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3 years ago

What kind of things rust

Physics

2 answers:

Wewaii [24]3 years ago

6 0

Things that rust -metals, pipes , car bodies , bolts

natali 33 [55]3 years ago

6 0

Rust is iron oxide. Anything with iron in it that's exposed to air forms iron oxide and is then said to have "rusted".

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A man whose mass is 69 kg and a woman whose mass is 52 kg sit at opposite ends of a canoe 5 m long, whose mass is 20 kg. Suppose

dusya [7]

Answer:

the canoe moved 1.2234 m in the water

Explanation:

Given that;

A man whose mass = 69 kg

A woman whose mass = 52 kg

at opposite ends of a canoe 5 m long, whose mass is 20 kg

now let;

x1 = position of the man

x2 = position of canoe

x3 = position of the woman

Now,

Centre of mass = [m1x1 + m2x2 + m3x3] / m1 + m2 + m3

= ( 69×0 ) + ( 52×5) + ( 20× 5/2) / 69 + 52 + 20

= (0 + 260 + 50 ) / ( 141 )

= 310 / 141

= 2.19858 m

Centre of mass is 2.19858 m

Now, New center of mass will be;

52 × 2.5 / ( 69 + 52 + 20 )

= 130 / 141

= 0.9219858 m { away from the man }

To get how far, the canoe moved;

⇒ 2.5 + 0.9219858 - 2.19858

= 1.2234 m

Therefore, the canoe moved 1.2234 m in the water

5 0

3 years ago

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:

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

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 energy of a photon whose frequency is 6.0 x 10^20?

omeli [17]

Answer:

3.75 MeV

Explanation:

The energy of the photon can be given in terms of frequency as:

E = h * f

Where h = Planck's constant

The frequency of the photon is 6 * 10^20 Hz.

The energy (in Joules) is:

E = 6.63 x10^(-34) * 6 * 10^(20)

E = 39.78 * 10^(-14) J = 3.978 * 10^(-13) J

We are given that:

1 eV = 1.06 * 10^(-19) Joules

This means that 1 Joule will be:

1 J = 1 / (1.06 * 10^(-19)

1 J = 9.434 * 10^(18) eV

=> 3.978 * 10^(-13) J = 3.978 * 10^(-13) * 9.434 * 10^(18) = 3.75 * 10^(6) eV

This is the same as 3.75 MeV.

The correct answer is not in the options, but the closest to it is option C.

6 0

3 years ago

We know that the magnitudes of the negative charge on the electron and the positive charge on the proton are equal. Suppose, how

TEA [102]

Answer:3,45 x 10^9 N

Explanation: We have considered the total charge for each coin , this is the total atoms x 29 electrons for cooper and multiplier by electron charge, the total charge for each coin is 0,464 C

Finally we use the Coulomb law,

F=k Q/ (r)^2

6 0

3 years ago

The vapor pressure of ethanol at 293 K is 5.95 kPa and at 336.5 K it is 53.3 kPa. Calculate the enthalpy of vaporization of etha

denis-greek [22]

Answer:

$H=41.3kJmol^{-1}$

Explanation:

The equation relating the the enthalphy, pressure and temperature is expressed as

$ln(\frac{P_{2}}{P_{1}} )=\frac{H}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}} ) \\$

Where P is the pressure, H is the enthalphy, and T is the temperature.

since the given values are

$T_{1}=293k, \\T_{2}=336.5k\\P_{1}=5.95kPA\\p_{2}=53.3kPA\\and R=8.314J.K^{-1}mol_{-1}$

if we insert values, we arrive at

$ln(\frac{53.3}{5.95} )=\frac{H}{8.314}(\frac{1}{293}-\frac{1}{336.5} )\\2.19=\frac{H}{8.314}(0.00044)\\H=(2.19*8.314)/0.00044\\H=41,268.8Jmol^{-1}\\H=41.3kJmol^{-1}$

4 0

3 years ago