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3 years ago

Calculate the acceleration of a 25 kg object that has move the force of 300 n

Physics

1 answer:

Dimas [21]3 years ago

5 0

acceleration=
a=F/m 300N/25kg=12m/s^2

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A 13 kg hanging sculpture is suspended by a 95-cm-long, 5.0 g steel wire. When the wind blows hard, the wire hums at its fundame

Artyom0805 [142]

Answer:

$f=81.96 \ Hz$

Explanation:

Givens

$L=95cm$

$m_{sculpture} =13kg$

$m_{wire}=5g$

The frequency is defined by

$f=\frac{v}{\lambda}$

Where $v$ is the speed of the wave in the string and $\lambda$ is its wave length.

The wave length is defined as $\lambda = 2L = 2(0.95m)=1.9m$

Now, to find the speed, we need the tension of the wire and its linear mass density

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

Where $\mu=\frac{0.005kg}{0.95m}= 5.26 \times 10^{-3}$ and the tension is defined as $T=m_{sculpture} g=13kg(9.81 m/s^{2} )=127.53N$

Replacing this value, the speed is

$v=\sqrt{\frac{127.53N}{5.26 \times 10^{-3} } }=155.71 m/s$

Then, we replace the speed and the wave length in the first equation

$f=\frac{v}{\lambda}\\f=\frac{155.71 m/s}{1.9m}\\ f=81.96Hz$

Therefore, the frequency is $f=81.96 \ Hz$

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3 years ago

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How fast must a bug swim to keep up with the waves it produces? How fast must it move to produce a bow wave?

Elden [556K]

Answer:

A bug must swim as fast as the wave speed to keep up with the waves it produces. Moreso, a boat must be moving faster than the waves it creates to produce a bow wave.

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3 years ago

If Joey ran 60 meters in 10 seconds, then his velocity is ________ m/s.

BabaBlast [244]

Answer:

6 m/s

Explanation:

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If you have a 1.0 m aqueous solution of NaCl, by how much will it increase the water’s boiling point, if KB = 0.512 °C/m? In oth

fgiga [73]

Answer: The elevation in boiling point is 1.024°C.

Explanation:

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=ik_b\times m$

where,

i = Van't Hoff factor = 2 (for NaCl)

$\Delta T_b$ = change in boiling point = ?

$k_b$ = boiling point constant = $0.512^oC/m$

m = molality = 1.0 m

Putting values in above equation, we get:

$\Delta Tb=2\times 0.512^oC/m\times 1.0m\\\\\Delta Tb=1.024^oC$

Hence, the elevation in boiling point is 1.024°C.

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3 years ago

What kind of weather is signaled by cumulus clouds

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