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3 years ago

20 points please answer

1 answer:

6 0

Another way to test your question is to build your own miniature buildings. Depending on how in-depth you go, building could get a little pricey, but if you keep it basic there shouldn't be a problem. Decide on a certain number of foundations to test [maybe 3 or so] and try simulating an earthquake.
<span>Hope this helps! </span>

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Grains of fine california beach sand are approximately spheres with an average radius of 50 μm and are made of silicon dioxide,

kykrilka [37]

The average radius(r) of each grain is r = 50 nanometers = 50*10^-6 meters

Since it is spherical, so

Volume=(4/3)*pi*r^3

V= (4/3)*pi*(50*10^-6)^3

V=5.23599*10^-13 m^3

We are given the Density(ρ) =2600kg/m^3

We know that:

Density(p) = mass(m)/volume(V)

m = ρV

So the mass of a single grain is:

m = 5.23599*10^-13 * 2600 = 1.361357*10^-9 kg

The surface area of a grain is:

a = 4*pi*r^2

a = 4*pi*(50*10^-6)^2

a = 3.14*10^-8 m^2

Since we know the surface area and mass of a grain, the conversion factor is:

1.361357*10^-9 kg / 3.14*10^-8 m^2

Find the Surface area of the cube:

cube = 6a^2

cube = 6*1.1^2 = 7.26m^2

multiply this by the converions ratio to get:

total mass of sand grains = (7.26 m^2 * 1.361357*10^-9 kg) / (3.14*10^-8 m^2)

total mass of sand grains = 0.3148 kg = 314.80 g

3 0

3 years ago

What is an astronomer? Answer in the simpliest way please.

Sphinxa [80]

They would be an expert in astronomy
astronomy:the branch of science that deals with celestial objects

8 0

3 years ago

Which of the following is the flow of electrons through a wire or a conductor

Hatshy [7]

Wires or silver and copper

8 0

3 years ago

The entropy of an isolated system must be conserved, so it never changes.a. Trueb. Fasle

Snowcat [4.5K]

Answer:

B: False

Explanation:

The second law of thermodynamics states that: the entropy of an isolated system will never decrease because isolated systems always tend to evolve towards thermodynamic equilibrium which is a state with maximum entropy.

Thus, it means that the entropy change will always be positive.

Therefore, the given statement in the question is false.

6 0

3 years ago

Usain Bolt's world-record 100 m sprint on August 16, 2009, has been analyzed in detail. At the start of the race, the 94.0 kg Bo

ZanzabumX [31]

a) 893 N

b) 8.5 m/s

c) 3816 W

d) 69780 J

e) 8030 W

Explanation:

The net force acting on Bolt during the acceleration phase can be written using Newton's second law of motion:

$F_{net}=ma$

where

m is Bolt's mass

a is the acceleration

In the first 0.890 s of motion, we have

m = 94.0 kg (Bolt's mass)

$a=9.50 m/s^2$ (acceleration)

So, the net force is

$F_{net}=(94.0)(9.50)=893 N$

And according to Newton's third law of motion, this force is equivalent to the force exerted by Bolt on the ground (because they form an action-reaction pair).

Since Bolt's motion is a uniformly accelerated motion, we can find his final speed by using the following suvat equation:

$v=u+at$

where

v is the final speed

u is the initial speed

a is the acceleration

t is the time

In the first phase of Bolt's race we have:

u = 0 m/s (he starts from rest)

$a=9.50 m/s^2$ (acceleration)

t = 0.890 s (duration of the first phase)

Solving for v,

$v=0+(9.50)(0.890)=8.5 m/s$

First of all, we can calculate the work done by Bolt to accelerate to a speed of

v = 8.5 m/s

According to the work-energy theorem, the work done is equal to the change in kinetic energy, so

$W=K_f - K_i = \frac{1}{2}mv^2-0$

where

m = 94.0 kg is Bolt's mass

v = 8.5 m/s is Bolt's final speed after the first phase

$K_i = 0 J$ is the initial kinetic energy

So the work done is

$W=\frac{1}{2}(94.0)(8.5)^2=3396 J$

The power expended is given by

$P=\frac{W}{t}$

where

t = 0.890 s is the time elapsed

Substituting,

$P=\frac{3396}{0.890}=3816 W$

First of all, we need to find what is the average force exerted by Bolt during the remaining 8.69 s of motion.

In the first 0.890 s, the force exerted was

$F_1=893 N$

We know that the average force for the whole race is

$F_{avg}=820 N$

Which can be rewritten as

$F_{avg}=\frac{0.890 F_1 + 8.69 F_2}{0.890+8.69}$

And solving for $F_2$ , we find the average force exerted by Bolt on the ground during the second phase:

$F_{avg}=\frac{0.890 F_1 + 8.69 F_2}{0.890+8.69}\\F_2=\frac{(0.890+8.69)F_{avg}-0.890F_1}{8.69}=812.5 N$

The net force exerted by Bolt during the second phase can be written as

$F_{net}=F_2-D$ (1)

where D is the air drag.

The net force can also be rewritten as

$F_{net}=ma$

where

$a=\frac{v-u}{t}$ is the acceleration in the second phase, with

u = 8.5 m/s is the initial speed

v = 12.4 m/s is the final speed

t = 8.69 t is the time elapsed

Substituting,

$a=\frac{12.4-8.5}{8.69}=0.45 m/s^2$

So we can now find the average drag force from (1):

$D=F_2-F_{net}=F_2-ma=812.5 - (94.0)(0.45)=770.2 N$

So the increase in Bolt's internal energy is just equal to the work done by the drag force, so:

$\Delta E=W=Ds$

where

d is Bolt's displacement in the second part, which can be found by using suvat equation:

$s=\frac{v^2-u^2}{2a}=\frac{12.4^2-8.5^2}{2(0.45)}=90.6 m$

And so,

$\Delta E=Ds=(770.2)(90.6)=69780 J$

The power that Bolt must expend just to voercome the drag force is given by

$P=\frac{\Delta E}{t}$

where

$\Delta E$ is the increase in internal energy due to the air drag

t is the time elapsed

Here we have:

$\Delta E=69780 J$

t = 8.69 s is the time elapsed

Substituting,

$P=\frac{69780}{8.69}=8030 W$

And we see that it is about twice larger than the power calculated in part c.

3 0

3 years ago