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mars1129 [50]
2 years ago
10

When a hot metal cylinder is dropped into a sample of water, the water molecules _____________________ to reach the final temper

ature.
Physics
2 answers:
ASHA 777 [7]2 years ago
5 0

Answer: I believe the answer is speed up.

Explanation: This is because when water heats up the molecules move farther apart from eachother they speed up, eventually causing the water to boil.

Anna11 [10]2 years ago
5 0

Answer:

When a hot metal cylinder is dropped into a sample of water, the water molecules <u>will move with greater speed</u> to reach the final temperature.

Explanation:

As we know that internal energy of the system of gas or any fluid or any matter is only function of its temperature.

Which means if the internal energy of the system is increased then its temperature must have to increase which means the kinetic energy of the molecules of that system will have to increase

So in the given case if hot cylinder is dropped in water then due to energy exchange between cylinder and water the temperature of water will increase and due to this the molecules of water will have more internal energy.

Due to increase in internal energy of molecules of water the kinetic energy of the molecules will increase and this leads to speed up of the water molecules.

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Gasoline burns in the cylinder of an automobile engine. During the combustion reaction, the production of gas forces the piston
serg [7]

Answer:

\Delta U = 1640 J

Explanation:

As we know by first law of thermodynamics that for ideal gas system we have

Heat given = change in internal energy + Work done

so here we will have

Heat given to the system = 2.2 kJ

Q = 2200 J

also we know that work done by the system is given as

W = 560 J

so we have

\Delta U = Q - W

\Delta U = 2200 - 560

\Delta U = 1640 J

6 0
3 years ago
A 2.00 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0300 m . The spring has
iogann1982 [59]

Answer:

v = 0.41 m/s

Explanation:

  • In this case, the change in the mechanical energy, is equal to the work done by the fricition force on the block.
  • At any point, the total mechanical energy is the sum of the kinetic energy plus the elastic potential energy.
  • So, we can write the following general equation, taking the initial and final values of the energies:

       \Delta K + \Delta U = W_{ffr}  (1)

  • Since the block and spring start at rest, the change in the kinetic energy is just the final kinetic energy value, Kf.
  • ⇒ Kf = 1/2*m*vf²  (2)
  • The change in the potential energy, can be written as follows:

       \Delta U = U_{f}  - U_{o}  = \frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (3)

       where k = force constant = 815 N/m

       xf = final displacement of the block = 0.01 m (taking as x=0 the position

      for the spring at equilibrium)

      x₀ = initial displacement of  the block = 0.03 m

  • Regarding the work done by the force of friction, it can be written as follows:

       W_{ffr} = - \mu_{k}* F_{n} * \Delta x  (4)

       where μk = coefficient of kinettic friction, Fn = normal force, and Δx =

       horizontal displacement.

  • Since the surface is horizontal, and no acceleration is present in the vertical direction, the normal force must be equal and opposite to the force due to gravity, Fg:
  • Fn = Fg= m*g (5)
  • Replacing (5) in (4), and (3) and (4) in (1), and rearranging, we get:

        \frac{1}{2} * m* v^{2} = W_{ffr} - \Delta U = W_{ffr} - (U_{f} -U_{o})  (6)

        \frac{1}{2} * m* v^{2} = (- \mu_{k}* m*g* \Delta x)  -\frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (7)

  • Replacing by the values of m, k, g, xf and x₀, in (7) and solving for v, we finally get:

    \frac{1}{2} * 2.00 kg* v^{2}  = (-0.4*2.00 kg*9.8m/s2*0.02m) +( (\frac{1}{2} *815 N/m)* (0.03m)^{2} - (0.01m)^{2}) = -0.1568 J + 0.326 J (8)

  • v =\sqrt{(0.326-0.1568}  =  0.41 m/s  (9)
7 0
3 years ago
What do you think would happen to the force of attraction of two interacting charges if their distance apart is halved?
sweet [91]

Answer:

The new force becomes 4 times the initial force.

Explanation:

The force of attraction or repulsion is given by the relation as follows :

F=k\dfrac{q_1q_2}{d^2}

Where

d is the distance between the interacting charges

F is inversely proportional to the distance between charges.

If the distance is halved, d'=(d/2), new force is given by :

F'=k\dfrac{q_1q_2}{d'^2}\\\\=k\dfrac{q_1q_2}{(\dfrac{d}{2})^2}\\\\=k\dfrac{q_1q_2}{\dfrac{d^2}{4}}\\\\=4\times \dfrac{kq_1q_2}{d^2}\\\\F'=4F

So, the new force becomes 4 times the initial force.

4 0
3 years ago
A car has a force of 0.27 N and an acceleration of 3m/s2 What is the car's mass?​
lyudmila [28]

Answer:

0.09 kg

Explanation:

f=0.27

a=3

m=f/a

m=0.27/3

m=0.09

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3 years ago
20 points<br> Which BEST replaces the question mark?<br><br> In Saudi Arabia, __________?
dsp73
<span>The economy is based upon the production of oil. Saudi Arabi is one of the leading oil producing countries in the world. It is one of the founding counties of Opec, an oil cartel made up of middle eastern oil producing countries which try to manipulate the worldwide price of oil.</span>
3 0
3 years ago
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