All categories

4 years ago

Which change is more difficult to reverse, chemical or physical?

2 answers:

8 0

Chemical, because once you pour or add the chemicals to each other. Then it will be much harder to reverse those changes than making a physical change.<span />

Elena-2011 [213]4 years ago

4 0

Chemical because you cant reverse it

You might be interested in

Consider two displacements, one of magnitude 3 m and another of magnitude 4 m. Show how the displacement vectors may be combined

Rufina [12.5K]

Answer:

Explanation:

A = 3m

B = 4 m

let the angle between the two vectors is θ.

the resultant of two vectors is given by

$R=\sqrt{A^{2}+B^{2}+2ABCos\theta }$

(a) R = 7 m

So, $7=\sqrt{3^{2}+4^{2}+2\times 3\times 4 Cos\theta }$

49 = 9 + 16 + 24 Cosθ

Cosθ = 1

θ = 0°

Thus, the two vectors are inclined at 0°.

(b) R = 1 m

So, $1=\sqrt{3^{2}+4^{2}+2\times 3\times 4 Cos\theta }$

1 = 9 + 16 + 24 Cosθ

Cosθ = - 1

θ = 180°

Thus, the two vectors are inclined at 180°.

So, $5=\sqrt{3^{2}+4^{2}+2\times 3\times 4 Cos\theta }$

25 = 9 + 16 + 24 Cosθ

Cosθ = 0

θ = 90°

Thus, the two vectors are inclined at 90°.

3 0

3 years ago

An object is dropped on Earth from a height of 15 m. How long did it take for the object to reach the ground?

Inga [223]

x=gt^2/2=>t^2=2x/g=30/9.81=3.05=>t=sqrt(3.05)=1.8 s

6 0

3 years ago

A 9Ω resistance is cut into three equal parts and connected in parallel. Find the equivalent resistance of the combination.

sammy [17]

Answer:

the ans i 1 ohm

Explanation:

if i cut a resistor of 9 ohm into 3 equal parts then each resistor will have a resistance of 3 ohm and if they are in parallel combination the net resistance will be R

so 1/R=1/R1+1/R2+1/R3

1/R=1/3+1/3+1/3

R=1

7 0

3 years ago

In a single-slit diffraction experiment, a coherent light source illuminates a slit in a barrier, and the resulting pattern is p

UkoKoshka [18]

Answer:

hsjdsdddwqdqwdd

Explanation:

4 0

3 years ago

What force is required to push a block (mass m) up an inclined plane that makes an angle of θ with the horizon at a constant vel

Verizon [17]

Answer:

b. mg ( μ · cos θ + sin θ)

Explanation:

Hi there!

Please, see the attached figure for a graphical description of the problem.

We have the following parallel forces acting on the block (in parallel direction to the direction of movement):

F = applied force.

Fr = friction force.

wx = parallel component of the weight.

According to Newton´s second law:

∑F = m · a

Where "m" is the mass of the block and "a" its acceleration.

Then:

F - Fr - wx = m · a

Since the block is to be pushed at a constant velocity, the acceleration is zero. Then:

F - Fr - wx = 0

F = Fr + wx

The applied force has to be equal to the friction force plus the parallel component of the weight to push the block at constant velocity.

The friction force is calculated as follows:

Fr = μ · N

Where N is the normal force and μ is the coefficient of friction.

Notice that the normal force is of the same magnitude as the perpendicular component of the weight, wy.

Let´s apply Newton´s second law in the perpendicular direction to show this:

∑F = m · a

N - wy = m · a

The acceleration of the block in the perpendicular direction is zero. Then:

N - wy = 0

N = wy

And wy can be obtained by trigonometry (see figure):

wy = W · cos θ

N = wy = mg · cos θ

The parallel component of the weight is calculated using trigonometry (see figure):

wx = W · sin θ

wx = mg · sin θ

Then the applied force will be:

F = Fr + wx

F = μ · N + mg · sin θ (N = wy = mg · cos θ)

F = μ · mg · cos θ + mg · sin θ

F = mg ( μ · cos θ + sin θ)

The correct answer is the b.

8 0

3 years ago