Kinetic energy portfolio in part 2 the independent changes to----?

A cubical block of iron 10 cm on each side is floating on mercury in a vessel. (i) What is the height of the block above the mer

Gre4nikov [31]

Answer:

i 5.3 cm ii. 72 cm

Explanation:

i

We know upthrust on iron = weight of mercury displaced

To balance, the weight of iron = weight of mercury displaced . So

ρ₁V₁g = ρ₂V₂g

ρ₁V₁ = ρ₂V₂ where ρ₁ = density of iron = 7.2 g/cm³ and V₁ = volume of iron = 10³ cm³ and ρ₂ = density of mercury = 13.6 g/cm³ and V₂ = volume of mercury displaced = ?

V₂ = ρ₁V₁/ρ₂ = 7.2 g/cm³ × 10³ cm³/13.6 g/cm³ = 529.4 cm³

So, the height of iron above the mercury is h = V₂/area of base iron block

= 529.4 cm³/10² cm² = 5.294 cm ≅ 5.3 cm

ρ₁V₁g = ρ₂V₂g

ii

ρ₁V₁ = ρ₃V₃ where ρ₁ = density of iron = 7.2 g/cm³ and V₁ = volume of iron = 10³ cm³ and ρ₃ = density of water = 1 g/cm³ and V₃ = volume of water displaced = ?

V₃ = ρ₁V₁/ρ₃ = 7.2 g/cm³ × 10³ cm³/1 g/cm³ = 7200 cm³

So, the height of column of water is h = V₃/area of base iron block

= 7200 cm³/10² cm² = 72 cm

7 0

3 years ago

What is the common name for tropospheric ozone

Galina-37 [17]

Ozone in troposphere is also know as Bad Ozone, Evil Ozone and Ground Level Ozone.

5 0

3 years ago

An ac series circuit has an impedance of 60 Ohm and

Iteru [2.4K]

Answer:

Power factor of the AC series circuit is $cos\phi=0.5$

Explanation:

It is given that,

Impedance of the AC series circuit, Z = 60 ohms

Resistance of the AC series circuit, R = 30 ohms

We need to find the power factor of the circuit. It is given by :

$cos\phi=\dfrac{R}{Z}$

$cos\phi=\dfrac{30}{60}$

$cos\phi=\dfrac{1}{2}$

$cos\phi=0.5$

So, the power factor of the ac series circuit is $cos\phi=0.5$ . Hence, this is the required solution.

6 0

3 years ago

What do the mitochondria do??

Ksju [112]

Mitochondria breaks down sugar to release energy to the cell.

4 0

4 years ago

For a certain optical medium the speed of light varies from a low value of 1.90 × 10 8 m/s for violet light to a high value of 2

Dmitry_Shevchenko [17]

Answer:

a. The refractive index ranges from 1.5 - 1.56

b. 18.7° for violet light and 19.5° for red light.

c. 33.7° for violet light and 35.3° for red light.

Explanation:

a. The refractive index of an object is the ratio of the speed of light in a vacuum and the speed of light in the object.

Mathematically,

$n = \frac{c}{v}$

The speed of violet light in the object is $1.9 * 10^8 m/s$ .

The speed of red light in the object is $2 * 10^8 m/s$

Hence, the refractive index for violet light is:

$n = \frac{3 * 10^8 }{1.9 * 10^8} \\\\n = 1.56$

and for red light, it is:

$n = \frac{3 * 10^8 }{2 * 10^8} \\\\n = 1.5$

Hence, the refractive index ranges from 1.5 - 1.56.

b. The refractive index is also the ratio of the sine of the angle of incidence to the sine of the angle of refraction.

$n = \frac{sin(i)}{sin(r)}$

The angle of incidence is 30°.

The angle of refraction for violet light will be:

$1.56 = \frac{sin(30)}{sin(r)}\\ \\sin(r) = \frac{sin(30)}{1.56} = \frac{0.5}{1.56} \\\\sin(r) = 0.3205\\\\r = 18.7^o$

And the angle of refraction for red light will be:

$1.5 = \frac{sin(30)}{sin(r)}\\ \\sin(r) = \frac{sin(30)}{1.5} = \frac{0.5}{1.5} \\\\sin(r) = 0.3333\\\\r = 19.5^o$

The angle of refraction for red light is larger than that of violet light when the angle of incidence is 30°.

c. The angle of incidence is 60°.

The angle of refraction for violet light will be:

$1.56 = \frac{sin(60)}{sin(r)}\\ \\sin(r) = \frac{sin(60)}{1.56} = \frac{0.8660}{1.56} \\\\sin(r) = 0.5551\\\\r = 33.7^o$

And the angle of refraction for red light will be:

$1.5 = \frac{sin(60)}{sin(r)}\\ \\sin(r) = \frac{sin(60)}{1.5} = \frac{0.8660}{1.5} \\\\sin(r) = 0.5773\\\\r = 35.3^o$

The angle of refraction for red light is still larger than that of violet light when the angle of incidence is 60°.

6 0

3 years ago