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Alenkinab [10]
3 years ago
6

Can a compound be broken down into simpler substances

Physics
1 answer:
jonny [76]3 years ago
4 0
Yes.
Take the compound glucose.
C6H12O6 is its formula, and when combined with O2 (oxygen gas), it breaks down into carbon dioxide and water, which are simpler substances. Also, compounds can be broken down into distinguishable elements, which are the simplest unit.
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How are the electric field lines around a positive charge affected when a second positive charge is near it?
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What is the difference between distance and displacement? Give an example of a situation where distance and displacement both ha
Oksana_A [137]

Explanation :

Distance is total path travelled by an object during its entire journey. It is a scalar quantity i.e only magnitude.

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When an object is travelling in a straight line and stops at the end point, then both distance and displacement are same.

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4 years ago
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6 0
3 years ago
Ans of this question A test charge of 1 couloumb moved from 30cm against the field of intensity 50N/c find the energy store in i
UkoKoshka [18]

Answer:

A. Zero

Explanation:

Given data,

The charge of the test charge, q = 1 C

The distance the charge moved against the filed of intensity, x = 30 cm

                                                                                                        = 0.3 m

The electric field intensity, E = 50 N/C

The energy stored in the charge at 0.3 m is given by the formula,

                                V = k q/r

Where,                        

                                     = 9 x 10⁹ Nm²C⁻²

The charge is moved from the potential V₁ to V₂ at 30 cm

Substituting the given values in the above equation

                            V₁ = 9 x 10⁹ x 30 / 0.3

                                =  1.5 x 10¹² J

And,

                            V₂ = 1.5 x 10¹² J

The energy stored in it is,

                             W = V₂ - V₁

                                  = 0

Hence, the energy stored in the charge is, W = 0        

6 0
3 years ago
A 7 ft tall person is walking away from a 20 ft tall lamppost at a rate of 5 ft/sec. Assume the scenario can be modeled with rig
aleksandr82 [10.1K]

Answer:

The rate of change of the shadow length of a person is 2.692 ft/s

Solution:

As per the question:

Height of a person, H = 20 ft

Height of a person, h = 7 ft

Rate = 5 ft/s

Now,

From Fig.1:

b = person's distance from the lamp post

a = shadow length

Also, from the similarity of the triangles, we can write:

\frac{a + b}{20} = \frac{a}{7}

a = \frac{7}{13}b

Differentiating the above eqn w.r.t t:

\frac{da}{dt} = \frac{7}{13}.\frac{db}{dt}

Now, we know that:

Rate = \frac{db}{dt} = 5\ ft/s

Thus

\frac{da}{dt} = \frac{7}{13}.\times 5 = 2.692\ ft/s

5 0
3 years ago
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