This is what occurs when matter transitions between solid, liquid and gas.

What is the force the when does when Gravity pushes you

BARSIC [14]

Answer:

Image result for what is the force the when does when Gravity pushes you

The important thing to remember is that gravity is neither a push nor a pull; what we interpret as a “force” or the acceleration due to gravity is actually the curvature of space and time — the path itself stoops downward.

Explanation:

Image result for what is the force the when does when Gravity pushes you

The important thing to remember is that gravity is neither a push nor a pull; what we interpret as a “force” or the acceleration due to gravity is actually the curvature of space and time — the path itself stoops downward.

6 0

3 years ago

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It is claimed that if a lead bullet goes fast enough, it can melt completely when it comes to a halt suddenly, and all its kinet

TEA [102]

Answer:

$v=346.05\ m.s^{-1}$

Explanation:

Given:

initial temperature of the lead bullet, $T_i=43^{\circ}C$

latent heat of fusion of lead, $L_f=2.32\times 10^4\ J.kg^{-1}$

melting point of lead, $T_m=327.3^{\circ}C$

We have:

specific heat capacity of lead, $c=129\ J.kg^{-1}.K^{-1}$

<em>According to question the whole kinetic energy gets converted into heat which establishes the relation:</em>

$\rm KE=(heat\ of\ rising\ the\ temperature\ from\ 43\ to\ 327.3\ degree\ C)+(heat\ of\ melting)$

$\frac{1}{2} m.v^2=m.c.\Delta T+m.L_f$

$\frac{1}{2} m.v^2=m(c.\Delta T+L_f)$

$\frac{v^2}{2} =129\times(327.3-43)+23200$

$v=346.05\ m.s^{-1}$

3 0

3 years ago

A rod has a radius of 10 mm is subjected to an axial load of 15 N such that the axial strain in the rod is ????௫ = 2.75*10-6, de

EleoNora [17]

Answer:

Knowing we only have one load applied in just one direction we have to use the Hooke's law for one dimension

ex = бx/E

бx = Fx/A = Fx/π $r^{2}$

Using both equation and solving for the modulus of elasticity E

E = бx/ex = Fx / π $r^{2}$ ex

E = $\frac{15}{pi (10 * 10^{-3})^{2} * 2.75 * 10^{-6} } = 17.368 * 10^{9} Pa = 17.4 GPa$

Apply the Hooke's law for either y or z direction (circle will change in every direction) we can find the change in radius

ey = $\frac{1}{E}$ (бy - v (бx + бz)) = $-\frac{v}{E}$ бx

= $\frac{vFx}{Epir^{2} }$ = $\frac{0.23 * 15}{pi (10 * 10^{-3)^{2} } * 17.362 * 10^{9} } = -0.63 *10^{-6}$

Finally

ey = Δr / r

Δr = ey * r = 10 * $-0.63* 10^{-6} mm = -6.3 * 10^{-6} mm$

Δd = 2Δr = $-12.6 * 10^{-6} mm$

Explanation:

5 0

3 years ago

In a bi-prism experiment the eye-piece was placed at a distance 1.5m from the source. The distance between the virtual sources w

Vladimir [108]

Answer:

λ = 1.4 × 10^(-7) m

Explanation:

We are given;

distance of eye piece from the source;D = 1.5 m

distance between the virtual sources;d = 7.5 × 10^(-4) m

To find the wavelength, we will use the formula for fringe width;

X = λD/d

Where X is fringe width, λ is wavelength, while d and D remain as before.

Now, fringe width = eye-piece distance moved transversely/number of fringes

Eye piece distance moved transversely = 1.88 cm = 1.88 × 10^(-2) m

Thus,

Fringe width = (1.88 × 10^(-2))/10 = 1.88 × 10^(-3) m

Thus;

1.88 × 10^(-3) = λ(1.5)/(7.5 × 10^(-4))

λ = [1.88 × 10^(-3) × (7.5 × 10^(-4))]/1.5

λ = 1.4 × 10^(-7) m

6 0

3 years ago

Kawas 200 osebanywhes he remembered he had to return some books they

tia_tia [17]

I think you need to add more.. but I may know where you are leading

Was he 200 m away and made the trip in 200 seconds?

If yes...
2 m/s was his speed and 0 velocity

6 0

3 years ago