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AveGali [126]
3 years ago
10

The radiator of a car is a type of heat exchanger. Hot fluid coming from the car engine, called the coolant, flows through alumi

num radiator tubes of thickness d that release heat to the outside air by conduction. The average temperature gradient between the coolant and the outside air is about 130 K/mm .
The term deltaT/dis called the temperature gradient which is the temperature difference deltaT between x coolant inside and the air outside per unit thickness of tube.
Cooling temperature gradient?
A) twice that of the old unit
B) three halves that of the old unit
C) same as that of the old unit
D) two third that of the old unit
E) one-half that of the old unit
Part B
If the old radiator is replaced with anew one made of a material that has twice the thermal conductivity of the material of the old radiator what should the total surface area available for heat exchange in the new radiator be to achieve the desired temperature gradient ?
Assume that the tubes of the new radiator are as thick as tubes in older unit.
A) twice that of the old unit
B) three halves that of the old unit
C) same as that of the old unit
D) two third that of the old unit
E) one-half that of the old unit
Physics
1 answer:
Alex777 [14]3 years ago
8 0

Answer:

(A) The correct answer is option (B) three halves that of the old unit.

(B) The answer is three fourth that of old unit

Explanation:

from the relation;

(A) Fromthe expression;

K = Qd/AΔT

Anew = 3/2 A(old)

(B)

K¹ = 2K(old), so we get

A(old) = A(old)/2

Combining with part A, we have;

Anew = 3/2 *A(old)/2

          = 3/4A(old)

The answer is three fourth that of old unit

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Some of the highest tides in the world occur in the Bay of Fundy on the Atlantic Coast of Canada. At Hopewell Cape the water dep
Nady [450]

Answer:

(a) 1.939 m/h

(b) 0.926 m/h

(c) -0.315 m/h

(d) -1.21 m/h

Explanation:

Here, we have the water depth given by the function of time;

D(t) = 7 + 5·cos[0.503(t-6.75)]

Therefore, to find the velocity of the depth displacement with time, we differentiate the given expression with respect to time as follows;

D'(t) = \frac{d(7 + 5\cdot cos[0.503(t-6.75)])}{dt}

= 5×(-sin(0.503(t-6.75))×0.503

= -2.515×(-sin(0.503(t-6.75))

= -2.515×(-sin(0.503×t-3.395))

Therefore we have;

(a) at 5:00 AM = 5 -  0:00 = 5

D'(5) =  -2.515×(-sin(0.503×5-3.395)) = 1.939 m/h

(b) at 6:00 AM = 6 -  0:00 = 6

D'(5) =  -2.515×(-sin(0.503×6-3.395)) = 0.926 m/h

(c) at 7:00 AM = 7 -  0:00 = 7

D'(5) =  -2.515×(-sin(0.503×7-3.395)) = -0.315 m/h

(d) at Noon 12:00 PM = 12 -  0:00 = 12

D'(5) =  -2.515×(-sin(0.503×12-3.395)) = -1.21 m/h.

4 0
3 years ago
Insulators have very high .
Vlad1618 [11]

Answer:

Resistance to electrical currents

Explanation:

Conductors have low resistance to electrical currents, and are used to "conduct" the flow of electricity.

Insulators have very high resistance and are used to protect us from the flow of electricity.

5 0
3 years ago
IN WHAT CONDITION DO SOUND ECHO
DerKrebs [107]

Answer:

The conditions necessary for hearing the echo. The distance between the sound source and the reflecting surface must not be less than 17 metres where the time period between hearing the original sound and its echo should not be less than 0.1 of a second.

6 0
2 years ago
Sharon the ant (Aaron’s sister) sits at the edge of a turntable of radius R that is spinning with period T. As she makes one-hal
Dmitry_Shevchenko [17]

Answer:

a = \dfrac{4\pi^2R}{T^2}

Explanation:

The acceleration of a circular motion is given by

a = \omega^2 R

where \omega is the angular velocity and R is the radius.

Angular velocity is related to the period, T, by

\omega=\dfrac{2\pi}{T}

Substitute into the previous formula.

a = (\dfrac{2\pi}{T})^2 R

a = \dfrac{4\pi^2R}{T^2}

This acceleration does not depend on the linear or angular displacement. Hence, the amount of rotation does not change it.

6 0
3 years ago
Relative formula mass of CuCO3
Elodia [21]

AnMolar mass of CuCO3 = 123.5549 g/mol

This compound is also known as Copper(II) Carbonate.

Convert grams CuCO3 to moles  or  moles CuCO3 to grams

Molecular weight calculation:

63.546 + 12.0107 + 15.9994*3

Percent composition by element

Element   Symbol   Atomic Mass   # of Atoms   Mass Percent

Copper Cu 63.546 1 51.431%

Carbon C 12.0107 1 9.721%

Oxygen O 15.9994 3 38.848%

Explanation:

5 0
2 years ago
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