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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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chubhunter [2.5K]

I learned the equation as P•V=k•T .

So x=1, y=1, and z= -1 .

3 0
3 years ago
A ball of mass 0.160 kg is dropped from a height of 2.25 m. When it hits the ground it compresses 0.087 m.
Studentka2010 [4]

A) 6.64 m/s downward

B) 0.026 s

C) -40.9 N

Explanation:

A)

We can solve this problem by using the law of conservation of energy.

In fact, since the total mechanical energy of the ball must be conserved, this means that the initial gravitational potential energy of the ball before the fall is entirely converted into kinetic energy just before it reaches the floor.

So we can write:

PE=KE\\mgh = \frac{1}{2}mv^2

where

m = 0.160 kg is the mass of the ball

g=9.8 m/s^2 is the acceleration due to gravity

h = 2.25 m is the initial height of the ball

v is the final velocity of the ball before hitting the ground

Solving for v, we find:

v=\sqrt{2gh}=\sqrt{2(9.8)(2.25)}=6.64 m/s

And the direction of the velocity is downward.

B)

The motion of the ballduring the collision is a uniformly accelerated motion (= with constant acceleration), so the time of impact can be found by using a suvat equation:

s=(\frac{u+v}{2})t

where:

v is the final velocity

u is the initial velocity

s is the displacement of the ball during the impact

t is the time

Here we have:

u = 6.64 m/s is the velocity of the ball before the impact

v = 0 m/s is the final velocity after the impact (assuming it comes to a stop)

s = 0.087 m is the displacement, as the ball compresses by 0.087 m

Therefore, the time of the impact is:

t=\frac{2s}{u+v}=\frac{2(0.087)}{0+6.64}=0.026 s

C)

The force exerted by the floor on the ball can be found using the equation:

F=\frac{\Delta p}{t}

where

\Delta p is the change in momentum of the ball

t is the time of the impact

The change in momentum can be written as

\Delta p = m(v-u)

So the equation can be rewritten as

F=\frac{m(v-u)}{t}

Here we have:

m = 0.160 kg is the mass of the ball

v = 0 is the final velocity

u = 6.64 m/s is the initial velocity

t = 0.026 s is the time of impact

Substituting, we find the force:

F=\frac{(0.160)(0-6.64)}{0.026}=-40.9 N

And the sign indicates that the direction of the force is opposite to the direction of motion of the ball.

4 0
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anzhelika [568]
<span>The magnitude of the gravitational force between two bodies is the product of their masses divided by the square of the distance between them. So we have F = M1*M2 / r^2. M1 = 7.503 * 10e24 and M2 = 2.703 * 10e22 and r= 2.803 * 10e8; r^2 = 5.606 *10e16. So we have 7.503 *2.703 *10^(24+22) = 20.280 * 10^(46). Then we divide our answer by 5.606 * 10e16 which is the distance ; then we have 3.6175 * 10 e (46- 16) = 3.6175 * 10e30. To find the acceleration we use Newton's second law F = ma. F is 3.6175 * 10e30 and M is 7.503 * 10e24 so a = F/M and then we have 3.6175/7.503 * 10e (30-24) = 0.48 * 10e6. Similarly for moon, we have a = 3.6715/2.703 * 10e(30-22). = 1.358 * 10e8</span>
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