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3 years ago

An electron in a mercury atom changes from energy level b to a higher energy level when the

Physics

1 answer:

emmainna [20.7K]3 years ago

4 0

The energy is 3.06 electronvolts, E = 3.06eV

1eV = 1.6 * 10^-19 J

3.06 eV = 3.06* 1.6 * 10^-19 J = 4.896 * 10^-19 J

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According to newtons law of action reaction what is the most likely to occur if to ice skaters with approximately the same mass

Free_Kalibri [48]

Answer:

They would keep on moving but unless being acted upon or stop slowly because of the friction

Explanation:

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3 years ago

ADP binds to platelets in order to intiate the activation process. Two binding sites were identified on platelets, one with a Kd

12345 [234]

Answer:B) The binding site with a kid of7.9uM is the low affinity site.

Explanation:

The rate at which macro molecules like protein bind together is called affinity. The higher the affinity site the more readily these macro molecules fuses together. 7.9uM is a low affinity value than 359uM.

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4 years ago

Hinge Joints-These jointscan only move in one direction, like a door hinge. One bone works against another. Movement is back and

prisoha [69]

Hinge joints would be your elbow, your knee and, believe it or not, your ankle as well!

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3 years ago

Objects with masses of 165 kg and a 465 kg are separated by 0.340 m. (a) Find the net gravitational force exerted by these objec

Dmitrij [34]

Answer:

(a) $F_{net}$ = 4.19 x $10^{-5}$ N, and its direction is towards $m_{2}$ .

(b) It must be placed inside a hollow shell.

Explanation:

Let, $m_{1}$ = 165 kg, $m_{2}$ = 465 kg, $m_{3}$ = 60 kg, and the distance between $m_{1}$ and $m_{2}$ is 0.340 m.

(a) Since $m_{3}$ is placed midway between $m_{1}$ and $m_{2}$ , then its distance to both masses is 0.170 m.

From the Newton's law of universal gravitation,

F = $\frac{Gm_{1}m_{2} }{r^{2} }$

Where all variables have their usual meaning.

Then,

a. $F_{net}$ = $F_{23}$ - $F_{13}$

$F_{13}$ = $\frac{6.67*10^{-11}*165*60 }{(0.17)^{2} }$

= 2.25 x $10^{-5}$ N

$F_{23}$ = $\frac{6.67*10^{-11}*465*60 }{(0.17)^{2} }$

= 6.44 x $10^{-5}$ N

∴ $F_{net}$ = = 6.44 x $10^{-5}$ - 2.25 x $10^{-5}$

= 4.19 x $10^{-5}$ N

The net force exerted by the two masses on the 60 kg object is 4.19 x $10^{-5}$ N.

(ii) / $F_{net}$ / = / $F_{23}$ / - / $F_{13}$ /

= 6.44 x $10^{-5}$ - 2.25 x $10^{-5}$

= 4.19 x $10^{-5}$ N

(iii) The direction of the net force is to the right i.e towards $m_{2}$ .

(b) For the net force experienced by the 60 kg object to be zero, it must be placed inside a hollow shell.

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3 years ago

Why does the car stop? Where did the energy go?

Viefleur [7K]

Answer:

Because you hit the break?

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3 years ago