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Rudiy27
1 year ago
4

Please somebody help me with this project! It's due tomorrow and I'm a bit stuck with it! I'd really appreciate it if anyone is

willing to help.
Mass (kg) Volume (L) Density (Kg/L) Substance
Block A
Block B
Block C
Block D
Block E

Densities of Various Materials
Material Density (kg/L)
Wood - 0.40
Apple - 0.64
Plastic - 0.70
Ice - 0.92
Water - 1.00
Aluminum - 2.70
Diamond - 3.53
Lead - 11.3
Gold - 19.3

Questions:
1) Which block has the greatest density?
2) Which block has the lowest density?
3) Submit your data table and identify the substance each block is made of.


I tried to match them, not sure if they're correct:
Answer:
Block A - 19.3 Kg/L
Block B - 0.64 Kg/L
Block C - 0.70 Kg/L
Block D - 0.917 Kg/L
Block E - 3.53 Kg/L
Physics
1 answer:
OverLord2011 [107]1 year ago
7 0
Block d hehehehe I’m so smart
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3 years ago
Select three different examples of accelerated motion. a body traveling in a straight line and increasing in speed a body travel
IrinaVladis [17]
This is the same question that I just answered.

Have present the definition of acceleration:

         a = Δv / Δt, this is change in velocity per unit of time.

a and v are in bold to mean that they are vectors.

1) a body traveling in a straight line and increasing in speed: CORRECT:

Acceleration is the change in velocity, either magnitude or direction or both. So, a body increasing in speed is accelerated.

2) a body traveling in a straight line and decreasing in speed: CORRECT

A decrease in speed is a change in velocity, so it means acceleration.

3) a body traveling in a straight line at constant speed: FALSE.

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6 0
4 years ago
A 63-kg hiker is climbing the 828-m-tall Burj Khalifa in Dubai. If the efficiency of converting the energy content of the bars i
konstantin123 [22]

Answer:

T_f=5854.76 °C

Explanation:

Given:

mass of hiker, m= 63 kg

height to be climbed, h= 828 m

energy produced by an energy bar, E= 1.10\times 10^6 J

heat capacity of the hiker, c=75.3 J.mol^{-1}.K^{-1}= 4.184 J.kg^{-1}.K^{-1}

initial body temperature of hiker, T_i=36.6 \degree C

<em>The efficiency of converting the energy content of the bars into the work of climbing is 25%, the remaining 75% of the energy released through metabolism is heat released to her body.</em>

We find the energy required for climbing 828 m height:

W=m.g.h

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W= 511207.2 J

∵Hike eats 2 energy bars= 2\times 1.1\times 10^{6} J

Energy produced= 2.2\times 10^{6} J

Now, according to her efficiency:

Total energy required for producing the work of W= 511207.2 J which is required to climb the given height will be (say, E):

25\% of E= 511207.2

\Rightarrow E= 511207.2\times \frac{100}{25}

E=2044828.8 J

&

Amount of total energy (E) converted into heat(Q) is:

Q=2044828.8-511207.2\\Q=1533621.6J

As we know that:

heat, Q=m.c. (T_f-T_i).................(1)

where:

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1533621.6= 63\times 4.184\times (T_f-36.6)

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