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Elena-2011 [213]
3 years ago
7

How much does a 1 pound rock weigh

Physics
1 answer:
Paraphin [41]3 years ago
5 0

1 pound

0.453592 KG

16 ounces

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A torque of 50 N*m acts on a wheel of moment of inertia 25 kg*m^2 for 4 s and then is removed.
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You have to find the calculate<span> the circumference first then you can just multiply the diameter by π, which is about 3.142. That gives you the distance for each </span>revolution<span>. Then you can multiply by the </span>number of revolutions<span> per minute.

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3 years ago
Forces are never isolated because they come in... *
Fudgin [204]

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different shapes

Explanation:

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3 years ago
Why is the same side of the Moon always visible from Earth?
son4ous [18]

Answer:

Explanation:

Answer

The true fact is that C is what happens in outer space.  Both rotations take 27.3 days.  

A: The exact opposite is true. It does rotate about it's axis.

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8 0
2 years ago
A thick steel sheet of area 100 in.2 is exposed to air near the ocean. After a one-year period it was found to experience a weig
vladimir1956 [14]

Answer:

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

Explanation:

The corrosion rate is the rate of material remove.The formula for calculating CPR or corrosion penetration rate is

CPR=\frac{KW}{DAT}

K= constant depends on the system of units used.

W= weight =485 g

D= density =7.9 g/cm³

A = exposed specimen area =100 in² =6.452 cm²

K=534 to give CPR in mpy

K=87.6  to give CPR in mm/yr

mpy

CPR=\frac{KW}{DAT}

        =\frac{534\times( 485g)\times( 10^3mg/g)}{(7.9g/cm^3) \times (100in^2)\times (24h/day)\times (365day/yr)\times 1yr}

        =37.4mpy

mm/yr

CPR=\frac{KW}{DAT}

        =\frac{87.6\times (485g)\times (10^3 mg/g)}{(7.9g/cm^3)\times (100in^2)\times(2.54cm/in)^2\times (24h/day)\times (365day/yr)\times 1yr}

       =0.952 mm/yr

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

3 0
4 years ago
Help please, I need it
love history [14]

Answer:

1.97×10⁻²¹ J

Explanation:

Use ideal gas law to find temperature.

PV = nRT

(9 atm) (9 L) = (83.3 mol) (0.0821 L·atm/mol/K) T

T = 11.9 K

The average kinetic energy per atom is:

KE = 3/2 kT

KE = 3/2 (1.38×10⁻²³ J/K) (11.9 K)

KE = 2.46×10⁻²² J

For a mass of 5.34×10⁻²⁶ kg, the kinetic energy is:

KE = (5.34×10⁻²⁶ kg) (1 mol / 0.004 kg) (6.02×10²³ atom/mol) (2.46×10⁻²² J)

KE = 1.97×10⁻²¹ J

5 0
4 years ago
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