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Degger [83]
2 years ago
5

A cup of hot coffee can be cooled by placing a cold spoon in it. A spoon of which of the following materials would be most effec

tive for this purpose: aluminum, copper, iron, or glass? Assume all spoons have the same mass.
Physics
1 answer:
arsen [322]2 years ago
8 0

Answer: copper

Explanation: when a spoon is placed in a hot coffee, the coffee will cool off based on the concept of heat transfer.

The coffee is at a higher temperature compared to the spoon, hence there is conduction of heat by the spoon from the coffee.

Heat is now transfered from the coffee to the spoon, the rate of heat transfer ( for this case coffee to spoon) is dependent on some properties but for this case, the thermal conductivity is of the most important.

A material of high thermal conductivity have a very fast rate of heat transfer and a material of low thermal conductivity have a slow rate of heat transfer.

Below is the value of thermal conductivity for copper, aluminum, iron and glass.

Copper (385 w/m.K)

Aluminum (205 W/m.K)

Iron (79.5 W/m.K)

Glass (0.8 W/m.K)

It can been seen that copper has the highest value for thermal conductivity which implies that it will conduct heat faster compared to others thus cooling the coffee faster.

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A fan is to accelerate quiescent air to a velocity of 12.5 m/s at a rate of 9 m3/s. Determine the minimum power that must be sup
Reika [66]

Answer:

= 829.69 Watt

≅ 830 Watt

Explanation:

Given that,

Velocity of air flow = 12.5m/s

Rate of flow of air = 9m³/s

Density of air = 1.18kg/m³

power by kinetic energy = 1/2(mv²)

mass = density × volume

m = 1.18 × 9

  = 10.62 kg/s

power = 1/2 mV²

           = 1/2 (10.62 × 12.5²)

           = 829.69 Watt

           ≅ 830 Watt

Flow rate  

u

=

9

 

m

3

/

s

Velocity of the air  

V

=

8

 

m/s

Density of the air  

ρ

=

1.18

 

kg

/

m

3

5 0
2 years ago
When numbers are very small or very large, it is convenient to either express the value in scientific notation and/or by using a
Oxana [17]

Answer:

5 mg, 5\cdot 10^{-3}g

Explanation:

First of all, let's rewrite the mass in grams using scientific notation.

we have:

m = 0.005 g

To rewrite it in scientific notation, we must count by how many digits we have to move the dot on the right - in this case three. So in scientific notation is

m=5\cdot 10^{-3}g

If  we want to convert into milligrams, we must remind that

1 g = 1000 mg

So we can use the proportion

1 g : 1000 mg = 0.005 g : x

and we find

x=\frac{(1000 mg)(0.005 g)}{1 g}=5 mg

4 0
2 years ago
A solid metal sphere of diameter D is spinning in a gravity-free region of space with an angular velocity of ωi. The sphere is s
Leona [35]

Answer:

0.6

Explanation:

The volume of a sphere = \frac{4}{3} \pi (\frac{D}{2})^3

Therefore \pi * r^2 * (\frac{D}{2} ) = \frac{4}{3} \pi (\frac{D}{2})^3

r of the disc = 1.15(\frac{ D}{2} )

Using conservation of angular momentum;

The M_i of the sphere = \frac{2}{5} m \frac{D}{2}^2

M_i of the disc = m*\frac{   \frac{1.15*D}{2}^2 }{2}

\frac{wd}{ws} = \frac{\frac{2}{5}m * \frac{D}{2}^2}{  m * \frac{(\frac{`.`5*D}{2})^2 }{2} }

= 0.6

5 0
2 years ago
You want to examine the hairy details of your favorite pet caterpillar, using a lens of focal length 8.9 cm 8.9 cm that you just
Zepler [3.9K]

Answer:

The angular magnification is M = 2.808

Explanation:

From the question we are told

           The focal length is  f = 8.9cm

          The near point is n_p = 25.0cm

The angular magnification is mathematically represented as

                          M = \frac{n_p}{f}

Substituting values

                        M = \frac{25}{8.9}

                           = 2.808

4 0
2 years ago
A truck is traveling down a road with a 4-percent grade at a speed of 75 mi/h when its brakes are applied to slow it down to 22.
kvasek [131]

Answer:

3.964 s

Explanation:

Metric unit conversion:

1 miles = 1.6 km = 1600 m.

1 hour = 60 minutes = 3600 seconds

75 mph = 75 * 1600 / 3600 = 33.3 m/s

22.5 mph = 22.5 * 1600/3600 = 10 m/s

Let g = 9.81 m/s2

Friction is the product of coefficient and normal force, which equals to the gravity

F_f = \mu N = \mu mg

The deceleration caused by friction is friction divided by mass according to Newton 2nd law.

a = F_f / m = \mu mg / m = \mu g = 0.6 *9.81 = 5.886 m/s^2

So the time required to decelerate from 33.3 m/s to 10 m/s so the wheels don't slide, with the rate of 5.886 m/s2 is

t = \frac{\Delta v}{a} = \frac{33.3 - 10}{5.886} = 3.964 s

3 0
2 years ago
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