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3 years ago

In what type of change is matter not destroyed

Physics

2 answers:

Natalka [10]3 years ago

7 0

Matter can not ne destroyed but both physical and chemical

Margaret [11]3 years ago

6 0

There are two types of change in matter: physical change and chemical change. ... This is called the Law of Conservation of Matter. It states that matter can never be created or destroyed, only changed and rearranged.

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What does this quote mean?

Svet_ta [14]

Answer:

Basically, it means that you never know until you try.

Explanation:

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3 years ago

In a video game a flying coconut moves at a cost of the last few 20 m/s a positive direction the coconut he stops going to move

Liula [17]

Answer:

30 m/s

Explanation:

30 m/s I think.

I am too lazy to explain

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3 years ago

Why is cloning done before an investigation begins??

svlad2 [7]

Answer:Purpose of Cloning

A forensic image of a hard drive captures everything on the hard drive, from the physical beginning to the physical end. ... Hard drives are susceptible to failure. Having two clones gives an investigator one to examine and one to fall back on.Therapeutic cloning involves creating a cloned embryo for the sole purpose of producing embryonic stem cells with the same DNA as the donor cell. These stem cells can be used in experiments aimed at understanding disease and developing new treatments for disease.

Explanation:

hope this helps have a nice night lo❤️❤️❤️l

3 0

3 years ago

Suppose your friend claims to have discovered a mysterious force in nature that acts on all particles in some region of space. H

kirill [66]

Answer:

U = 1 / r²

Explanation:

In this exercise they do not ask for potential energy giving the expression of force, since these two quantities are related

F = - dU / dr

this derivative is a gradient, that is, a directional derivative, so we must have

dU = - F. dr

the esxresion for strength is

F = B / r³

let's replace

∫ dU = - ∫ B / r³ dr

in this case the force and the displacement are parallel, therefore the scalar product is reduced to the algebraic product

let's evaluate the integrals

U - Uo = -B (- / 2r² + 1 / 2r₀²)

To complete the calculation we must fix the energy at a point, in general the most common choice is to make the potential energy zero (Uo = 0) for when the distance is infinite (r = ∞)

U = B / 2r²

we substitute the value of B = 2

U = 1 / r²

5 0

3 years ago

Block B (of mass m) is initially at rest. Block A (of mass 3m) travels toward B with an initial speed v0 (vee nought) and collid

NeTakaya

Answer:

The maximum height reached by the two blocks is approximately 0.1147959 × v₀²

Explanation:

The mass of block B = m

The mass of block A = 3·m

The initial velocity of block B, v₂ = 0 m/s

The initial velocity of block A, v₁ = v₀

The amount of friction between the blocks and the surface = Negligible friction

By the Law of conservation of linear momentum, we have;

Total initial momentum = Total final momentum

3·m·v₁ + m·v₂ = (3·m + m)·v₃ = 4·m·v₃

Plugging in the values for the velocities gives;

3·m × v₀ + m × 0 = (3·m + m)·v₃ = 4·m·v₃

∴ 3·m × v₀ = 4·m·v₃

$\therefore v_3 = \dfrac{3}{4} \times v_0 = 0.75 \times v_0$

The kinetic energy, K.E. of the combined blocks after the collision is given as follows;

K.E. = 1/2 × mass × v²

$\therefore K.E. = \dfrac{1}{2} \times 4\cdot m \times \left (\dfrac{3}{4} \cdot v_0 \right )^2 = \dfrac{9}{8} \cdot m\cdot v_0^2$

The potential energy, P.E., gained by the two blocks at maximum height = The kinetic energy, K.E., of the two blocks before moving vertically upwards

The potential energy, P.E. = m·g·h

Where;

m = The mass of the object at the given height

g = The acceleration due to gravity

h = The height at which the object of mass, 'm', is located

Therefore, for h = The maximum height reached by the two blocks, we have;

P.E. = K.E.

$m \cdot g \cdot h = \dfrac{9}{8} \cdot m\cdot v_0^2$

$h = \dfrac{\dfrac{9}{8} \cdot m\cdot v_0^2}{m \cdot g } = \dfrac{9}{8} \cdot \dfrac{ v_0^2}{ g } = \dfrac{9}{8} \cdot \dfrac{ v_0^2}{ 9.8} = \dfrac{42}{392} \cdot v_0^2 \approx 0.1147959 \cdot v_0^2$

The maximum height reached by the two blocks, h ≈ 0.1147959·v₀².

7 0

3 years ago