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xxMikexx [17]
2 years ago
12

Imagine you derive the following expression by analyzing the physics of a particular system: M= (mv2r)(mGr2). Simplify the expre

ssion for M using the techniques mentioned above.
Physics
1 answer:
alex41 [277]2 years ago
3 0

Answer:

The simplified expression is M  =  \frac{v^2 r}{G}

Explanation:

From the question we are told that  

     M  = \frac{ \frac{m v^2}{r} }{\frac{ mG}{r^2 } }

So simplifying we have

    M  =    \frac{m v^2}{r} *  \frac{r^2 }{ mG }

    M  =  \frac{v^2 r}{G}

Thus the simplified formula is M  =  \frac{v^2 r}{G}

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Object A with a mass of 500 kilograms hits stationary object B with a mass of 920 kilograms. If the collision is elastic, what h
Vika [28.1K]
In elastic collision, both the kinetic energy and momentum are conserved. Conservation means that both the kinetic energy and momentum will have the same values before and after elastic collision.

<span>As the object A has low mass than object B. Hence upon collision, object B moves forward, while object A will move backward. So option "C" is correct. </span>

5 0
2 years ago
Which of the following statements about horizons is true?
nalin [4]
<span>All soils have completely different horizon patterns.</span>
6 0
1 year ago
Read 2 more answers
The density of nuclear matter is about 1018 kg/m3. Given that 1 mL is equal in volume to 1 cm3, what is the density of nuclear m
Sonbull [250]

Answer:

density is 10^{6} Mg/µL

Explanation:

given data

density of nuclear = 10^{18} kg/m³

1 ml = 1 cm³

to find out

density of nuclear matter in Mg/µL

solution

we know here

1 Mg = 1000 kg

so

1 m³ is equal to 10^{6} cm³

and here 1 cm³ is equal to  1 mL

so we can say 1 mL is equal to 10³ µL

so by these we can convert density

density = 10^{18} kg/m³

density = 10^{18} kg/m³ × \frac{10^{-3} }{10^{6} }  Mg/µL

density =  10^{6} Mg/µL

8 0
2 years ago
Read 2 more answers
A plane wall with constant properties is initially at a uniform temperature To. Suddenly, the surface at x = L is exposed to a c
Rzqust [24]

Answer:

The distribution is as depicted in the attached figure.

Explanation:

From the given data

  • The plane wall is initially with constant properties is initially at a uniform temperature, To.
  • Suddenly the surface x=L is exposed to convection process such that T∞>To.
  • The other surface x=0 is maintained at To
  • Uniform volumetric heating q' such that the steady state temperature exceeds T∞.

Assumptions which are valid are

  1. There is only conduction in 1-D.
  2. The system bears constant properties.
  3. The volumetric heat generation is uniform

From the given data, the condition are as follows

<u>Initial Condition</u>

At t≤0

T(x,0)=T_o

This indicates that initially the temperature distribution was independent of x and is indicated as a straight line.

<u>Boundary Conditions</u>

<u>At x=0</u>

<u />T(0,t)=T_o<u />

This indicates that the temperature on the x=0 plane will be equal to To which will rise further due to the volumetric heat generation.

<u>At x=L</u>

<u />-k\frac{\partial T}{\partial x}]_{x=L}=h[T(L,t)-T_{\infty}]<u />

This indicates that at the time t, the rate of conduction and the rate of convection will be equal at x=L.

The temperature distribution along with the schematics are given in the attached figure.

Further the heat flux is inferred from the temperature distribution using the Fourier law and is also as in the attached figure.

It is important to note that as T(x,∞)>T∞ and T∞>To thus the heat on both the boundaries will flow away from the wall.

3 0
2 years ago
Consider a variety of colors of visible light (say 400 nm to 700 nm) falling onto a pair of slits.
babymother [125]

Answer:

Explanation:

The relationship between angle and wavelength for maxima and minima in Young's double slit experiment is given by

For constructive interference

d\sin \theta =m\lambda

For Destructive interference

d\sin \theta =(m+\frac{1}{2})\lambda

where \lambda =wavelength

d=slit\ width

m=order of maxima and minima

for second order maxima i.e. m=2

For smallest separation taking \lambda =400 nm, \theta =90^{\circ}

d\sin 90=2\times 400\times 10^{-9}

d=0.8\times 10^{-6}

d=0.8\mu m

   

6 0
2 years ago
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