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just olya [345]
3 years ago
8

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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Two ropes have equal length and are stretched the same way. The speed of a pulse on rope 1 is 1.4 times the speed on rope 2. Par
kondor19780726 [428]

Answer:

m1/m2 = 0.51

Explanation:

First to all, let's gather the data. We know that both rods, have the same length. Now, the expression to use here is the following:

V = √F/u

This is the equation that describes the relation between speed of a pulse and a force exerted on it.

the value of "u" is:

u = m/L

Where m is the mass of the rod, and L the length.

Now, for the rod 1:

V1 = √F/u1 (1)

rod 2:

V2 = √F/u2 (2)

Now, let's express V1 in function of V2, because we know that V1 is 1.4 times the speed of rod 2, so, V1 = 1.4V2. Replacing in the equation (1) we have:

1.4V2 = √F/u1 (3)

Replacing (2) in (3):

1.4(√F/u2) = √F/u1 (4)

Now, let's solve the equation 4:

[1.4(√F/u2)]² = F/u1

1.96(F/u2) =F/u1

1.96F = F*u2/u1

1.96 = u2/u1 (5)

Now, replacing the expression of u into (5) we have the following:

1.96 = m2/L / m1/L

1.96 = m2/m1 (6)

But we need m1/m2 so:

1.96m1 = m2

m1/m2 = 1/1.96

m1/m2 = 0.51

5 0
3 years ago
A distant large asteroid is detected that might pose a threat to Earth. If it were to continue moving in a straight line at cons
Vlada [557]

Answer:

The minimum speed required is 5.7395km/s.

Explanation:

To escape earth, the kinetic energy of the asteroid must be greater or equal to its gravitational potential energy:

K.E\geq P.E

or

\dfrac{1}{2}mv^2 \geq  G\dfrac{Mm}{R}

where m is the mass of the asteroid, R= 24,000,000\:m is its distance form earth's center, M = 5.9*10^{24}kg is the mass of the earth, and G = 6.7*10^{-11}m^3/kg\: s^2 is the gravitational constant.

Solving for v we get:

v \geq \sqrt{\dfrac{2GM}{R} }

putting in numerical values gives

v \geq \sqrt{\dfrac{2(6.7*10^{-11})(5.9*10^{24})}{(24,000,000)} }

\boxed{v\geq 5739.5m/s}

in kilometers this is

v\geq5.7395m/s.

Hence, the minimum speed required is 5.7395km/s.

5 0
3 years ago
List all the forces that act on a hot air ballon when it is in the sky
nikdorinn [45]
Gravity, acceleration, kinetic energy, the atmosphere
3 0
3 years ago
Read 2 more answers
Why do astronauts' weights differ in space?
OLga [1]

Answer:

There is less gravitational force in space.

Explanation:

Gravity doesn't exist as it does on Earth. Earth's gravity is as 6 times stronger as it is on the moon.

7 0
3 years ago
A small current element carrying a current of I = 1.00 A is placed at the origin given by d → l = 4.00 m m ^ j Find the magnetic
xxTIMURxx [149]

Answer:

the magnitude and direction of d → B on the x ‑axis at x = 2.50 m is -6.4 × 10⁻¹¹T(Along z direction)

the magnitude and direction of d → B on the z ‑axis at z = 5.00 m is 1.6 × 10⁻¹¹T(Along x direction)

Explanation:

Use Biot, Savart, the magnetic field

d\bar{B}=\frac{U}{4\pi } \frac{i(d\bar{l}\times r)}{r^2}

Given that,

i = 1.00A

d → l = 4.00 m m ^ j

r = 2.5m

Displacement vector is

\bar{r}=x\hat i+y\hat j+z \hat k\\

\bar{r}= (2.5m) \hat i +(0m)^2 + (0m)^2

 =2.5m

on the axis of x at x = 2.5

r = \sqrt{(2.5)^2 + (0)^2 + (0)^2}

r = 2.5m

And unit vector

\hat r =\frac{\bar{r}}{r}

= \frac{2.5 \hat i}{2.5}\\\\= 1\hat i

Therefore, the magnetic field is as follow

d\bar{B}=\frac{U}{4\pi } \frac{i(d\bar{l}\times r)}{r^2}

d\bar{B} = \frac{(10^-^7)(1)(4\times10^-^3j\times i}{(2.50)^2} \\\\d\bar{B} = -6.4\times10^{-11} T

(Along z direction)

B)r = 5.00m

Displacement vector is

\bar{r}=x\hat i+y\hat j+z \hat k\\

\bar{r}= (5.00m) \hat i +(0m)^2 + (0m)^2

 =5.00m

on the axis of x at x = 5.0

r = \sqrt{(5.00)^2 + (0)^2 + (0)^2}

r = 5.00m

And unit vector

\hat r =\frac{\bar{r}}{r}

= \frac{5.00 \hat i}{5.00}\\\\= 1\hat i\\

Therefore, the magnetic field is as follow

d\bar{B}=\frac{U}{4\pi } \frac{i(d\bar{l}\times r)}{r^2}

d\bar{B} = \frac{(10^-^7)(1)(4\times10^-^3j\times i}{(5.00)^2} \\\\d\bar{B} = 1.6\times10^{-11} T

(Along x direction)

7 0
3 years ago
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