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3 years ago

15

Are basins a collection of smaller watersheds

1 answer:

mihalych1998 [28]3 years ago

7 0

Explanation:

both are areas of land that drain to particular water bodies such as lakes

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Why is it important to use democratic method of choosing someone for a position of authority

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Read 2 more answers

A proton (mass=1.67x10^-27 kg, charge= 1.60x10^-19 C) moves from point A to point under the influence of an electrostatic force

Tom [10]

Answer:

$VB - VA = - 33.4$

Explanation:

Generally the workdone in moving the proton is mathematically represented as

$W = KE_f - KE_i$

Where $KE_i \ and \ KE_f \ are\ the\ initial \ and \ final \ kinetic \ energy$

So

$KE_i = \frac{1}{2} m v_a^2$

Here $v_a$ is the velocity at A with value 50 m/s

So

$KE_i = \frac{1}{2} (1.67*10^{-27}) * 50^2$

$KE_i = 2.09 *10^{-24} \ J$

Also

$KE_f = \frac{1}{2} m v_b^2$

Here $v_a$ is the velocity at A with value $80 km/s = 80000 m/s$

=> $KE_f = \frac{1}{2} (1.67*10^{-27}) * 80000^2$

=> $KE_f = 5.34 *10^{-18} \ J$

So

$W = 5.34 *10^{-18} - 2.09 *10^{-24}$

$W = 5.34 *10^{-18} m/s$

Now this workdone is also mathematically represented as

$W = q * V$

So

$q * V = 5.34 *10^{-18}$

Here $q = 1.60*10^{-19} C$

So

$V = \frac{5.34 *10^{-18} }{1.60*10^{-19}}$

$V = 33.4 \ V$

Generally proton movement is in the direction of the electric field it means that $VA>VB$

So

$VB - VA = - 33.4$

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3 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

3 years ago

Calculate the de broglie wavelength (in picometers) of a hydrogen atom traveling at 440 m/s.

Aleonysh [2.5K]

De broglie wavelength, $\lambda = \frac{h}{mv}$ , where h is the Planck's constant, m is the mass and v is the velocity.

$h = 6.63*10^{-34}$

Mass of hydrogen atom, $m = 1.67*10^{-27}kg$

v = 440 m/s

Substituting

Wavelength $\lambda = \frac{h}{mv} = \frac{6.63*10^{-34}}{1.67*10^{-27}*440} = 0.902 *10^{-9}m = 902 *10^{-12}m$

$1 pm = 10^{-12}m\\ \\ So , \lambda =902 pm$

So the de broglie wavelength (in picometers) of a hydrogen atom traveling at 440 m/s is 902 pm

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3 years ago