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AveGali [126]
2 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]2 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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A 50-kg load is suspended from a steel wire of diameter 1.0 mm and length 11.2 m. By what distance will the wire stretch? Young'
lbvjy [14]

Answer:

3.5 cm

Explanation:

mass, m = 50 kg

diameter = 1 mm

radius, r = half of diameter = 0.5 mm = 0.5 x 10^-3 m

L = 11.2 m

Y = 2 x 10^11 Pa

Area of crossection of wire = π r² = 3.14 x 0.5 x 10^-3 x 0.5 x 10^-3  

                                              = 7.85 x 10^-7 m^2

Let the wire is stretch by ΔL.

The formula for Young's modulus is given by

Y =\frac{mgL}{A\Delta L}

\Delta L =\frac{mgL}{A\times Y}

ΔL = 0.035 m = 3.5 cm

Thus, the length of the wire stretch by 3.5 cm.

5 0
2 years ago
A reflecting telescope is built with a 20-cm-diameter mirror having a 1.00 m focal length. it is used with a 10× eyepiece.
kondaur [170]
I really wish I could be helping you. I don't know.
3 0
2 years ago
a plane travels 204 km, northeast in 15.0 minutes. It also increases elevation by 1.6 km, upward in the same amount of time. Wha
mina [271]

Answer:

230 m/s northeast, 1.8 m/s up

Explanation:

204 kilometres = 204000 metres

15.0 minutes = 900 seconds

Velocity = Distance / Time

= 204000 / 900

= 230 m/s northeast (to 2 sf.)

1.6km = 1600 metres

Velocity = 1600 / 900

= 1.8 m/s up (to 2 sf.)

7 0
2 years ago
The absolute pressure, in kilopascals, a depth 10m below sea level is most nearly?
saul85 [17]

Answer:

option A

Explanation:

given,

depth of the sea level = 10 m

g = 10 m/s²

Pressure underwater = ?

we know,

P = ρ g h

where ρ is the density of water which is equal to 1000 kg/m³

h is the depth of sea level

P = ρ g h

P = 1000 x 10 x 10

P = 100000 Pa

P = 100 kPa

Hence, the correct answer is option A

8 0
2 years ago
What's the diameter of a dish antenna that will receive 10−20W of power from Voyager at this time? Assume that the radio transmi
Murrr4er [49]

Complete Question:

The Voyager 1 spacecraft is now beyond the outer reaches of our solar system, but earthbound scientists still receive data from the spacecraft s 20-W radio transmitter. Voyager is expected to continue transmitting until about 2025, when it will be some 25 billion km from Earth.

What s the diameter of a dish antenna that will receive 10−20W of power from Voyager at this time? Assume that the radio transmitter on Voyager transmits equally in all directions(isotropically).  In fact, the antenna on Voyager focuses the signal in a beam aimed at the earth, so this problem over-estimates the size of the receiving dish needed.

Answer:

d = 2,236 m.

Explanation:

The received power on Earth, can be calculated as the product of the intensity (or power density) times the area that intercepts the power radiated.

As we assume that  the transmitter antenna is ominidirectional, power is spreading out over a sphere with a radius equal to the distance to the source.

So, we can get the power density as follows:

I = P /A = P / 4*π*r², where P = 20 W, and r= 25 billion km = 25*10¹² m.

⇒ I = 20 W / 4*π* (25*10¹²)² m²

The received power, is just the product of this value times the area of the receiver antenna, which we assumed be a circle of diameter d:

Pr = I. Ar =( 20W / 4*π*(25*10¹²)² m²) * π * (d²/4) = 10⁻²⁰ W

Simplifying common terms, we can solve for d:

d= √(16*(25)²*10⁴/20) = 2,236 m.

3 0
2 years ago
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