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larisa [96]
3 years ago
13

Interconversion of states of matter

Physics
2 answers:
ss7ja [257]3 years ago
8 0

Answer:

interconversion of matter refers to change of one state to another. it is a process by which matter changes from one state to another back to its original State without any change in its chemical composition

Lady bird [3.3K]3 years ago
7 0

Explanation:

hope it helpss.............

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The velocity v of a particle moving in the xy plane is given by v = (6.0t -4.0t2 )i + 8.5j, in m/s. Here v is in m/s and t (for
atroni [7]

Answer:

Explanation:

Given

Velocity of the particle in vector form is given by

v=\left ( 6t-4t^2\right )\hat{i}+8.5\hat{j}

acceleration is rate of change of velocity thus acceleration is

a=\frac{\mathrm{d} v}{\mathrm{d} t}

a=(6-8t)\hat{i}+0\hat{j}

at t=3\ s

a=6-8\times 3

a=-18\ m/s^2

7 0
4 years ago
Calculate the potential V(r) for r>rb. (Hint: The net potential is the sum of the potentials due to the individual spheres.)
Marat540 [252]

Answer:

The potential for r > rb is equal to zero.

Explanation:

For r > rb, the potential is:

V=\frac{Kq}{r}

Then, the net potential is:

V_{(r)} =\frac{K(+\epsilon )}{r} +\frac{K(-\epsilon )}{r}

K=\frac{1}{4\pi \epsilon _{o}  }

V_{(r)} =\frac{K(+\epsilon )}{r} -\frac{K(\epsilon )}{r}\\V_{(r)}=0

8 0
4 years ago
Two very small spheres are initially neutral and separated by a distance of 0.30 m. Suppose that 2.50 1013 electrons are removed
quester [9]

Answer:

F = - 1,598 10⁻³ N

Explanation:

Electic strength is given by Coulomb's law

          F = k q₁ q₂  / r²

Where k is the Coulomb constant that is worth 8.99 10⁸ N m²/C², q₁ and q₂ are the charges and r is the distance that separates the electric charges

In this case the charge of the two spheres is the same and of a different sign since when you remove the charge of a sphere that was initially neutral, it is left with that charge removed but of the opposite sign

       q₁ = q₂ = 2.50 10¹³ electrons = 2.50 10¹³ 1.6 10⁻¹⁹

       q₀ = 4.0 10⁻⁶ C

Let's calculate

       F = - 8.99 10⁸ (4.0 10⁻⁶)² / 0.30²

       F = - 1,598 10⁻³ N

4 0
4 years ago
A 0.400-kg object is swung in a circular path and in a vertical plane on a 0.500-m-length string. If the angular speed at the bo
Talja [164]

Answer:

T = 16.72 N

Explanation:

When the object is swung in a circular path, and in a vertical plane, there are two forces external to the object acting on it at any time: the gravity (which is always downward) and the tension in the string (which always points towards the center of the circle).

At the bottom of the circle, the tension is directly upward, so these two forces, are opposite each other, and the difference between them is the centripetal force , which at this point, keeps the object swinging in a circle.

This is the point of the trajectory where T is maximum.

We can apply Newton's 2nd Law, choosing an axis vertical (y-axis) being the upward direction the positive one, as follows:

T- m*g = m*a

The acceleration, at the bottom of the circle, is only normal (as there are no forces in the horizontal direction) , and is equal to the centripetal acceleration, as follows:

ac =  v² / r = ω²*r⇒ T- m*g = m*ω²*r

Replacing by the givens, we can solve for T as follows:

T = m* (ω²*r+g) = 0.4 kg*((8.00)² rad/sec²*0.5m)+9.8 m/s²) = 16.72 N

5 0
3 years ago
Consider the following. (Let C1 = 20.80 µF and C2 = 14.80 µF.) A rectangular circuit contains a battery and four capacitors. The
umka21 [38]

Find the below attachment

7 0
4 years ago
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