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3 years ago

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A mass accelerates uniformly when the resultant force on it:

1 answer:

WARRIOR [948]3 years ago

3 0

F = ma
So if you want the "a" to stay constant (as it is then uniform acceleration), the F must also be constant to apply a constant force.

2 is the answer

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A horse does 910 J of work in 380 seconds while pulling a wagon. What is the power output of the horse? Round your answer to two

Alekssandra [29.7K]

2.39 Watts roughly since watts is joules per second it’s just 910j/380s

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3 years ago

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A spherical balloon has a radius of 6.55 m and is filled with helium. The density of helium is 0.179 kg/m3, and the density of a

Ksju [112]

Answer:

$M_1 = 317.7 kg$

Explanation:

Mass of the helium gas filled inside the volume of balloon is given as

$m = \rho V$

$m = 0.179(\frac{4}{3}\pi R^3)$

$m = 0.179(\frac{4}{3}\pi 6.55^3)$

$m = 210.7 kg$

now total mass of balloon + helium inside balloon is given as

$M = 210.7 + 990$

$M = 1200.7 kg$

now we know that total weight of balloon + cargo = buoyancy force on the balloon

so we will have

$(M + M_1)g = \rho_{air} V g$

$(1200.7 + M_1) = (\frac{4}{3}\pi 6.55^3) (1.29)$

$1200.7 + M_1 = 1518.4$

$M_1 = 317.7 kg$

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3 years ago

two cars are traveling in opposite directions on the thruway. car a is traveling at 55mi/h and car b is traveling at 70 mi/h the

Tresset [83]

a minute or less than a minute

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3 years ago

A person throws a ball with a force of 140 N that accelerates at 15 m/s? What is the mass of the

mr_godi [17]

Newton's 2nd law

$\tt \sum F=m.a$

input the value:

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3 years ago

An 80-cm-long steel string with a linear density of 1.0 g/m is under 200 N tension. It is plucked and vibrates at its fundamenta

icang [17]

Answer:

Wavelength of the sound wave that reaches your ear is 1.15 m

Explanation:

The speed of the wave in string is

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

where T= 200 N is tension in the string , $\mu$ =1.0 g/m is the linear mass density

$v=\sqrt{\frac{200}{1\times 10^{-3} }$

$v=447.2 m/s$

Wavelength of the wave in the string is

$\lambda =2L=2\times 0.8=1.6 m$

The frequency is

$f=\frac{v}{\lambda} \\f=\frac{447.2}{1.6}\\f=298.25 Hz$

The required wavelength pf the sound wave that reaches the ear is( take velocity of air v=344 m/s)

$\lambda=\frac{v_{air}}{f} \\\lambda=\frac{344}{298.25} \\\lambda=1.15 m$

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3 years ago