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Ksivusya [100]
3 years ago
5

Fill a Styrofoam cup with very hot water ( or Coffee). Take objects made of different materials, such as a metal spoon, a wooden

pencil, a plastic pen or a glass rod. Put one end of each object into the water and answer following questions.
Required:
a. How close to the surface of the water do you have to hold the object for it to feel noticeably warm to the touch?
b. From observations, which material would you judge to be the best conductor of heat? Which is the worst?
c. Rank the materials according to their ability to conduct heat from best to worst.
Physics
1 answer:
torisob [31]3 years ago
6 0

Answer:

Explanation:

a ) A good conducting material conducts heat easily from one point to another . So the metal spoon which is good conductor will become hot easily by conducting heat from hot water to tip of the spoon . So even after touching the far end of spoon , we can feel the heat of the hot water in the cup . For other material like wood or plastic we have to hold the part which is  very close to water to feel heat of water .

b ) The best conductor will be that which touches hot even when we touch its farthest  end . Hence metal spoon will be best conductor . Plastic pen will be worst conductor because even touching its part close to hot water , we do not feel much heat .

c ) metal spoon > glass rod  > wooden pencil > plastic pen .

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Answer:

Corpuscular theory of light

Explanation:

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The United States spends over $20 billion a year on space exploration through NASA. Do you think that this has been worth the co
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2 years ago
Continuous and aligned fiber-reinforced composite with cross-sectional area of 340 mm2 (0.53 in.2) is subjected to a longitudina
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(a) 23.4

The fiber-to-matrix load ratio is given by

\frac{F_f}{F_m}=\frac{E_f V_f}{E_m V_m}

where

E_f = 131 GPa is the fiber elasticity module

E_m = 2.4 GPa is the matrix elasticity module

V_f=0.3 is the fraction of volume of the fiber

V_m=0.7 is the fraction of volume of the matrix

Substituting,

\frac{F_f}{F_m}=\frac{(131 GPa)(0.3)}{(2.4 GPa)(0.7)}=23.4 (1)

(b) 44,594 N

The longitudinal load is

F = 46500 N

And it is sum of the loads carried by the fiber phase and the matrix phase:

F=F_f + F_m (2)

We can rewrite (1) as

F_m = \frac{F_f}{23.4}

And inserting this into (2):

F=F_f + \frac{F_f}{23.4}

Solving the equation, we find the actual load carried by the fiber phase:

F=F_f (1+\frac{1}{23.4})\\F_f = \frac{F}{1+\frac{1}{23.4}}=\frac{46500 N}{1+\frac{1}{23.4}}=44,594 N

(c) 1,906 N

Since we know that the longitudinal load is the sum of the loads carried by the fiber phase and the matrix phase:

F=F_f + F_m (2)

Using

F = 46500 N

F_f = 44594 N

We can immediately find the actual load carried by the matrix phase:

F_m = F-F_f = 46,500 N - 44,594 N=1,906 N

(d) 437 MPa

The cross-sectional area of the fiber phase is

A_f = A V_f

where

A=340 mm^2=340\cdot 10^{-6}m^2 is the total cross-sectional area

Substituting V_f=0.3, we have

A_f = (340\cdot 10^{-6} m^2)(0.3)=102\cdot 10^{-6} m^2

And the magnitude of the stress on the fiber phase is

\sigma_f = \frac{F_f}{A_f}=\frac{44594 N}{102\cdot 10^{-6} m^2}=4.37\cdot 10^8 Pa = 437 MPa

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The cross-sectional area of the matrix phase is

A_m = A V_m

where

A=340 mm^2=340\cdot 10^{-6}m^2 is the total cross-sectional area

Substituting V_m=0.7, we have

A_m = (340\cdot 10^{-6} m^2)(0.7)=238\cdot 10^{-6} m^2

And the magnitude of the stress on the matrix phase is

\sigma_m = \frac{F_m}{A_m}=\frac{1906 N}{238\cdot 10^{-6} m^2}=8.0\cdot 10^6 Pa = 8.0 MPa

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Katarina [22]
The west constituent of their sequence needs to cancel out 58 mph crosswind. Subsequently a northwest direction is a 45-degree angle up to even with the destination. That is the third point out of the triangle and the right angle is at the destination. The top side is the west constituent of their flight the vertical side is their resultant travel and the hypotenuse is their definite distance flown. Since the 58 mph crosswind was negated by flying northwest, the distance from the beginning to the destination must be the same distance as the west component of their travel. The hypotenuse is square root of twice the side since it has 2 identical sides.

c = sqrt (58^2 + 58^2) = sqrt (6728) = 82.02 
 
Alternative solution:

c = sqrt (2) * 58 = 1.414 * 58 = 82.02

Therefore, they have to fly 82.02 mph
5 0
3 years ago
Read 2 more answers
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