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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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Two charges are placed on the x axis. One of the charges (q1 = +7.7 µC) is at x1 = +3.1 cm and the other (q2 = -19 µC) is at x2
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Answer:

a)E=50.53\times 10^{6}\ N/C

The direction will be negative direction.

b)E=268.22\times 10^{6}\ N/C

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Given that

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q2 = -19 µC is at x2 = +8.9 cm

We know that electric filed due to a charge given as

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E_1=K\dfrac{q_1}{r_1^2}

E_2=K\dfrac{q_2}{r_2^2}

Now by putting the va;ues

a)

E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.031^2}\ N/C

E_1=72.11\times 10^{6}\ N/C

E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.089^2}\ N/C

E_2=21.58\times 10^{6}\ N/C

The net electric field

E=E_1-E_2

E=50.53\times 10^{6}\ N/C

The direction will be negative direction.

As we know that electric filed line emerge from positive charge and concentrated at negative charge.

b)

Now

distance for charge 1 will become =5.5 - 3.1 = 2.4 cm

distance for charge 2 will become =8.9-5.5 = 3.4 cm

E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.024^2}\ N/C

E_1=120.3\times 10^{6}\ N/C

E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.034^2}\ N/C

E_2=147.92\times 10^{6}\ N/C

The net electric field

E=E_1+E_2

E=268.22\times 10^{6}\ N/C

The direction will be positive direction.

   

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