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TiliK225 [7]
4 years ago
13

A 56 kg boat that is 7.6 m in length is initially 7.6 m from the pier. A 39 kg child stands at the end of the boat closest to th

e pier. The child then notices a turtle on a rock at the far end of the boat and proceeds to walk to the far end of the boat to observe the turtle. 7.6 m 7.6 m How far is the child from the pier when she reaches the far end of the boat?

Physics
1 answer:
svlad2 [7]4 years ago
6 0

Answer:

Explanation:

Given

mass of boat M=56\ kg

mass of Person m=39\ kg

Length of boat L=7.6\ m

Center of mass of boat and child

x_{cm}=\frac{M\times 0.5L+m\times L}{M+m}

x_{cm}=\frac{56\times 0.5\times 7.6+39\times 7.6}{56+39}

x_{cm}=\frac{509.2}{95}

x_{cm}=5.36---1

when child reaches another end of boat, boat is shifted by x cm

Center of mass of the system will remain the same

x'_{cm}=\frac{M\times (x+0.5L)+m\times x}{M+m}

x'_{cm}=\frac{56\times (x+0.5\times 7.6)+39\times x}{56+39}----2      

Equating 1 and 2 we get

509.2=95x+212.8

x=3.12\ m  

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the common attributes are positive and negatively charged

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2 years ago
A 500.0 kg module is attached to a 440.0 kg shuttle craft, which moves at 1050. m/s relative to the stationary main spaceship. T
nika2105 [10]

Answer:

0.955286  j

Explanation:

A 500.0 kg module is attached to a 440.0 kg shuttle craft, which moves at 1050. m/s relative to the stationary main spaceship. Then a small explosion sends the module backward with speed 100.0 m/s relative to the new speed of the shuttle craft. As measured by someone on the main spaceship, by what fraction did the kinetic energy of the module and shuttle craft, Ki, increase because of the explosion?

M=500 kg,  m=440 kg

V=1000 m/s, v = 100 m/s

Let relative speed =Vs

Momentum rule says

(M+m)V=mVs+M(Vs-v)

 940(1000)=500(Vs-100)+440Vs

 940000=500Vs-50000+440Vs

 940Vs=940000+50000

 940Vs=990000

 Vs= 990000/940=1053.19 m/s

So, the module speed = Vs-v=1053.19-100=953.19 m/s

Fractional increase in KE is given by;

Total KE after explosion / He before explosion

=500(953.19)2+ 400(1053.19)2/ 940(1000)2= 0.955286

5 0
4 years ago
Discuss Joule-Thompson effect with relevant examples and formulae.
Delicious77 [7]

Answer:

\mu _j=\dfrac{1}{C_p}\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp

Explanation:

Joule -Thompson effect

 Throttling phenomenon is called Joule -Thompson effect.We know that throttling is a process in which pressure energy will convert in to thermal energy.

Generally in throttling exit pressure is low as compare to inlet pressure but exit temperature maybe more or less or maybe remains constant depending upon flow or fluid flow through passes.

Now lets take Steady flow process  

Let

 P_1,T_1 Pressure and temperature at inlet and

 P_2,T_2 Pressure and temperature at exit

We know that Joule -Thompson coefficient given as

\mu _j=\left(\frac{\partial T}{\partial p}\right)_h

Now from T-ds equation

dh=Tds=vdp

So

Tds=C_pdt-\left [T\left(\frac{\partial v}{\partial T}\right)_p\right]dp

⇒dh=C_pdt-\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp

So Joule -Thompson coefficient

\mu _j=\dfrac{1}{C_p}\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp

This is Joule -Thompson coefficient for all gas (real or ideal gas)

We know that for Ideal gas Pv=mRT

\dfrac{\partial v}{\partial T}=\dfrac{v}{T}

So by putting the values in

\mu _j=\dfrac{1}{C_p}\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp

\mu _j=0 For ideal gas.

6 0
3 years ago
You can analogize the photoelectric energy to
Anon25 [30]

Answer:

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4 years ago
A 75.0 W bulb is connected to a
adoni [48]

Answer:

Explanation:

W = 75 watts

V = 110 volts

Formula

W = V * I

Solution

75 = 110 * I          Divide by 110

75 / 110 = I

I = 0.6818 Amperes

6 0
2 years ago
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