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svetoff
2 months ago
3

You would like to know whether silicon will float in mercury and you know that can determine this based on their densities. unfo

rtunately, you have the density of mercury in units of kilogrammeter3 and the density of silicon in other units: 2.33 gramcentimeter3. you decide to convert the density of silicon into units of kilogrammeter3 to perform the comparison. by which combination of conversion factors will you multiply 2.33 gramcentimeter3 to perform the unit conversion?

Physics
2 answers:
inna [3.1K]2 months ago
3 0

Here's my answer; is it correct?

Maru [3.3K]2 months ago
3 0

The correct answer is to multiply by 1000.

Since 1 g/cm³ equals 1000 kg/m³, converting 2.33 g/cm³ to kg/m³ requires multiplying by 1000.

Calculation: 2.33 × 1000 = 2330

Therefore, 2.33 g/cm³ is equivalent to 2330 kg/m³.


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In mammals, the weight of the heart is approximately 0.5% of the total body weight. Write a linear model that gives the heart we
Yuliya22 [3333]

Answer:

The typical weight of a human heart is approximately 0.93 lbs.

Explanation:

Based on this,

the heart's weight constitutes about 0.5% of total body mass.

Total human weight = 185 lbs

Let the entire body weight be represented as w and the heart's weight as w_{h}.

We aim to determine the heart's weight for a human

Using the provided information

w_{h}=0.5\times w

Where, h = heart weight

w = human weight

w_{h}=\dfrac{0.5}{100}\times 185

w_{h}=0.93\ lbs

The final weight of a human heart is 0.93 lbs.

8 0
2 months ago
An imaginary cubical surface of side L has its edges parallel to the x-, y- and z-axes, one corner at the point x = 0, y = 0, z
Sav [3153]

Refer to the attached file for the solution

7 0
2 months ago
Modern wind turbines generate electricity from wind power. The large, massive blades have a large moment of inertia and carry a
Yuliya22 [3333]

Answer:

Explanation:

a )

Each blade resembles a rod with its axis positioned near one end.

The moment of inertia for one blade is:

= 1/3 x m l²

where m stands for the mass of the blade

l represents the length of each blade.

Total moment of inertia for 3 blades is:

= 3 x\frac{1}{3}  x m l²

ml²

2 )

Details provided include:

m = 5500 kg

l = 45 m

Substituting these values produces:

moment of inertia of one blade:

= 1/3 x 5500 x 45 x 45

= 37.125 x 10⁵ kg.m²

Moment of inertia for 3 blades:

= 3 x 37.125 x 10⁵ kg.m²

= 111.375 x 10⁵ kg.m²

c )

Angular momentum

= I x ω

I denotes the moment of inertia of the turbine

ω symbolizes angular velocity

ω = 2π f

f indicates the rotational frequency of the blades

d )

We have I = 111.375 x 10⁵ kg.m² (Calculated)

f = 11 rpm (revolutions per minute)

= 11 / 60 revolutions per second

ω = 2π f

=  2π x  11 / 60 rad / s

Calculating angular momentum yields

= I x ω

111.375 x 10⁵ kg.m² x  2π x  11 / 60 rad / s

= 128.23 x 10⁵  kgm² s⁻¹.

4 0
1 month ago
A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x1 from its relaxed posit
serg [3582]

Answer:

x₂=2×1

Explanation:

According to the work-energy theorem, we can assume that the gravitational potential energy at the lowest point of compression is zero since the kinetic energy change is 0;

mgx-(kx)²/2 =0 where m refers to the object's mass, g indicates the acceleration due to gravity, k denotes spring constant, and x represents the spring's compression.

mgx=(kx)²/2

x=2mg/k----------------compression when the object is at rest

However, ΔK.E =-1/2mv²⇒kx²=mv² -----------where v symbolizes the object's velocity and K.E signifies kinetic energy

Thus, if kx²=mv² then

v=x *√(k/m) ----------------where v=0

<pDoubling v results in multiplying x *√(k/m) by 2, leading to x₂ being double x₁

7 0
2 months ago
Mosses don't spread by dispersing seeds; they disperse tiny spores. The spores are so small that they will stay aloft and move w
Keith_Richards [3271]

Solution:

Em_{f} / Em₀ = 0.30

Explanation:

In this problem, we apply the connection between mechanical energy, kinetic energy, and gravitational potential energy.

      K = ½ m v²

      U = mgh

We assess the mechanical energy at two positions:

Initial. Lower

    Em₀ = K = ½ m v²

At its highest point

    Em_{f} = U = mg and

Now let's compute

    Em₀ = ½ m 3.6²

    Em₀ = m 6.48

    Em_{f} = m 9.8 × 0.2

    Em_{f} = m 1.96

Thus the energy lost is given by:

    Em_{f} / Em₀ = m 1.96 / m 6.48

   Em_{f} / Em₀ = 0.30

This means that 30% of the sun's energy is transformed into potential energy.

There are various conversion possibilities.

This energy changes into thermal energy affecting the spores and air, since it cannot be regained.

8 0
1 month ago
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