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3 years ago

1.05 MEASURING PHYSICAL PROPERTIES LAB REPORT

Physics

1 answer:

Nana76 [90]3 years ago

4 0

Answer:

Objective(s): In your own words, what was the purpose of this lab?

The purpose of this lab is to get us familiar with the physical properties of different types of materials that are used for building.

Hypothesis: In this section, include the if/then statements you developed during your lab activity. These statements reflect your predicted outcomes for the experiment.

If we use clay brick to build the roof of the house, then the temperature inside the house will remain cooler.

If we use wood to build the walls and floors of the house, then the temperature inside the house will remain cooler.

If we use nickel to bring electricity into the home, then it will allow electricity to flow into the home at a faster rate.

If we use iron to construct the latches on the windows and doors, then the magnetism will keep the latches secure.

Procedure: The materials and procedures are listed in your virtual lab. You do not need to repeat them here. However, you should note if you experienced any errors or other factors that might affect your outcome.

Using the summary questions, clearly define the dependent and independent variables of the experiment.

Data: Record the results of each of your physical property tests in the table below.

Wood _ 3/10 _0 _0.12 W/(m·K) _0 S/m _ 4 g/cm3

Clay Brick _ 2/10 _1 _0.6 W/(m·K) _ 0 S/m _ 5.88235 g/cm3

Iron _ 4/10 _10 _80 W/(m·K) _1x107 S/m _9.09091 g/cm3

Aluminum_ 7/10 _ 0 _235 W/(m·K) _ 3.8x107 S/m _ 6.66667 g/cm3

Copper _ 6/10 _0 _400 W/(m·K) _3.8x107 S/m _9.52381 g/cm3

Nickel _ 5/10 _7 _ 91 W/(m·K) _1.4x107 S/m _9.43396 g/cm3

Conclusion: Your conclusion will include a summary of the lab results and an interpretation of the results. Please answer all questions in complete sentences using your own words.

Using two to three sentences, summarize what you investigated and observed in this lab.

In this, I investigated the effect that different types of physical properties had on six different materials. I also investigated which materials would be the best to build a house.

2. What building material did you use to build your house? Did your results support or fail to support your hypotheses?

For the roof of my house, I used clay bricks. For the walls and floors of the house, I used wood. To bring electricity into the home I used nickel. To construct the latches on the windows and doors, I used iron. My results supported my hypotheses because clay bricks and wood had the lowest thermal conductivity rates, nickel had the second highest electricity conductivity rate, and iron had the highest magnetism rate.

3. What were the densities of the materials you chose for the walls and floor for the home in Tiny World? Why do you think a building material's density is important when building homes or architectural structures?

The density of the material I chose for the walls and floors for the home in Tiny World was 4 g/cm3 (Wood). I think a buildings material density is important when building homes or architectural structures because in order for that structure to stay stable the materials that are used must be dense. If the materials are not dense, the structure would most likely collapse if it is placed under too much pressure.

4. Why wouldn’t you choose wood or aluminum for the latches on your house?

I wouldn’t choose wood or aluminum for the latches on my house because neither of those two materials are magnetic.

5. Which material has the highest thermal conductivity? Which material has the highest electrical conductivity? Explain why thermal and electrical conductivity is so high with this material.

Copper has the highest thermal conductivity. Aluminum and copper have the highest electrical conductivity. Thermal conductivity is high in copper because copper is a very thin type of material, and this allows heat to pass through it easily. Electric Conductivity is high in both aluminum and copper because they are both a very thin type of metal and because they are both so thin, electricity will pass through them quickly.

I'm sorry if any of the answers are wrong, this was just assigned to me and my teacher hasn't graded it yet.

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Answer:

The BMX lands 5.4 m from the end of the ramp.

Explanation:

Hi there!

The position of the BMX is given by the position vector "r":

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

Where:

r = position vector at time t

x0 = initial horizontal position

v0 = initial velocity

α = jumping angle

y0 = initial vertical position

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive)

Please, see the attached graphic for a better understanding of the situation. At final time, when the bicycle reaches the ground, the vector position will be "r final" (see figure). The y-component of the vector "r final" is - 2.4 m (placing the origin of the frame of reference at the jumping point). With that information, we can use the equation of the y-component of the vector "r" (see above) to calculate the time of flight. With that time, we can then obtain the x-component (rx in the figure) of the vector "r final". Then:

y = y0 + v0 · t · sin α + 1/2 · g · t²

-2.4 m = 0 m + 5.9 m/s · t · sin 40° - 1/2 · 9.8 m/s² · t²

0 = -4.9 m/s² · t² + 5.9 m/s · t · sin 40° + 2.4 m

Solving the quadratic equation:

t = 1.2 s

Now, we can calculate the x-component of the vector "r final" that is the horizontal distance traveled by the bicycle:

x = x0 + v0 · t · cos α

x = 0 m + 5.9 m/s · 1.2 s · cos 40°

x = 5.4 m

The BMX lands 5.4 m from the end of the ramp.

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Answer:

40m/s

Explanation:

The horizontal component of velocity remains constant because there are no external forces in that direction

By applying motion equations, V= U+ at

where ,

v - final velocity
u - initial velocity
a-acceleration
t - time

v = u +at

As no force act on the ball ( we neglect air resistance here) no acceleration is seen,

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Answer:

Current, I = 0.0011 A

Explanation:

It is given that,

Diameter of rod, d = 2.56 cm

Radius of rod, r = 1.28 cm = 0.0128 m

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Voltage, $V=1\times 10^3\ V$

The resistivity of the rod is given by :

$R=\rho\dfrac{L}{A}$

$R=2300\ \Omega-m\dfrac{0.2\ m}{\pi (0.0128\ m)^2}$

R = 893692.30 ohms

Current flowing in the rod is calculated using Ohm's law as :

V = I R

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So, the current flowing in the rod is 0.0011 A. Hence, this is the required solution.

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