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krok68 [10]
2 years ago
7

He measures 100ml of hot water (90 degrees) using a graduated cylinder he places this sample in the refrigerator to cool it down

to 4 degrees he notices that the cylinder indicates 96ml. If water is incompressible why did the volume change
Physics
1 answer:
Rasek [7]2 years ago
4 0

The volume of the water, which was initially 100mL changed to 96mL after refrigeration because the volume of water reduces when cool.

<h3>Why does water change in volume?</h3>

Water is a substance that is known for its unique characteristics that makes it stand out among all other liquids.

One notable feature of water is it's ability to become denser when subjected to cooler temperatures i.e. water is more dense when cold.

As water cools or becomes less warm, it contracts and decreases in volume. When water decreases in volume, it becomes more dense.

Therefore, according to this question, it can be said that the volume of the water, which was initially 100mL changed to 96mL after refrigeration because the volume of water reduces when cool.

Learn more about water volume at:brainly.com/question/17322215

#SPJ1

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What’s the difference between expressed/implied powers, concurrent powers, and reserved powers?.
Lisa [10]

Answer:

A reserved power is a power specifically reserved to the states. Powers include setting up local governments and determining the speed limit. A concurrent power is a power that is given to both the states and the federal government.

Explanation:

3 0
3 years ago
A 310-g air track cart is traveling at 1.25 m/s and a 260-g cart traveling in the opposite direction at 1.33 m/s. What is the sp
UNO [17]

Answer:

v_{CM}=0.0732\ m/s

Explanation:

given,

mass of the cart 1, m₁ = 310 g

speed of car 1 , v₁ = 1.25 m/s

mass of cart 2, m₂ = 260 g

speed of cart 2, v₂ = -1.33 m/s

speed of center of mass

v_{CM}=\dfrac{m_1v_1 + m_2 v_2}{m_1 + m_2}

v_{CM}=\dfrac{0.31\times 1.25 +0.26\times (-1.33)}{0.31+0.26}

v_{CM}=\dfrac{0.0417}{0.57}

v_{CM}=0.0732\ m/s

Hence, speed of center of mass of the system is equal to 0.0732 m/s

7 0
3 years ago
Two metal disks, one with radius R1 = 2.45 cm and mass M1 = 0.900 kg and the other with radius R2 = 5.00 cm and mass M2 = 1.60 k
larisa86 [58]

Answer:

part (a) a_1\ =\ 2.9\ kg

Part (b) a_2\ =\ 6.25\ kg

Explanation:

Given,

  • Mass of the larger disk = M_2\ =\ 1.60\ kg
  • Mass of the smaller disk = M_1\ =\ 0.900\ kg
  • Radius of the larger disk = R_2\ =\ 5.00\ cm\ =\ 0.05\ m
  • Radius of the smaller disk = R_1\ =\ 2.45\ cm\ =\ 0.0245\ m
  • Mass of the block = M = 1.60 kg

Both the disks are welded together, therefore total moment of inertia of the both disks are the summation of the individual moment of inertia of the disks.

\therefore I\ =\ I_1\ +\ I_2\\\Rightarrow I\ =\ \dfrac{1}{2}M_1R_1^2\ +\ \dfrac{1}{2}M_2R_2^2\\\Rightarrow I\ =\ \dfrac{1}{2} (0.9\times 0.0245^2\ +\ 1.60\times 0.05^2)\\\Rightarrow I\ =\ 2.27\times 10^{-3}\ kgm^2

part (a)

Given that a block of mass m which is hanging with the smaller disk,

Let 'T' be 'a' be the tension in the string and acceleration of the block.

From the free body diagram of the smaller block,

mg\ -\ T\ =\ ma\\\Rightarrow T\ =\ mg\ -\ ma\,\,\,\,eqn (1)

From the pulley,

\sum \tau\ =\ I\alpha\\\Rightarow T\times R_1\ =\ I\alpha\ =\ \dfrac{Ia}{R_1}\\\Rightarrow T\ =\ \dfrac{I\alpha}{R_1^2}\,\,\,eqn(2)

From the equation (1) and (2),

mg\ -\ ma\ =\ \dfrac{Ia}{R_1^2}\\\Rightarrow a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}{0.0245^2}}\ +\ 1.60}\\\Rightarow a\ =\ 2.91\ m/s^2

part (b)

Above expression for the acceleration of the block is only depended on the radius of the pulley.

Radius of the larger pulley = R_2\ =\ 0.05\ m

Let a_2 be the acceleration of the block while connecting to the larger pulley.\therefore a\ =\ \dfrac{mg}{\dfrac{I}{R_2^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}{0.05^2}\ +\ 1.60}}\\\Rightarow a\ =\ 6.25\ m/s^2

4 0
3 years ago
A circuit is shown in the picture above. The chemical energy in the battery is used to light up the light bulb on the other end.
Ainat [17]
To be honest, the picture is so far above that I can't see it at all.
But reading the information in the question's statement, I'd say
the blank should be filled in so that it says:

<span>   The chemical energy in the battery is used to light up the light bulb
   on the other end. Chemical energy in the battery is transformed into
   electrical energy which runs through the circuit wire. (A)</span>
8 0
4 years ago
Read 2 more answers
A parallel-plate air capacitor is made from two plates 0.190 m square, spaced 0.770 cm apart. It is connected to a 120 V battery
Naddik [55]

Answer:

aaksj

Explanation:

a) the capacitance is given of a plate capacitor is given by:

C = \epsilon_0*(A/d)

Where \epsilon_0 is a constant that represents the insulator between the plates (in this case air, \epsilon_0 = 8.84*10^(-12) F/m), A is the plate's area and d is the distance between the plates. So we have:

The plates are squares so their area is given by:

A = L^2 = 0.19^2 = 0.0361 m^2

C = 8.84*10^(-12)*(0.0361/0.0077) = 8.84*10^(-12) * 4.6883 = 41.444*10^(-12) F

b) The charge on the plates is given by the product of the capacitance by the voltage applied to it:

Q = C*V = 41.444*10^(-12)*120 = 4973.361 * 10^(-12) C = 4.973 * 10^(-9) C

c) The electric field on a capacitor is given by:

E = Q/(A*\epsilon_0) = [4.973*10^(-9)]/[0.0361*8.84*10^(-12)]

E = [4.973*10^(-9)]/[0.3191*10^(-12)] = 15.58*10^(3) V/m

d) The energy stored on the capacitor is given by:

W = 0.5*(C*V^2) = 0.5*[41.444*10^(-12) * (120)^2] = 298396.8*10^(-12) = 0.298 * 10 ^6 J

6 0
3 years ago
Read 2 more answers
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