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3 years ago

Help me with this problem please

Physics

1 answer:

zaharov [31]3 years ago

5 0

Answer:

Total moment of inertia when arms are extended: 1.613 $kg\,m^2$

Explanation:

This second part of the problem could be a pretty complex one, but if they expect you to do a simple calculation, which is what I imagine, the idea is just adding another moment of inertia to the first one due to the arms extended laterally and use the moment of inertia for such as depicted in the image I am attaching.

In that image:

L is the length from one end to the other of the extended arms (each 0.75m from the center of the body) which gives 1.5 meters.

m is the mass of both arms. That is: twice 5% of the mass of the person: which mathematically can be written as: 2 * 0.05 * 56.5 = 5.65 kg

Therefore this moment of inertia to be added can be obtained using the formula shown in the image:

$I_z=\frac{1}{12} \,m\,L^2\\\\I_z=\frac{5.65\,*\,1.5^2}{12} \\I_z=1.05937\,kg\,m^2$

Now, one needs to add this to the previous moment that you calculated, resulting in:

0.554 + 1.059 = 1.613 $kg\,m^2$

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Alecsey [184]

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5 0

3 years ago

A(n) 7.7-kg object is sliding across the ice at 2.34 m/s in the positive x direction. An internal explosion occurs, splitting th

-BARSIC- [3]

Answer:

Average acceleration on first part of the chunk is given as

$a_1 = 13.125 m/s^2$

Average acceleration on second part of the chunk is given as

$a_2 = -13.125 m/s^2$

Explanation:

By momentum conservation along x direction we will have

$mv_i = \frac{m}{2}v_1 + \frac{m}{2}v_2$

so we have

$v_1 + v_2 = 2v$

$v_1 + v_2 = 4.68$

also by energy conservation

$\frac{1}{2}(\frac{m}{2})v_1^2 + \frac{1}{2}(\frac{m}{2})v_2^2 - \frac{1}{2}mv^2 = 17 J$

$\frac{1}{4}m(v_1^2 + v_2^2) - \frac{1}{2}mv^2 = 17$

$(v_1^2 + v_2^2) - 2v^2 = \frac{4}{7.7}(17)$

$(4.68 - v_2)^2 + v_2^2 - 2v^2 = 8.83$

$21.9 + 2v_2^2 - 9.36 v_2 - 10.95 = 8.83$

$2v_2^2 - 9.36v_2 + 2.12 = 0$

by solving above equation we will have

$v_1 = 4.44 m/s$

$v_2 = 0.24 m/s$

Average acceleration on first part of the chunk is given as

$a_1 = \frac{4.44 - 2.34}{0.16}$

$a_1 = 13.125 m/s^2$

Average acceleration on second part of the chunk is given as

$a_2 = \frac{0.24 - 2.34}{0.16}$

$a_2 = -13.125 m/s^2$

5 0

4 years ago

Dina has a mass of 50 kilograms and is waiting at the top of a ski slope that’s 5.0 meters high.

Over [174]

All of Dina's potential energy Ep is converted into kinetic energy Ek so Ep=Ek, where Ep=m*g*h and Ek=(1/2)*m*v². m is the mass of Dina, h is the height of ski slope, g=9.8 m/s² and v is the maximal velocity.

So we solve for v:

m*g*h=(1/2)*m*v², masses cancel out,

g*h=(1/2)*v², we multiply by 2,

2*g*h=v² and take the square root to get v

√(2*g*h)=v, we plug in the numbers and get:

v=9.9 m/s.

So Dina's maximum velocity on the bottom of the ski slope is v=9.9 m/s.

8 0

3 years ago

Read 2 more answers

How much current is in a circult that includes a 9.0-volt battery and a bulb with a resistance of 4.0 ohms?

SVETLANKA909090 [29]

Answer:

Current, I = 2.3 A

Explanation:

We have,

Voltage of the battery in a circuit is 9 volts

Resistance of the circuit is 4 ohms

It is required to find the current in a circuit. When the voltage and the resistance of the circuit is given then we can find the current in it is given by Ohm's law as :

$V=IR$

I is electric current

$I=\dfrac{V}{R}\\\\I=\dfrac{9}{4}\\\\I=2.25\ A$

I = 2.3 A

So, the current in the circuit is 2.3 A.

6 0

3 years ago

A deuterium atom is a hydrogen atom with a neutron added to its nucleus. Approximate the binding energy of this nucleus, given t

Mademuasel [1]

Answer:

c. 2 MeV.

Explanation:

The computation of the binding energy is shown below

$= [Zm_p + (A - Z)m_n - N]c^2\\\\=[(1) (1.007825u) + (2 - 1 ) ( 1.008665 u) - 2.014102 u]c^2\\\\= (0.002388u)c^2\\\\= (.002388) (931.5 MeV)\\\\=2.22 MeV$

= 2 MeV

As 1 MeV = (1 u) c^2

hence, the binding energy is 2 MeV

Therefore the correct option is c.

We simply applied the above formula so that the correct binding energy could come

And, the same is to be considered

8 0

3 years ago