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Fed [463]
3 years ago
5

How does lightning occer in the sky​

Physics
2 answers:
Ivenika [448]3 years ago
8 0

Answer:

Explanation:

Lightning happens in the sky when the negative charges electrons in the bottom of the cloud are attracted to the positive charges protons in the ground.

victus00 [196]3 years ago
7 0

Answer:

Lightning happens when the negative charges (electrons) in the bottom of the cloud are attracted to the positive charges (protons) in the ground.

Explanation:

Hope it helps u

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A) Charge q1 = +5.60 nC is on the x-axis at x = 0 and an unknown charge q2 is on the x-axis at x = -4.00 cm. The total electric
jeka94

Answer:

a) F₃₁ = 63.0 μN  

b) F₃₂ = - 14.0 μN

c) q₂ = - 5.0 nC

Explanation:

a)

  • Assuming that the three charges can be taken as point charges, the forces between them must obey Coulomb's Law, and can be found independent each other, applying the superposition principle.
  • So, we can find the force that q₁ exerts along the x-axis on q₃, as follows:

       F_{31} =\frac{k*q_{1}*q_{3} }{r_{13}^{2}} = \frac{9e9Nm2/C2*5.6e-9C*2.0e-9C}{(0.04m)^{2}}  = 63.0 \mu N   (1)

b)

  • Since total force exerted by q₁ and q₂ on q₃ is 49.0 μN, we can find the force exerted only by q₂ (which is along the x-axis only too) just by difference, as follows:

      F_{32} = F_{3} - F_{31}  = 49.0\mu N  - 63.0\mu N = -14.0 \mu N  (2)

c)

  • Finally, in order to find the value of q₂, as we know the value and sign of F₃₂, we can apply again the Coulomb's Law, solving for q₂, as follows:

      q_{2}  = \frac{F_{32} * r_{23}^{2} }{k*q_{3}} = \frac{(-14\mu N)*(0.08m)^{2}}{9e9Nm2/C2* 2 nC} = - 5 nC  (3)

6 0
3 years ago
Why might an electromagnet be used to pick up old cars in junkyards?
siniylev [52]
B) Hope it helps ,Have a nice day :)
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4 years ago
Read 2 more answers
Main method of energy transfer​
tatyana61 [14]

Answer:

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6 0
3 years ago
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| A T-ball with a mass of 0.6 kg travels in the
r-ruslan [8.4K]

Answer:

|I|=6\ Kg.m/s

F=120\ N

Explanation:

Impulse and Momentum

They are similar concepts since they deal with the dynamics of objects having their status of motion changed by the sudden application of a force. The momentum at a given initial time is computed as

p_o=m.v_o

When a force is applied, the speed changes to v_1 and the new momentum is

p_1=m.v_1

The change of momentum is

\Delta p=p_1-p_0=m(v_1-v_o)

The impulse is equal to the change of momentum of an object and it's defined as the average net force applied times the time it takes to change the object's motion

I=F.t=\Delta p

Part 1

The T-ball initially travels at 10 m/s and then suddenly it's stopped by the glove. The final speed is zero, so

\Delta p=0.6\ Kg(0-10\ m/s)=-6\ Kg.m/s

The impulse is

I=\Delta p

I=-6\ Kg.m/s

The magnitude is

|I|=6\ Kg.m/s

Part 2

The force can be computed from the formula

I=F.t

The direction of the impulse the T-ball receives is opposite to the direction of the force exerted by the ball on the glove, thus I_b=6\ kg.m/s

\displaystyle F=\frac{I}{t}=\frac{6\ kg.m/s}{0.05\ s}

\boxed{F=120\ N}

7 0
3 years ago
Convert 65 mi/h to m/s. 1 mi = 1609 m. Answer in units of m/s.
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The answer is 29.0576 m/s 
7 0
3 years ago
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