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Simora [160]
3 years ago
7

Is a single property such as density sufficient to always correctly identify a substance?

Physics
1 answer:
bazaltina [42]3 years ago
7 0
Not at all. Density on its own is not sufficient
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Do we include signs if we calculate Electric field strength?
expeople1 [14]
Yes because if not people wouldn't understand how did you calculate electric field strength.
6 0
3 years ago
A man goes for a walk, starting from the origin of an xyz coordinate system, with the xy plane horizontal and the x axis eastwar
tekilochka [14]
<h2>a) Displacement of penny = 1300 i + 2400 j - 640 k</h2><h2>b) Magnitude of his displacement = 2729.47 m</h2>

Explanation:

a) He walks 1300 m east, 2400 m north, and then drops the penny from a cliff 640 m high.

1300 m east = 1300 i

2400 m north = 2400 j

Drops the penny from a cliff 640 m high = -640 k

Displacement of penny = 1300 i + 2400 j - 640 k

b) Displacement of man for return trip = -1300 i - 2400 j

    \texttt{Magnitude = }\sqrt{(-1300)^2+(-2400)^2}=2729.47m

    Magnitude of his displacement = 2729.47 m

3 0
3 years ago
Read 2 more answers
17.
Zolol [24]

Answer:

I don't know this answer at all

Explanation:

I don't know about these problems

6 0
3 years ago
The Escape speed at the surface of a certain planet is twice that of the earth. what is its mass in unit of earth's mass?
VLD [36.1K]

Answer:

22Km/sec

Explanation:

6 0
2 years ago
Steam in a heating system flows through tubes whose outer diameter is 5 cm and whose walls are maintained at a temperature of 13
svet-max [94.6K]

Answer:

5945.27 W per meter of tube length.

Explanation:

Let's assume that:

  • Steady operations exist;
  • The heat transfer coefficient (h) is uniform over the entire fin surfaces;
  • Thermal conductivity (k) is constant;
  • Heat transfer by radiation is negligible.

First, let's calculate the heat transfer (Q) that occurs when there's no fin in the tubes. The heat will be transferred by convection, so let's use Newton's law of cooling:

Q = A*h*(Tb - T∞)

A is the area of the section of the tube,

A = π*D*L, where D is the diameter (5 cm = 0.05 m), and L is the length. The question wants the heat by length, thus, L= 1m.

A = π*0.05*1 = 0.1571 m²

Q = 0.1571*40*(130 - 25)

Q = 659.73 W

Now, when the fin is added, the heat will be transferred by the fin by convection, and between the fin and the tube by convection, thus:

Qfin = nf*Afin*h*(Tb - T∞)

Afin = 2π*(r2² - r1²) + 2π*r2*t

r2 is the outer radius of the fin (3 cm = 0.03 m), r1 is the radius difference of the fin and the tube ( 0.03 - 0.025 = 0.005 m), and t is the thickness ( 0.001 m).

Afin = 0.006 m²

Qfin = 0.97*0.006*40*(130 - 25)

Qfin = 24.44 W

The heat transferred at the space between the fin and the tube will be:

Qspace = Aspace*h*(Tb - T∞)

Aspace = π*D*S, where D is the tube diameter and S is the space between then,

Aspace = π*0.05*0.003 = 0.0005

Qspace = 0.0005*40*(130 - 25) = 1.98 W

The total heat is the sum of them multiplied by the total number of fins,

Qtotal = 250*(24.44 + 1.98) = 6605 W

So, the increase in heat is 6605 - 659.73 = 5945.27 W per meter of tube length.

5 0
3 years ago
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