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2 years ago

If the volume of an object is reported as 5.0 ft3 what is the volume in cubic meters

1 answer:

3 0

The problem statement is simply asking us to convert units. We convert from units of ft^3 to units of m^3. To do this, we need a conversion factor. For this case, we use 1 m is equal to 3.28084 ft. We do as follows:

5.0 ft^3 ( 1 m / 3.28084 ft )^3 = 0.1416 m^3

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A rigid, uniform bar with mass mmm and length bbb rotates about the axis passing through the midpoint of the bar perpendicular t

Pie

Answer:

$I = \frac{mvb}{6}$

Explanation:

we know angular velocity in terms of moment of inertia and angular speed

$L = I$ ω .... (1)

moment of inertia of rod rotating about its center of length b

$I = \frac{ mb^2}{12}$ ........ .(2)

using v = ωr

where w is angular velocity

and r is radius of rod which is equal to b

so we get 2v = ωb

ω = 2v/b ................. (3)

here velocity is two time because two opposite ends are moving opposite with a velocity v so net velocity will be 2v

put second and third equation in ist equation

$L = \frac{mb^2}{12}$ × $\frac{2v}{b}$

so final answer will be $L = \frac{mvb}{6}$

7 0

2 years ago

Two movers use a rope system to lift a box to a third-story apartment. They do 1,200 J of work on the rope system, and the rope

xeze [42]

Efficiency is defined as the measure of the amount of work or energy is conserved in a certain process. At all times, in every process, work or energy is always lost or wasted due to certain interference. Not all work given is converted to useful work or energy. Thus , efficiency is calculated by dividing the energy or work output to the energy or work input then the value is multiplied by 100 to express efficiency as percentage.

Efficiency = work output / work input
Efficiency = (1020 J / 1200 J) = 85%

8 0

2 years ago

Read 2 more answers

An object that weighs 2.450 N is attached to an ideal massless spring and undergoes simple harmonic oscillations with a period o

Viktor [21]

Answer:

Spring constant, k = 24.1 N/m

Explanation:

Given that,

Weight of the object, W = 2.45 N

Time period of oscillation of simple harmonic motion, T = 0.64 s

To find,

Spring constant of the spring.

Solution,

In case of simple harmonic motion, the time period of oscillation is given by :

$T=2\pi\sqrt{\dfrac{m}{k}}$

m is the mass of object

$m=\dfrac{W}{g}$

$m=\dfrac{2.45}{9.8}$

m = 0.25 kg

$k=\dfrac{4\pi^2m}{T^2}$

$k=\dfrac{4\pi^2\times 0.25}{(0.64)^2}$

k = 24.09 N/m

k = 24.11 N/m

So, the spring constant of the spring is 24.1 N/m.

6 0

1 year ago

Calculate the buoyant force in air on a kilogram of titanium (whose density is about 4.5 grams per cubic centimeter). compare wi

aleksklad [387]

1) The buoyant force acting on an object immersed in a fluid is:
$B=d_f V_d g$
where $d_f$ is the density of the fluid, $V_d$ is the volume of displaced fluid, and $g=9.81~m/s^2$ is the gravitational acceleration.

2) We must calculate the volume of displaced fluid. Since the titanium object is completely immersed in the fluid (air), this volume corresponds to the volume of 1 Kg of titanium, whose density is $d=4.5~g/cm^3 = 4.5\cdot10^3~Kg/m^3$ . Using the relationship between density, volume and mass, we find
$V_d= \frac{m}{d}= \frac{1~Kg}{4.5\cdot10^3Kg/m^3}=2.22\cdot10^{-4}~m^3$

3) Now we can recall the formula written at step 1) and calculate the buoyant force. The air density is $d_f = 1~Kg/m^3$ , so we have
$B=d_f V_d g=1~Kg/m^3 \cdot 2.22\cdot10^{-4}~m^3 \cdot 9.81~m/s^2=2.22\cdot10^{-3}~N$

4) The weight of 1 Kg of titanium is instead:
$W=mg=1~Kg \cdot 9.81~m/s^2=9.81~N$
So, the buoyant force is negligible compared to the weight.

7 0

1 year ago

Para proteger un computador de sobrecargas eléctricas, Juan coloca un filamento delgado de cobre llamado fusible en su circuito,

Lynna [10]

Answer:

Los fusibles están diseñados de tal forma que estos se "rompen" o se funden, cuando la demanda eléctrica supera un dado valor (cuando demasiada electricidad pasa a través de el).

Una vez el filamento se rompe, la corriente ya no puede circular por el (podes pensar en esta situación como un cable roto, la electricidad no puede circular por este cable)

Entonces, al romperse el filamento, en caso de una sobrecarga eléctrica, el flujo de electricidad se corta, y de esta forma se protege al computador de posibles sobrecargas.

7 0

1 year ago