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Mekhanik [1.2K]
3 years ago
15

I am sitting on a train car traveling horizontally at a constant speed of 50 m/s. I throw a ball straight up into the air.

Physics
1 answer:
drek231 [11]3 years ago
8 0

Answer:

Explanation:

I am sitting on a train car traveling horizontally at a constant speed of 50 m/s. I throw a ball straight up into the air. Before , the ball gets separated from my hand , both me the ball will be moving with velocity of 50 m /s in horizontal direction .

As soon as ball is separated from the hand , it acquires addition velocity in upward direction and acceleration in downward direction . This will give relative velocity to the ball with respect to me . So I will see the ball going in upward direction under  gravitational acceleration . It appears as if I am sitting at rest and ball is going in upward direction under deceleration . My motion at 50 m/s will have no effect on the motion of ball in upward direction , according to first law of Newton . It is so because ball too will be moving in forward direction with the same speed which will not be visible to me because I too am moving with the same speed.

If I  am  sitting at rest at home and I threw a ball straight up into the air , I will have the same experience of seeing ball going in similar way as described above.

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A uranium and iron atom reside a distance R = 44.10 nm apart. The uranium atom is singly ionized; the iron atom is doubly ionize
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Answer:

distance r from the uranium atom is 18.27 nm

Explanation:

given data

uranium and iron atom distance R = 44.10 nm

uranium atom = singly ionized

iron atom = doubly ionized

to find out

distance r from the uranium atom

solution

we consider here that uranium electron at distance = r

and electron between uranium and iron so here

so we can say electron and iron  distance = ( 44.10 - r ) nm

and we know single ionized uranium charge q2= 1.602 × 10^{-19} C

and charge on iron will be q3 = 2 × 1.602 × 10^{-19} C

so charge on electron is q1 =  - 1.602 × 10^{-19} C

and we know F = k\frac{q*q}{r^{2} }  

so now by equilibrium

Fu = Fi

k\frac{q*q}{r^{2} }  =  k\frac{q*q}{r^{2} }

put here k = 9*10^{9} and find r

9*10^{9}\frac{1.602 *10^{-19}*1.602 *10^{-19}}{r^{2} }  =  9*10^{9}\frac{1.602 *10^{-19}*1.602 *10^{-19}}{(44.10-r)^{2} }

\frac{1}{r^{2} } = \frac{2}{(44.10 -r)^2}

r = 18.27 nm

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The kinetic energy of the ejected electrons will be 2782.5 ×10⁻²² eV.

<h3>What is threshold frequency?</h3>

The threshold frequency of incoming radiation is the lowest frequency at which photoelectric emission or electron emission is impossible.

The threshold frequency is the light frequency that causes an electron to dislodge and emit from the metal's surface.

From the photoelectric effect, the equation obtained as;

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