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3 years ago

IF a rock has a mass of 92,224 kg, what is it's weight?

Physics

1 answer:

KATRIN_1 [288]3 years ago

5 0

Answer:

904 717.44 N

Explanation:

The mass of an object is a fundamental property of the object.

The weight of an object is the force of gravity on the object.

This is defined as follows

Weight (W)= mass(m) × weight of a unit mass(g)

( gravitational field intensity = g)

assume g =9.81 m/s2

weight = 92 224 * 9.81

= 904 717.44 N

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Read 2 more answers

Alcohol of mass 33.2g and density 0.79kg/m³ or 790kg/m³ is mixed with water of 9g. What is the density of the resulting mixture?

KATRIN_1 [288]

Answer:

0.83 g/cm³

Explanation:

The volume of the alcohol is ...

(33.2 g)/(0.79 g/cm³) ≈ 42.0253 cm³

The density of water is about 1 g/cm³, so the volume of 9 g of it is ...

(9 g)/(1 g/cm³) = 9 cm³

The total volume is ...

42.0253 cm³ +9 cm³ = 51.0253 cm³

The total mass is ...

33.2 g + 9 g = 42.4 g

So, the resulting density is ...

(42.4 g)/(51.0253 cm³) ≈ 0.83 g/cm³

The resulting mixture has a density of about 0.83 g/cm³.

_____

<em>Additional comment</em>

Alcohol dissolves in water, so the total volume will be slightly less than 51.0253 cm³. The attached curve shows the result of mixing ethanol and water.

The weight of a mole of ethanol is about 46 g, of water, about 18.02 g. Then the mole fraction of alcohol is ...

(33.2/46)/(33.2/46 +9/18.02) ≈ 0.59

The volume of the mix is then estimated to be (-1.05 cc/mol)(1.221 mol), or about 1.28 cm³ less than the volume indicated above. That brings the density up to about 0.85 g/cm³.

We're not completely sure of the relevance of this calculation, since many of the applicable parameters are not specified. The point is that <em>the density of the mix will probably be slightly more than the value calculated above</em>. YMMV

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3 years ago

An object of mass m = 5.0 kg hangs from a cord around a light pulley: The length of the cord between the oscillator and the pull

puteri [66]

Answer:

$\mu=0.0049Kg/m$

Explanation:

When a standing wave is formed with six loops means the normal mode of the wave is n=6, the frequency of the normal mode is given by the expression:

$f_n=\frac{nv}{2L}$

Where $L$ is the length of the string and $v$ the velocity of propagation. Use this expression to find the value of $v$ .

$f_6=\frac{6v}{2L}\\(150)=\frac{6v}{2(2)} \\150=\frac{3v}{2} \\3v=150(2)\\ v=\frac{300}{3} \\v=100m/s$

The velocity of propagation is given by the expression:

$v=\sqrt{\frac{T}{\mu }$

Where $\mu$ is the desirable variable of the problem, the linear mass density, and $T$ is the tension of the cord. The tension is equal to the weight of the mass hanging from the cord: