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Mumz [18]
3 years ago
10

A twig from a tree drops from a 200m high cliff on to a beach below.during its fall 40% twig's energy is converted into thermal

energy.what is the speed with which the twig hits the beach?
Physics
1 answer:
kolezko [41]3 years ago
7 0
<span>Calculating for potential energy P.E=mgh
m=mass
g=acceleration due to gravity
h=height
</span>Let the twig's mass be M<span> 
</span><span>
therefore at the start, the twig has 200*9.8*M potential energy. 

</span>During its fall 40% twig's energy is converted into thermal energy and the remaining <span>60% of this will turn into kinetic energy during the fall, giving us 1176M
Now the equation for kinetic energy, K.E is 0.5*m*v^2
m is mass
v is velocity

Therefore 1176M=0.5*M*v^2 (mass will divide out)
velocity is </span><span>48.49 m/s</span>
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A 20 cm-radius ball is uniformly charged to 71 nC.
artcher [175]

Answer:

Part a)

\rho = 2.12\mu C/m^3

Part b)

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q_2 = 8.88 nC

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Part c)

E_1 = 3996 N/C

E_2 = 7992 N/C

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Explanation:

Part a)

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V = \frac{4}{3}\pi R^3

V = \frac{4}{3}\pi(0.20)^3

V = 0.0335 m^3

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\rho = \frac{71 nC}{0.0335} = 2.12\mu C/m^3

Part b)

Now the charge enclosed by the surface is given as

q = \rho V

at radius of 5 cm

q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.05)^3

q = 1.11 nC

at radius of 10 cm

q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.10)^3

q = 8.88 nC

at radius of 20 cm

q = 71 nC

Part c)

As we know that electric field is given as

E = \frac{kq}{r^2}

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E_1 = \frac{(9\times 10^9)(1.11 nC)}{0.05^2}

E_1 = 3996 N/C

electric field at r = 10 cm

E_2 = \frac{(9\times 10^9)(8.88 nC)}{0.10^2}

E_2 = 7992 N/C

electric field at r = 20 cm

E_3 = \frac{(9\times 10^9)(71 nC)}{0.20^2}

E_3 = 15975 N/C

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