years ago
Explain how and why water travels through the four Earth systems as it moves through the water cycle
Nostrana [21]
Water in the Earth system is influencing all aspects of life on Earth. ... The returning water falls directly back into the oceans, or onto land as snow or rain. It soaks into the soil to move into the groundwater or runs off the Earth's surface in streams, rivers and lakes, which drain back into the oceans.
6 0
3 years ago
You place a large pebble in a slingshot, pull the elastic band back to your chin, and release it, launching the pebble horizonta
sladkih [1.3K]

Answer:

The larger pebble has 25 times more mass.

Explanation:

To solve the exercise it is necessary to apply the work and energy conservation equations.

For the case described, the work done must be preserved and must be the same, that is,

W = 0

By definition work linked to the conservation of kinetic energy would be given by

\Delta KE = W = 0

\Delta KE = 0

KE_{larger}-KE_{smaller} = 0

KE_{larger}=KE_{smaller}

\frac{1}{2}m_lv_l^2 = \frac{1}{2}m_sv_s^2

m_lv_l^2 = m_sv_s^2

The ratio between the mass and the velocity would be,

\frac{m_l}{m_s}=\frac{v_s^2}{v_l^2}

\frac{m_l}{m_s} = \frac{500^2}{100^2}

\frac{m_l}{m_s} = 25

Therefore the answer is: The larger pebble has 25 times more mass.

4 0
3 years ago
A uniform electric field is produced due to the charge distribution inside the closed cylindrical surface. (a) What type of char
DiKsa [7]

Answer with Explanation:

a. Option d is true.

a negatively charged plane parallel to the end faces of the cylinder

b. Radius of cylinder, r=0.66m

Magnitude of electric field, E=300 N/C

We have to find the net flux through the closed surface.

Net electric flux,\phi=-2 EA=-2E(\pi r^2)

\phi=-2\times 300\times (3.14\times (0.66)^2)

\phi=-820.67 Nm^2/C

c.

Net charge,Q=\epsilon_0\times \phi

Where

\epsilon_0=8.85\times 10^{-12}

Q=-820.67\times 8.85\times 10^{-12}

Q=-7.26\times 10^{-9} C

Q=-7.26nC

Where 1nC=10^{-9}C

7 0
3 years ago
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