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Strike441 [17]
3 years ago
9

What happens to the movement of particles in a liquid when it is heated?

Physics
2 answers:
Margarita [4]3 years ago
5 0

Answer:

molecules and atoms vibrate faster.

Explanation:

Yanka [14]3 years ago
3 0

Answer:

When heat is added to a substance, the molecules and atoms vibrate faster. As atoms vibrate faster, the space between atoms increases. The motion and spacing of the particles determines the state of matter of the substance. The end result of increased molecular motion is that the object expands and takes up more space.

Explanation:

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A motorboat traveling 4 m/s, East encounters a current traveling 3.0 m/s, North. a) What is the resultant velocity of the motorb
Nostrana [21]

Explanation:

It is given that,

Velocity in East, v_1=4\ m/s

Velocity in North, v_2=3\ m/s

(a) The resultant velocity is given by :

v=\sqrt{4^2+3^2}=5\ m/s

(b) The width of the river is, d = 80 m

Let t is the time taken by the boat to travel shore to shore. So,

t=\dfrac{d}{v}

t=\dfrac{80\ m}{5\ m/s}

t = 16 seconds

(c) Let x is the distance covered by the boat to reach the opposite shore. So,

x=v_2\times t

x=3\ m/s\times 16\ s

x = 48 meters

Hence, this is the required solution.

4 0
3 years ago
Two objects carry initial charges that are q1 and q2, respectively, where |q2| > |q1|. They are located 0.160 m apart and beh
mart [117]

Answer:

\rm |q_1|=8.0\times 10^{-7}\ C,\ \ \ |q_2| = 4.6\times 10^{-6}\ C.

Explanation:

According to the Coulomb's law, the magnitude of the electrostatic force between two static point charges  \rm q_1 and \rm q_1, separated by a distance \rm r, is given by

\rm F = \dfrac{kq_1q_2}{r^2}.

where k is the Coulomb's constant.

Initially,

\rm r = 0.160\ m\\F_i = -1.30\ N.\\\\and \ \ |q_2|>|q_1|.

The negative sign is taken with force F because the force is attractive.

Therefore, the initial electrostatic force between the charges is given by

\rm F_i = \dfrac{kq_1q_2}{r^2}.\\-1.30=\dfrac{kq_1q_2}{0.160^2}\\\rm\Rightarrow q_2 = \dfrac{-1.30\times 0.160^2}{q_1k}\ \ \ ..............\ (1).

Now, the objects are then brought into contact, so the net charge is shared equally, and then they are returned to their initial positions.

The force is now repulsive, therefore, \rm F_f = +1.30\ N.

The new charges on the two objects are

\rm q_1'=q_2' = \dfrac{q_1+q_2}{2}.

The new force is given by

\rm F_f = \dfrac{kq_1'q_2'}{r^2}\\+1.30=\dfrac{k\left (\dfrac{q_1+q_2}{2}\right )\left (\dfrac{q_1+q_2}{2}\right )}{0.160^2}\\\Rightarrow \left (\dfrac{q_1+q_2}{2}\right )^2=\dfrac{+1.30\times 0.160^2}{k}\\(q_1+q_2)^2=\dfrac{4\times 1.30\times 0.160^2}{k}\\q_1^2+q_2^2+2q_1q_2=\dfrac{4\times 1.30\times 0.160^2}{k}\\\\

Using (1),

\rm q_1^2+\left ( \dfrac{-1.30\times 0.160^2}{q_1k}\right )^2+2\left (\dfrac{-1.30\times 0.160^2}{k} \right )=\dfrac{4\times 1.30\times 0.160^2}{k}\\q_1^2+\dfrac 1{q_1^2}\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0\\q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0

\rm q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0\\q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{9\times 10^9}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{9\times 10^9} \right )=0\\q_1^4-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{9\times 10^9} \right )+\left ( \dfrac{-1.30\times 0.160^2}{9\times 10^9}\right )^2=0

\rm q_1^4-q_1^2\left (2.22\times 10^{-11} \right )+\left ( 1.37\times 10^{-23}\right ) =0\\\Rightarrow q_1^2 = \dfrac{-(-2.22\times 10^{-11})\pm \sqrt{(-2.22\times 10^{-11})^2-4\cdot (1)\cdot (1.37\times 10^{-23})}}{2}\\=1.11\times 10^{-11}\pm 1.046\times 10^{-11}.\\=6.4\times 10^{-13}\ \ \ or\ \ \ 2.156\times 10^{-11}\\\Rightarrow q_1 = \pm 8.00\times 10^{-7}\ C\ \ \ or\ \ \ \pm 4.64\times 10^{-6}\ C.

Using (1),

When \rm q_1 = \pm 8.00\times 10^{-7}\ C,

\rm q_2=\dfrac{-1.30\times 0.160^2}{\pm 8.00\times 10^{-7}\times 9\times 10^9}=\mp4.6\times 10^{-6}\ C.

When \rm q_1=\pm 4.6\times 10^{-6}\ C,

\rm q_2=\dfrac{-1.30\times 0.160^2}{\pm 4.64\times 10^{-6}\times 9\times 10^9}=\mp7.97\times 10^{-7}\ C\approx 8.0\times 10^{-7}\ C.

Since, \rm |q_2|>|q_1|

Therefore, \rm |q_1|=8.0\times 10^{-7}\ C,\ \ \ |q_2| = 4.6\times 10^{-6}\ C.

7 0
2 years ago
3. Why is a subducting plate boundary considered a type of<br> convergent plate boundary?
JulsSmile [24]

Answer:

✔ convergent

Explanation:

5 0
3 years ago
The Canadian bobsled team hit the brakes on the sled they are traveling East in so that it decelerates at a rate of 0.43
mafiozo [28]

Answer:

Explanation:

85 = ½(0.43)t²

t = √(2(85)/0.43)

t = 19.883380...

t = 20 s

v→ 8.55 m/s initial, 0 m/s final

a← 0.43 m/s²

8 0
2 years ago
Can anything block a magnet's force feild?
SVEN [57.7K]
Something that is not magnetic
7 0
3 years ago
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