1 g = 1 ÷ 1000 kg
= 0.001 kg

1 cm³ = 1 ÷ 100 ÷ 100 ÷ 100 m³
= 0.000001 m³

1 g/cm³ = 1 g / 1 cm³
= 0.001 kg / 0.000001 m³
= 1000 kg/m³

The density is 1000 kg/m³.

3 0

3 years ago

[O.04H]The table below shows the use of some energy production methods over time.

mr Goodwill [35]

I think The coastal areas were highly polluted

8 0

3 years ago

Read 2 more answers

As an intern with an engineering firm, you are asked to measure the moment of inertia of a large wheel, for rotation about an ax

AysviL [449]

Answer:

$I=2.766\ kg.m^2$

Explanation:

We have:

diameter of the wheel, $d=0.88\ m$

weight of the wheel, $w_w=280\ N$

mass of hanging object to the wheel, $m_o=6.32\ kg$

speed of the hanging mass after the descend, $v_o=4\ m.s^{-1}$

height of descend, $h=2.5\ m$

(a)

moment of inertia of wheel about its central axis:

$I=\frac{1}{2} m.r^2$

$I=\frac{1}{2} \frac{w_w}{g}.r^2$

$I=\frac{1}{2} \times \frac{280}{9.8}\times 0.44^2$

$I=2.766\ kg.m^2$

3 0

4 years ago

Multiple-Concept Example 7 and Interactive LearningWare 26.1 provide some helpful background for this problem. The drawing shows

Vlada [557]

Answer:

n = 1.4266

Explanation:

Given that:

refractive index of crystalline slab n = 1.665

let refractive index of fluid is n.

angle of incidence θ₁ = 37.0°

Critical angle $\theta _c = sin^{-1} (\frac{n}{n_{slab}} )$

$sin \theta _ c =\frac{n}{n_{slab}}$

According to Snell's law of refraction:

$n sin \theta _1 = n_{slab} \ sin \ (90- \theta_c)$

At point P ; $90 - \theta _2 \leq \theta _c$

$\theta _2 = 90 - \theta _c$

Therefore:

$n \ sin \theta_1 = n_{slab} \sqrt{(1-sin^2 \theta _c)} \\ \\ n \ sin \theta_1 = n_{slab} \sqrt{(1- \frac{n}{n_{slab}} )}$

Then maximum value of refractive index n of the fluid is:

$n = \frac{n_{slab}}{\sqrt{1+ sin^2 \theta _1 } }$

$n = \frac{1.665}{\sqrt{1+ sin^2 \ 37} }$

n = 1.4266

3 0

3 years ago