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3 years ago

Proof of conservation of energy

1 answer:

6 0

If you compare the energy at two different times in an equation, you'll see no difference, because it's been conserved. The total energy of a system is the sum of its energy in motion - its kinetic energy - and its energy due to its position, which is its potential energy

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Satuarated solution changes into unsatuarated by heating give reason

Crazy boy [7]

Answer:

Due to application of heat on a saturated solution, the interparticle space increases due to increase in the kinetic energy of the particles which allows more solutes to dissolve in the solution thereby making it unsaturated.

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3 years ago

The basic unit of time is second - is it true or false ??

Lorico [155]

It is true. Time is always measured in seconds.

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3 years ago

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In the chemical formula 3NH4, how many total atoms of nitrogen (N

Rus_ich [418]

One atom on nitrogen because there isn’t a number directly after it’s the 3 signifies that there is three groups of those atoms as a total

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3 years ago

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A river 500 ft wide flows with a speed of 8 ft/s with respect to the earth. A woman swims with a speed of 4 ft/s with respect to

White raven [17]

Answer:

1) $\Delta s=1000\ ft$

2) $\Delta s'=998.11\ ft.s^{-1}$

3) $t\approx125\ s$

$t'\approx463.733\ s$

Explanation:

Given:

width of river, $w=500\ ft$

speed of stream with respect to the ground, $v_s=8\ ft.s^{-1}$

speed of the swimmer with respect to water, $v=4\ ft.s^{-1}$

Now the resultant of the two velocities perpendicular to each other:

$v_r=\sqrt{v^2+v_s^2}$

$v_r=\sqrt{4^2+8^2}$

$v_r=8.9442\ ft.s^{-1}$

Now the angle of the resultant velocity form the vertical:

$\tan\beta=\frac{v_s}{v}$

$\tan\beta=\frac{8}{4}$

$\beta=63.43^{\circ}$

Now the distance swam by the swimmer in this direction be d.

so,

$d.\cos\beta=w$

$d\times \cos\ 63.43=500$

$d=1118.034\ ft$

Now the distance swept downward:

$\Delta s=\sqrt{d^2-w^2}$

$\Delta s=\sqrt{1118.034^2-500^2}$

$\Delta s=1000\ ft$

On swimming 37° upstream:

The velocity component of stream cancelled by the swimmer:

$v'=v.\cos37$

$v'=4\times \cos37$

$v'=3.1945\ ft.s^{-1}$

Now the net effective speed of stream sweeping the swimmer:

$v_n=v_s-v'$

$v_n=8-3.1945$

$v_n=4.8055\ ft.s^{-1}$

The component of swimmer's velocity heading directly towards the opposite bank:

$v'_r=v.\sin37$

$v'_r=4\sin37$

$v'_r=2.4073\ ft.s^{-1}$

Now the angle of the resultant velocity of the swimmer from the normal to the stream:

$\tan\phi=\frac{v_n}{v'_r}$

$\tan\phi=\frac{4.8055}{2.4073}$

$\phi=63.39^{\circ}$

Now let the distance swam in this direction be d'.

$d'\times \cos\phi=w$

$d'=\frac{500}{\cos63.39}$

$d'=1116.344\ ft$

Now the distance swept downstream:

$\Delta s'=\sqrt{d'^2-w^2}$

$\Delta s'=\sqrt{1116.344^2-500^2}$

$\Delta s'=998.11\ ft.s^{-1}$

Time taken in crossing the rive in case 1:

$t=\frac{d}{v_r}$

$t=\frac{1118.034}{8.9442}$

$t\approx125\ s$

Time taken in crossing the rive in case 2:

$t'=\frac{d'}{v'_r}$

$t'=\frac{1116.344}{2.4073}$

$t'\approx463.733\ s$

7 0

4 years ago

If i have a kinematic equation vf^2=vi^2-2*a(xf-xi), how can i solve for xi step by step

azamat

Answer:

Explanation:

$v_f^2 = v_i^2-2a(x_f-x_i)$

Subtract both sides by $v_i^2$ :

$- v_i^2+v_f^2 = -2a(x_f-x_i)$

Divide both sides by -2*a:

$\frac{v_i^2 - v_f^2}{2a} =x_f-x_i$

Add both sides by $x_i$ :

$x_i+\frac{v_i^2 - v_f^2}{2a} =x_f$

Subtract both sides by $\frac{v_i^2 - v_f^2}{2a}$ :

$x_i=x_f-\frac{v_i^2 - v_f^2}{2a}$

8 0

4 years ago