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belka [17]
3 years ago
6

g An airplane is flying through a thundercloud at a height of 1500 m. (This is a very dangerous thing to do because of updrafts,

turbulence, and the possibility of electric discharge.) If a charge concentration of 25.0 C is above the plane at a height of 3000 m within the cloud and a charge concentration of -40.0 C is at height 850 m, what is the electric field at the aircraft
Physics
1 answer:
Stolb23 [73]3 years ago
4 0

Answer:

523269.9\ \text{N/m}

Explanation:

q = Charge

r = Distance

q_1=25\ \text{C}

r_1=3000\ \text{m}

q_2=40\ \text{C}

r_2=850\ \text{m}

The electric field is given by

E=E_1+E_1\\\Rightarrow E=k(\dfrac{q_1}{r_1^2}+\dfrac{q_2}{r_2^2})\\\Rightarrow E=9\times 10^9\times (\dfrac{25}{3000^2}+\dfrac{40}{850^2})\\\Rightarrow E=523269.9\ \text{N/m}

The electric field at the aircraft is 523269.9\ \text{N/m}

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The mass of the skateboard is 1.8kg and the mass of the skateboarder is 42kg.
Murljashka [212]
Total momentum must be conserved since there are no external forces in the horizontal direction on the skateboarder or the skateboard.

Momentum before the jump:
P = m_1v_1 + m_2v_2 = (1.8kg)(0) + (42kg)(0) = 0

Momentum after the jump:
P =  m_1v_1 + m_2v_2 = (1.8kg)(v_1) + (42kg)(0.3 \frac{m}{s} ) = 0

Solving for the velocity v:
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4 0
4 years ago
A hot-air balloon is rising straight up with a speed of 4.6 m/s. a ballast bag is released from rest relative to the balloon at
kherson [118]

The time taken by the ballast bag to reach the ground is 2.18 s

The ballast bag at rest with respect to the balloon has the upward velocity (u) of 4.6 m/s , which is the velocity of the balloon. When it is dropped from the balloon, its motion is similar to an object thrown upwards with an initial velocity <em>u </em>and it falls under the acceleration due to gravity<em> g.</em>

Taking the upward direction as positive and the downward direction as negative, the following equation of motion may be used.

s=ut+\frac{1}{2}at^2

The bag makes a net displacement <em>s</em> of 13.4 m downwards, hence

s=-13.4 m

Its initial velocity is

u=+4.6 m/s

The acceleration due to gravity acts downwards and hence it is negative.

g=-9.8 m/s^2

Use the values in the equation of motion and write an equation for t.

s=ut+\frac{1}{2} at^2 \\ -13.4=4.6t-\frac{1}{2}(9.8)t^2\\ 4.9t^2-4.6t-13.4=0

Solving the equation for t and taking only the positive value for t,

t=2.18 s

4 0
3 years ago
An apple falls out of a tree from a height of 2.3m. what is the impact speed of the apple
Anuta_ua [19.1K]

Answer:6.71 m/s

Explanation:

Given

Apple fall from a height of h=2.3 m  

We need to find the impact speed of apple which can be given by using

v^2-u^2=2gh  

where v=final velocity

u=initial velocity

h=Displacement

Assuming initial velocity to be zero

substituting the value we get

v^2-0=2\times 9.8\times 2.3  

v=\sqrt{2\times 9.8\times 2.3}  

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5 0
3 years ago
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