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3 years ago

A bin has a volume of 1.5m^3, what is its volume in ft^3?

Physics

1 answer:

il63 [147K]3 years ago

6 0

Answer:

$1.5m^3=1.5\times 3.28^3ft^3=52.931ft^3$

Explanation:

We have given volume V = $1.5m^3$

We have to convert this volume into $ft^3$

We know that 1 m = 3.28 feet

So for converting $1.5m^3$ to $ft^3$ we have to multiply $3.28^3$ to 1.5

So $1.5m^3=1.5\times 3.28^3ft^3=52.931ft^3$

So in cubic meter 1.5 will be equal to 82.923 cubic feet

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If a small rock is dropped from a height of 3.1 m how fast will it be moving when it reaches the ground 0.80 seconds later

garik1379 [7]

X=1/2 at^2
3.1=1/2 a *0.64
a=9.68
v=at
v=0.8*9.6875=7.75

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3 years ago

You go to the hardware store to buy a new 50 ft garden hose. You find you can choose between hoses of ½ inch and 5/8 inch inner

omeli [17]

To solve this problem it is necessary to consider two concepts. The first of these is the flow rate that can be defined as the volumetric quantity that a channel travels in a given time. The flow rate can also be calculated from the Area and speed, that is,

Q = V*A

Where,

A= Cross-sectional Area

V = Velocity

The second concept related to the calculation of this problem is continuity, which is defined as the proportion that exists between the input channel and the output channel. It is understood as well as the geometric section of entry and exit, defined as,

$Q_1 = Q_2$

$V_1A_1=V_2A_2$

Our values are given as,

$A_1=\frac{1}{2}^2*\pi=0.785 in^2$

$A_2=\frac{5}{8}^2*\pi=1.227 in^2$

Re-arrange the equation to find the first ratio of rates we have:

$\frac{V_1}{V_2}=\frac{A_2}{A_1}$

$\frac{V_1}{V_2}=\frac{1.227}{0.785}$

$\frac{V_1}{V_2}=1.56$

The second ratio of rates is

$\frac{V2}{V1}=\frac{A_1}{A2}$

$\frac{V2}{V1}=\frac{0.785}{1.227}$

$\frac{V2}{V1}=0.640$

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3 years ago

How would you define physics class in your own words ?

Andrei [34K]

Answer:

honestly i dont like physics class but for you im gonna write somethin' good but for me tho its B O R I N G

Explanation:

<em>Physics is the branch of science that deals with the structure of matter and how the fundamental constituents of the universe interact. It studies objects ranging from the very small using quantum mechanics to the entire universe using general relativity.</em>

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3 years ago

A steel wire of length 31.0 m and a copper wire of length 17.0 m, both with 1.00-mm diameters, are connected end to end and stre

Brut [27]

Answer:

The time taken is $t = 0.356 \ s$

Explanation:

From the question we are told that

The length of steel the wire is $l_1 = 31.0 \ m$

The length of the copper wire is $l_2 = 17.0 \ m$

The diameter of the wire is $d = 1.00 \ m = 1.0 *10^{-3} \ m$

The tension is $T = 122 \ N$

The time taken by the transverse wave to travel the length of the two wire is mathematically represented as

$t = t_s + t_c$

Where $t_s$ is the time taken to transverse the steel wire which is mathematically represented as

$t_s = l_1 * [ \sqrt{ \frac{\rho * \pi * d^2 }{ 4 * T} } ]$

here $\rho_s$ is the density of steel with a value $\rho_s = 8920 \ kg/m^3$

$t_s = 31 * [ \sqrt{ \frac{8920 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]$

$t_s = 0.235 \ s$

And

$t_c$ is the time taken to transverse the copper wire which is mathematically represented as

$t_c = l_2 * [ \sqrt{ \frac{\rho_c * \pi * d^2 }{ 4 * T} } ]$

here $\rho_c$ is the density of steel with a value $\rho_s = 7860 \ kg/m^3$

$t_c = 17 * [ \sqrt{ \frac{7860 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]$

$t_c =0.121$

$t = t_c + t_s$

$t = 0.121 + 0.235$

$t = 0.356 \ s$

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3 years ago

Do heavier objects fall more slowly than lighter objects?

vredina [299]

Depending on the surface area and weight, either can fall slower.

<u>Explanation: </u>

If you were to drop a feather and a brick off a building at the same height, the brick would reach the ground first. This is due to air resistance and how it slows the feather down compared to the brick. The surface area of the feather creates more resistance compared to its weight, therefore it hits the ground slower. It really all depends on the surface area, weight and the air resistance it creates.

If all air resistance were to be removed, then what would happen is that the two objects would fall and reach the ground at the same time. This is because the objects are in free fall, and because they are in free fall, the two objects would be falling at an acceleration of 9.81 m/s². This is true for all objects in free fall.

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3 years ago