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aleksandrvk [35]
3 years ago
10

During a circus act, an elderly performer thrills the crowd by catching a cannon ball shot at him. The cannon ball has a mass of

29.0 kg and its horizontal component of velocity is 8.00 m/s when the 65.0 kg performer catches it. If the performer is on nearly frictionless roller skates, what is his recoil velocity in meters per second?
Physics
1 answer:
Phantasy [73]3 years ago
3 0

Answer:

2.4681 m/s

Explanation:

m_1 = Mass of cannon ball = 29 kg

m_2 = Mass of performer = 65 kg

u_1 = Initial Velocity of cannon ball = 8 m/s

u_2 = Initial Velocity of performer = 0 m/s

v = Velocity of combined mass

As the linear momentum of the system conserved

m_1u_1 + m_2u_2 =(m_1 + m_2)v\\\Rightarrow v=\frac{m_1u_1 + m_2u_2}{m_1 + m_2}\\\Rightarrow v=\frac{29\times 8 + 65\times 0}{29 + 65}\\\Rightarrow v=2.4681\ m/s

The recoil velocity is 2.4681 m/s

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Here in this case we can use work energy theorem

As per work energy theorem

Work done by all forces = Change in kinetic Energy of the object

Total kinetic energy of the solid sphere is ZERO initially as it is given at rest.

Final total kinetic energy is sum of rotational kinetic energy and translational kinetic energy

KE = \frac{1}{2}Iw^2 +\frac{1}{2} mv^2

also we know that

I = \frac{2}{5}mR^2

w= \frac{v}{R}

Now kinetic energy is given by

KE = \frac{1}{2}(\frac{2}{5}mR^2)w^2 +\frac{1}{2} m(Rw)^2

KE = \frac{1}{5}mR^2w^2 +\frac{1}{2} mR^2w^2

KE = \frac{7}{10}mR^2w^2

KE = \frac{7}{10}*150*(0.200)^2(50)^2

KE = 10500 J

Now by work energy theorem

Work done = 10500 - 0 = 10500 J

So in the above case work done on sphere is 10500 J

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3 years ago
Most of the Earth's volcanoes occur
timofeeve [1]

Answer:

A.

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NEAR THE CENTER OF TECTONIC PLATES.

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3 years ago
Consider a container of oxygen gas at a temperature of 23°C that is 1.00 m tall. Compare the gravitational potential energy of a
Sergio039 [100]

Answer:

Yes, it is reasonable to neglect it.

Explanation:

Hello,

In this case, a single molecule of oxygen weights 32 g (diatomic oxygen) thus, the mass of kilograms is (consider Avogadro's number):

m=1molec*\frac{1mol}{6.022x10^{23}molec} *\frac{32g}{1mol}*\frac{1kg}{1000g}=5.31x10^{-26}kg

After that, we compute the potential energy 1.00 m above the reference point:

U=mhg=5.31x10^{-26}kg*1.00m*9.8\frac{m}{s^2}=5.2x10^{-25}J

Then, we compute the average kinetic energy at the specified temperature:

K=\frac{3}{2}\frac{R}{Na}T

Whereas N_A stands for the Avogadro's number for which we have:

K=\frac{3}{2} \frac{8.314\frac{J}{mol*K}}{6.022x10^{23}/mol}*(23+273)K\\ \\K=6.13x10^{-21}J

In such a way, since the average kinetic energy energy is about 12000 times higher than the potential energy, it turns out reasonable to neglect the potential energy.

Regards.

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3 years ago
What is a loose icy body with a long narrow orbit?
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Answer:

Explanation:

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