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3 years ago

What is the minimum speed must a salmon jumping at an angle of 35.2 leave the water in m/s?

Physics

1 answer:

yuradex [85]3 years ago

8 0

Height of the waterfall is 0.449 m

its horizontal distance will be 2.1 m

now let say his speed is v with which he jumped out so here the two components of his velocity will be

$v_x = vcos35.2 = 0.817 v$

$v_y = vsin35.2 = 0.576 v$

here the acceleration due to gravity is 9.81 m/s^2 downwards

now we can find the time to reach the other end by y direction displacement equation

$\Delta y = v_y * t + \frac{1}{2} at^2$

$-0.449 = 0.576 * v *t - \frac{1}{2}*9.81 * t^2$

also from x direction we can say

$\Delta x = v_x * t$

$2.1 = 0.817 v* t$

now we have

$v* t = 2.57$

we will plug in this value into first equation

$- 0.449 = 0.576 * 2.57 - 4.905 * t^2$

$1.93 = 4.905 * t^2$

$t = 0.63 s$

now as we know that

$v* t = 2.57$

t = 0.63 s

$v = \frac{2.57}{0.63}$

$v = 4.1 m/s$

so his minimum speed of jump is 4.1 m/s

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To solve this problem we will apply the concepts related to electric potential and electric potential energy. By definition we know that the electric potential is determined under the function:

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r = Radius

At the same time

$U = \frac{k_e q_1q_2}{r}$

The values of variables are the same, then if we replace in a single equation we have this expression,

$U = Vq$

If we replace the values, we have finally that the charge is,

$V = 800V$

$q = 1\mu C$

$U = (800V)(1*10^{-6}C)$

$U = 8*10^{-4}J$

Therefore the potential energy of the system is $8*10^{-4} J$

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3 years ago

Consider an electron with charge −e and mass m orbiting in a circle around a hydrogen nucleus (a single proton) with charge +e.

alexandr1967 [171]

Answer:

$v=\sqrt{k\frac{e^2}{m_e r}}$ , $2.18\cdot 10^6 m/s$

Explanation:

The magnitude of the electromagnetic force between the electron and the proton in the nucleus is equal to the centripetal force:

$k\frac{(e)(e)}{r^2}=m_e \frac{v^2}{r}$

where

k is the Coulomb constant

e is the magnitude of the charge of the electron

e is the magnitude of the charge of the proton in the nucleus

r is the distance between the electron and the nucleus

v is the speed of the electron

$m_e$ is the mass of the electron

Solving for v, we find

$v=\sqrt{k\frac{e^2}{m_e r}}$

Inside an atom of hydrogen, the distance between the electron and the nucleus is approximately

$r=5.3\cdot 10^{-11}m$

while the electron mass is

$m_e = 9.11\cdot 10^{-31}kg$

and the charge is

$e=1.6\cdot 10^{-19} C$

Substituting into the formula, we find

$v=\sqrt{(9\cdot 10^9 m/s) \frac{(1.6\cdot 10^{-19} C)^2}{(9.11\cdot 10^{-31} kg)(5.3\cdot 10^{-11} m)}}=2.18\cdot 10^6 m/s$

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3 years ago

Which describes the molecular motion of a solid nickel marble rolled across the floor?

Gemiola [76]

I can't decide between A and B, but B seems more likely to me. Even though the molecules don't look like they're moving, the area of contact is slightly more compressed.

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4 years ago

Fg =G m1m2/r2 solver for G

Blababa [14]

G = $\frac{F_{g} r^{2} }{m_{1} m_{2} }$

Explanation:

Solving for G simply implies that we make G the subject of the formula:

Given equation:

F $_{g}$ = G $\frac{m_{1} m_{2} }{ r^{2} }$

To make G the subject of this expression follow these steps:

Multiply both sides of the equation by $r^{2}$

F $_{g}$ x $r^{2}$ = G $\frac{m_{1} m_{2} }{ r^{2} }$ x $r^{2}$

This gives:

F $_{g}$ $r^{2}$ = G $m_{1} m_{2}$

Multiply both sides by $\frac{1}{m_{1} m_{2} }$

F $_{g}$ $r^{2}$ x $\frac{1}{m_{1} m_{2} }$ = $\frac{1}{m_{1} m_{2} }$ x G $m_{1} m_{2}$

Therefore:

G = $\frac{F_{g} r^{2} }{m_{1} m_{2} }$

Learn more:

Solving formula brainly.com/question/2998489

#learnwithBrainly

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Answer:

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