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Alona [7]
3 years ago
6

What is the escape speed of an electron launched from the surface of a 1.1-cm-diameter glass sphere that has been charged to 8.0

nc ?
Physics
1 answer:
zimovet [89]3 years ago
5 0
At surface,
v = kq/r

And potential energy of an electron is given by,
PE = -ev = -ekq/r

At escape velocity,
PE + KE = 0.
Therefore,
1/2mv^2 - ekq/r =0
1/2mv^2 = ekq/r
v = Sqrt [2ekq/mr], where v = escape velocity, e = 1.6*10^-19 C, k = 8.99*10^9 Nm^2/C^2, m = 9.11*10^-31 kg, r = 1.1*10^-2 m, q = 8*10^-9 C

Substituting;
v = Sqrt [(2*1.6*19^-19*8.99*10^9*8*10^-9)/(9.11*10^-31*1.1*10^-2)] = 47949357.23 m/s ≈ 4.795 *10^7 m/s
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a ball is projected upward at time t = 0.00 s from a point on a roof 70 m above the ground. The ball rises, then falls and strik
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Answer: 17.68 s

Explanation:

This problem is a good example of Vertical motion, where the main equation for this situation is:  

y=y_{o}+V_{o}t-\frac{1}{2}gt^{2} (1)  

Where:  

y=0 is the height of the ball when it hits the ground  

y_{o}=70 m is the initial height of the ball

V_{o}=82m/s is the initial velocity of the ball  

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Having this clear, let's find t from (1):  

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Rewritting (2):

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Where:

a=-\frac{1}{2}(9.8m/s^{2}

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Substituting the known values:

t=\frac{-82 \pm \sqrt{82^{2}-4(-\frac{1}{2}(9.8)(70)}}{2a}  (5)

Solving (5) we find the positive result is:

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