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4 years ago

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Calculate the amount of power in Watts) required to move an object weighing 525 N from point A to point B within 36 seconds. Dis

tance between point A and point B is 3.5 meters and they are at the same level.

1 answer:

MAVERICK [17]4 years ago

4 0

Answer:

Power required, P = 0 watts

Explanation:

It is given that,

Weight of the object, W = 525 N

Time taken to move from point A to B is, t = 36 s

Distance between A and B is 3.5 m.

Power required to move an object is given by work done divided by time taken.

$P=\dfrac{W}{t}$

Work done, $W=Fd\ cos\theta$

The weight of the object is acting in downward direction and object is moving in forward direction such that $\theta=90$

$W=Fd\ cos(90)$

Since, cos(90) = 0

This gives, work done = 0

As a result, the power required to move that object is 0 watts. Hence, this is the required solution.

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A. How many atoms of helium gas fill a spherical balloon of diameter 29.6 cm at 19.0°C and 1.00 atm? b. What is the average kine

Korolek [52]

Answer:

a) 3.39 × 10²³ atoms

b) 6.04 × 10⁻²¹ J

c) 1349.35 m/s

Explanation:

Given:

Diameter of the balloon, d = 29.6 cm = 0.296 m

Temperature, T = 19.0° C = 19 + 273 = 292 K

Pressure, P = 1.00 atm = 1.013 × 10⁵ Pa

Volume of the balloon = $\frac{4}{3}\pi(\frac{d}{2})^3$

or

Volume of the balloon = $\frac{4}{3}\pi(\frac{0.296}{2})^3$

or

Volume of the balloon, V = 0.0135 m³

Now,

From the relation,

PV = nRT

where,

n is the number of moles

R is the ideal gas constant = 8.314 kg⋅m²/s²⋅K⋅mol

on substituting the respective values, we get

1.013 × 10⁵ × 0.0135 = n × 8.314 × 292

or

n = 0.563

1 mol = 6.022 × 10²³ atoms

Thus,

0.563 moles will have = 0.563 × 6.022 × 10²³ atoms = 3.39 × 10²³ atoms

b) Average kinetic energy = $\frac{3}{2}\times K_BT$

where,

Boltzmann constant, $K_B=1.3807\times10^{-23}J/K$

Average kinetic energy = $\frac{3}{2}\times1.3807\times10^{-23}\times292$

or

Average kinetic energy = 6.04 × 10⁻²¹ J

c) rms speed = $\frac{3RT}{m}$

where, m is the molar mass of the Helium = 0.004 Kg

or

rms speed = $\frac{3\times8.314\times292}{0.004}$

or

rms speed = 1349.35 m/s

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3 years ago

Ramon and sally are observing a toy car speed up as it goes around a circular track. ramon says, "the car's speeding up, so ther

Anestetic [448]

The changing speed demonstrates the nearness of an increasing speed; this centripetal quickening is of enduring with significance and coordinated in any regard cases toward the hub of the pivot. This increasing speed is, thus, created by utilizing a centripetal anxiety that is furthermore reliable in significance and coordinated toward the hub of revolution. contained in regards to revolution around a settled hub of a firm build that is not irrelevantly little in contrast with the span of the course, each and every molecule of the constitution portrays a uniform around activity with the equivalent rakish speed, yet with speed and increasing speed different with the range with comprehend to the hub.

4 0

4 years ago

Need an answer in Meters

chubhunter [2.5K]

Answer:

• let's first get time

${ \rm{s = ut + \frac{1}{2}g {t}^{2} }} \\ \\ { \rm{78.4 = 30t + 5 {t}^{2} }} \\ \\ { \rm{ {t}^{2} + 6t - 15.68 = 0}} \\ \\ { \rm{time \approx \: 2 \: seconds}}$

• let's find horizontal displacement

${ \rm{s = ut + \frac{1}{2}g {t}^{2} }} \\ \\ { \rm{s = (30 \times 2) \times ( \frac{1}{2} \times 0 \times {2}^{2}) }} \\ \\ { \rm{displacement = 60 \: meters}}$

g is acceleration due to gravity [ vertically: g is 10 m/s². horizontally: g is 0 ]

6 0

3 years ago

A fan at a rock concert is 50.0 m from the stage, and at this point the sound intensity level is 114 dB. Sound is detected when

Marianna [84]

Answer:

A) $P=13.92\ J.s^{-1}$

B) $v=3730.9912\ m.s^{-1}$

C) $v=74.44\ mm.s^{-1}$

D) mosquitoes speed in part B is very much larger than that of part C.

Explanation:

Given:

Distance form the sound source, $s=50\ m$

sound intensity level at the given location, $\beta=114\ dB$

diameter of the eardrum membrane in humans, $d=8.4 \times 10^{-3}\ m$

We have the minimum detectable intensity to the human ears, $I_0=10^{-12}\ W.m^{-2}$

(A)

<u>Now the intensity of the sound at the given location is related mathematically as:</u>

$\beta=10\ log(\frac{I}{I_0} )$ ..........................................(1)

$114=10\ log\ (\frac{I}{10^{-12}} )$

$11.4=log\ I+12\ log\ 10$

$I=0.2512\ W.m^{-2}$

<em>As we know :</em>

$I=\frac{P}{A}$

$0.2512=\frac{P}{\pi\times \frac{8.4^2}{4} }$

$P=13.92\ J.s^{-1}$ is the energy transferred to the eardrums per second.

(B)

mass of mosquito, $m=2\times 10^{-6}\ kg$

<u>Now the velocity of mosquito for the same kinetic energy:</u>

$KE=\frac{1}{2} m.v^2$

$13.92=\frac{1}{2}\times 2\times 10^{-6}\times v^2$

$v=3730.9912\ m.s^{-1}$

(C)

Given:

Sound intensity, $\beta = 20\ dB$

<u>Using eq. (1)</u>

$20=10\ log\ (\frac{I}{10^{-12}} )$

$2=log\ I+12\ log\ 10$

$I=10^{-10}\ W.m^{-2}$

Now, power:

$P=I.A$

$P=10^{-10}\times \pi\times \frac{8.4^2}{4}$

$P=5.54\times 10^{-9}\ J.s^{-1}$

Hence:

$KE=\frac{1}{2} m.v^2$

$5.54\times 10^{-9}=0.5\times 2\times 10^{-6}\times v^2$

$v=0.07444\ m.s^{-1}$

$v=74.44\ mm.s^{-1}$

(D)

mosquitoes speed in part B is very much larger than that of part C.

7 0

3 years ago

An object of mass 12kg is held at a weight of 5m above the ground for 30s. Calculate the workdone during this period.

daser333 [38]

Answer:

0 J

Explanation:

Work = force × distance

Since the object isn't moved, the distance is 0. So the work is 0.

5 0

4 years ago