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3 years ago

Part ADoes the number of particles change as the substance changes its state?

Physics

1 answer:

vovangra [49]3 years ago

6 0

No, the number of particles does not change as the substance changes its state.

<h3><u>Explanation: </u></h3>

Change of state from one phase to another is achieved by providing or absorbing heat or pressure. For instance, liquid water if heated becomes vapour steam and if cooled becomes solid ice. Vapour can be compressed to form liquid water again and thus change of state is a reversible action.

The "chemical composition of the matter remains the same" irrespective of its state. Unless a chemical change is carried out, no change occurs with the number of particles. Phase change only affects the "arrangement of molecules", its structure and its motion.

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A disk of a radius 50 cm rotates at a constant rate of 100 rpm. What distance in meters will a point on the outside rim travel d

Dmitry [639]

Answer:

the distance in meters traveled by a point outside the rim is 157.1 m

Explanation:

Given;

radius of the disk, r = 50 cm = 0.5 m

angular speed of the disk, ω = 100 rpm

time of motion, t = 30 s

The distance in meters traveled by a point outside the rim is calculated as follows;

$\theta = \omega t\\\\\theta = (100 \frac{rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1\min}{60 s} ) \times (30 s)\\\\\theta = 100 \pi \ rad\\\\d = \theta r\\\\d = 100\pi \ \times \ 0.5m\\\\d = 50 \pi \ m = 157.1 \ m$

Therefore, the distance in meters traveled by a point outside the rim is 157.1 m

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3 years ago

A +2e charge is at the point (-1,0) mm in the x,y plane. A –e charge is at the point (0,1) mm. What is the electric field at the

Gennadij [26K]

Answer:

Let I and j be the unit vector along x and y axis respectively.

Electric field at origin is given by

E= kq1/r1^2 i + kq2/r2^2j

= 9*10^9*1.6*10^-19*/10^-6*(2i+ j)

= (2.88i + 1.44j)*10^-3 N/C

Force on charge= qE= 3*10^-19*1.6*(2.88i +1. 44 j) *10^-3

F= (1.382 i + 0.691 j) *10^-21

Goodluck

Explanation:

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1 year ago

After coming down a slope, a 60-kg skier is coasting northward on a level, snowy surface at a constant 15 m>s. Her 5.0-kg cat

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To solve this exercise, it is necessary to apply the concepts of conservation of the moment especially in objects that experience an inelastic colposition.

They are expressed as,

$m_1v_1+m_2v_2 = (m_1+m_2)v_f$

Where,

$m_1$ = mass of the skier

$m_2$ = mass of the cat

$v_1$ = initial velocity of skier

$v_2$ = initial velocity of cat

$v_f$ = final velocity of both

Re-arrange to find V_f we have,

$V_f = \frac{m_1v_1+m_2v_2}{(m_1+m_2)}$

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Once the final velocity is found it is possible to calculate the change in kinetic energy, so

$\Delta KE = KE_i-KE_f$

$\Delta KE = \frac{1}{2}(m_1v_1^2+m_2v^2_2)-\frac{1}{2}(m_1+m_2)v_f^2$

$\Delta KE = \frac{1}{2}((60)(15)^2+(5)(-3.8)^2)-\frac{1}{2}(60+5)(13.55)^2$

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2 years ago

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Answer:

Explanation:

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Now, the pilot is traveling in a circle of radius

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And the speed is

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Then, acceleration?

The acceleration of a circular motion can be determine using centripetal acceleration

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exis [7]

Answer:

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