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3 years ago

3. What is the main difference between a physical change and a chemical

Physics

1 answer:

Lana71 [14]3 years ago

3 0

Answer:

B and D could both be right as they are quit similar.

Consider two rods of the same length and diameter,

Increasing the diameter of one would change the expansion qualities of that rod even though there would be no chemical changes,

However, leaving the physical appearance of both rods the same while applying a reactive substance (acid or something) to one of the rods would not necessarily change the physical appearance of that rod but could make a considerable change in the physical properties of that rod.

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You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headligh

choli [55]

Answer:

They can be seen from a distance of 4.372 kilometers.

Explanation:

Using the Reyligh creterion for diffraction through a circular aperture we have

$\frac{x}{D}=\frac{1.22\lambda }{d}$

where symbol's have their usual meaning

thus applying values we get

$D=\frac{dx}{1.22\lambda }$

$\therefore D=\frac{0.633\times 4.61\times 10^{-3}}{1.22\times 547\times 10^{-9}}\\\\D=4372.77m\\=4.372km$

5 0

3 years ago

If the box weighs 40 newtons and is lifted a distance of 2 meters, how much work is done on the box?

Zarrin [17]

Answer:

The work done on the box is 80 J.

Explanation:

Given that,

Weight of box = 40 N

Distance = 2 meter

We need to calculate the work done

Using formula of work done

$W=F\times x$

$W=mg\times x$

Where, x = distance

mg = weight

Put the value into the formula

$W=40\times2$

$W= 80\ Nm$

$W=80\ J$

Hence, The work done on the box is 80 J.

5 0

3 years ago

Car A, moving in a straight line at a constant speed of 20. meters per second, is initially 200 meters behind car B, moving in t

MA_775_DIABLO [31]

First, let's express the movement of Car A and B in terms of their position over time (relative to car B)
For car A: y=20x-200 Car A moves 20 meters every second x, and starts 200 meters behind car B
For Car B: y= 15x Car B moves 15 meters every second and starts at our basis point

Set the two equations equal to one another to find the time x at which they meet:
20x - 200 = 15x
200 = 5x
x= 40
At time x=40 seconds, the cars meet. How far will Car A have traveled at this time?
Car A moves 20 meters every second:
20 x 40 = 800 meters

8 0

3 years ago

2. An alternating current is represented by the equation I=20sin 100mt.

Sladkaya [172]

Explanation:

The general equation of an AC current is given by :

$I=I_o sin\omega t$

Where

I₀ is the peak value of current

$\omega$ is angular frequency

$As\ \omega=2\pi f$

So,

$f=\dfrac{\omega}{2\pi}\\\\f=\dfrac{100\pi}{2\pi}\\\\f=50\ Hz$

We know that,

$I_{rms}=\dfrac{I_0}{\sqrt{2}}\\\\=\dfrac{20}{\sqrt{2}}\\\\I_{rms}=14.14\ A$

So, the frequency is 50 Hz and the maximum rms value of current is 14.14 A.

3 0

3 years ago

Which mathematical formula would you use to calculate the IMA of a wheel?

coldgirl [10]

W = F x D

The work you put into a machine -- the input force -- is the force you apply times the distance you apply it. The work done by the machine equals the resisting weight times the distance it moves when you perform the work.

7 0

3 years ago

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