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sp2606 [1]
3 years ago
7

What are the units of the following properties? Enter your answer as a sequence of five letters separated by commas, e.g., A,F,G

,E,D. Note that some properties listed may have the same units. 1) mass 2) heat 3) density 4) energy 5) molarity (A) g (B) J (C) mol (D) K (E) g/mol (F) mol/L (G) mol/K (H) g/mL (I) J/K (J) J/K*mol (K) J/K*g (L) kJ/L
Physics
1 answer:
Marysya12 [62]3 years ago
6 0

Answer:

The sequence is A,B,H,B,F

Explanation:

  • The Standard International unit is Kilogram (kg) and the mass of a body can also be expressed in gram (g).
  • Heat is a form of energy and the unit for energy is joule (J), thus the unit of heat is also joule (J).
  • Density is mass per unit volume where the unit of mass is gram (g) and the unit of volume can be taken as milli-liter (mL). Thus g/mL is the unit of density.
  • The unit of energy is joule (J).
  • Molarity is number of solute in mol dissolved in 1 liter of solution. Thus mol/L is the the unit of molarity.

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Consider the following. (Let C1 = 20.80 µF and C2 = 14.80 µF.) A rectangular circuit contains a battery and four capacitors. The
umka21 [38]

Find the below attachment

7 0
3 years ago
1. A piece of metal weighs 50.0 N in air, 36.0 N in water, and 41.0 N in an unknown
denis23 [38]

Answer:

a) 3.37 x 10^{3} kg/m^3

b) 6.42kg/m^{3}

Explanation:

a) Firstly we would calculate the volume of the metal using it`s weight in air and water , after finding the weight we would find the density .

Weight of metal in air = 50N = mg implies the mass of metal is 5kg.

Now the difference of weight of the metal in air and water = upthrust acting on it = volume (metal) p (liquid) g = V (1000)(10) = 14N. So volume of metal piece = 14 x 10^{-4}  kg/m^{3}. So density of metal = mass of metal / volume of metal = 5 / 14 x 10^{-4}  kg/m^{3} = 3.37 x 10^{3} kg/m^3

b) Water exerts a buoyant force to the metal which is 50−36 = 14N, which equals the weight of water displaced. The mass of water displaced is 14/10 = 1.4kg Since the density of water is 1kg/L, the volume displaced is 1.4L. Hence, we end up with 3.57kg/l. Moreover, the unknown liquid exerts a buoyant force of 9N. So the density of this liquid is 6.42kg/m^{3}

3 0
3 years ago
Capacitors C1 = 6.45 µF and C2 = 2.50 µF are charged as a parallel combination across a 250 V battery. The capacitors are discon
denpristay [2]

Answer:

For C1, Q =  1.6125×10⁻³ C

For C2, Q =  6.25×10⁻⁴ C

Explanation:

Note: Since the capacitors are connected in parallel, The voltage across each of them is equal.

From the question,

Q = CV........................ Equation 1

Where Q = Charge on the capacitor, V = Voltage across the capacitor, C = Capacitance of the capacitor.

For the first capacitor,

Q = C1V............. Equation 2

Where C1 = 6.45 μF= 6.45×10⁻⁶ F, V = 250 V

Substitute into equation 2

Q = (6.45×10⁻⁶ )(250)

Q = 1.6125×10⁻³ C.

For the the second capacitor,

Q = C2V............. Equation 3

Given: C2 = 2.50 μF = 2.5×10⁻⁶ F, V = 250 V

Q = (2.5×10⁻⁶ )(250)

Q = 6.25×10⁻⁴ C

4 0
3 years ago
Which labels best complete the diagram?
nlexa [21]

Answer: D

Explanation: Look at the figure, we can conclude that the correct answer is

X: Low potential energy

Y: High potential energy

Z: Flow of electrons

Because electrons flow where there is difference in potential energy. And electrons move from a region of high potential energy to a region of low potential energy.

Since the arrow is pointing to X, that means

Z is the flow of electrons, X is of low potential energy and Y is of high potential energy.

6 0
3 years ago
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