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BartSMP [9]
3 years ago
6

Volume is the space occupied by a quantity of matter. a.true b.false

Physics
2 answers:
Solnce55 [7]3 years ago
7 0
A. The answer is True
Leokris [45]3 years ago
3 0

Answer:

true

Explanation:

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A domestic water heater holds 189 L of water at 608C, 1 atm. Determine the exergy of the hot water, in kJ. To what elevation, in
Gekata [30.6K]

A.

The energy of the hot water is 482630400 J

Using Q = mcΔT where Q = energy of hot water, m = mass of water = ρV where ρ = density of water = 1000 kg/m³ and V = volume of water = 189 L = 0.189 m³,

c = specific heat capacity of water = 4200 J/kg-°C and ΔT = temperature change of water = T₂ - T₁ where T₂ = final temperature of water = 608 °C. If we assume the water was initially at 0°C, T₁ = 0 °C. So, the temperature change ΔT = 608 °C - 0 °C = 608 °C

Substituting the values of the variables into the  equation, we have

Q = mcΔT

Q = ρVcΔT

Q = 1000 kg/m³ × 0.189 m³ × 4200 J/kg-°C × 608 °C

Q = 482630400 J

So, the energy of the hot water is 482630400 J

B.

The elevation <u>the mass would have to be raised from zero elevation relative to the reference environment for its exergy to equal that of the hot water</u> is 49248 m.

Using the equation for gravitational potential energy ΔU = mgΔh where m = mass of object = 1000 kg, g = acceleration due to gravity = 9.8 m/s² and Δh = h - h' where h = required elevation and h' = zero level elevation = 0 m

Since the energy of the mass equal the energy of the hot water, ΔU = 482630400 J

So, ΔU = mgΔh

ΔU = mg(h - h')

making h subject of the formula, we have

h = h' + ΔU/mg

Substituting the values of the variables into the equation, we have

h = h' + ΔU/mg

h = 0 m + 482630400 J/(1000 kg × 9.8 m/s²)

h = 0 m + 482630400 J/(9800 kgm/s²)

h = 0 m + 49248 m

h = 49248 m

So, the elevation <u>the mass would have to be raised from zero elevation relative to the reference environment for its exergy to equal that of the hot water</u> is 49248 m.

Learn more about heat energy here:

brainly.com/question/11961649

5 0
2 years ago
a coin press creates a pressure of 3.20*10^8 Pa on a nickel of radius 0.0106 m. how much force does the press exert on the coin?
Naya [18.7K]

Answer:

1.13 x 10⁵N

Explanation:

Given parameters:

Pressure of the coin press = 3.2 x 10⁸ Pa

radius of the nickel coin = 0.0106m

Unknown:

Force of the press on coil = ?

Solution:

Our knowledge of pressure will help us solve this problem.

Pressure is defined as the force applied per unit area on a body.

              Pressure  = \frac{force}{area}

  Force = Pressure  x  Area

 Since the pressure is known;

Area of the coin = Area of a circle = π r²

 where r is the radius of the coin;

Area of the coin = π x 0.0106²  = 3.53 x 10⁻⁴m²

  Force = 3.2 x 10⁸ x  3.53 x 10⁻⁴  = 1.13 x 10⁵N

3 0
3 years ago
A train travels 160 km in 2 h. What is the train’s average speed in km/h?
loris [4]
The train’s average speed is 80km/h
3 0
3 years ago
Read 2 more answers
How much does it take a person to walk 12km north at a velocity of 6.5 km/h?
mote1985 [20]

Answer:

1.846....hours

Explanation:

since if we divide 12 km or just 12 by 6.5 we will get this number meaning that this is how long it would take someone

4 0
3 years ago
A small plastic bead has been charged to -15nC.
marishachu [46]

Answer:

Explanation:

q = - 15 n C = - 15 x 10^-9 c

(a) d = 0.9 cm = 0.009 m

electric field on proton, E = Kq/d²

E = 9 x 10^9 x 15 x 10^-9 / (0.009)²

E = 1.67 x 10^6 N/C

Force on proton, F = charge on proton x Electric field

F = 1.6 x 10^-19 x 1.67 x 10^6 = 2.67 x 10^-13 N

acceleration of proton, a = F / m

a = (2.67 x 10^-13) / (1.67 x 10^-27)

a = 1.6 x 10^14 m/s²

(B) the direction of acceleration is towards the bead, as the force is attractive.

(C) d = 0.9 cm = 0.009 m

electric field on proton, E = Kq/d²

E = 9 x 10^9 x 15 x 10^-9 / (0.009)²

E = 1.67 x 10^6 N/C

Force on electron, F = charge on electron x Electric field

F = 1.6 x 10^-19 x 1.67 x 10^6 = 2.67 x 10^-13 N

acceleration of proton, a = F / m

a = (2.67 x 10^-13) / (9.1 x 10^-31)

a = 3 x 10^17 m/s²

(D) the direction of acceleration is away from the bead, as the force is repulsive.  

8 0
3 years ago
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