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TEA [102]
2 years ago
9

Given that an electric field of 3×106V/m3×106V/m is required to produce an electrical spark within a volume of air, estimate the

length of a thundercloud lightning bolt.
Physics
1 answer:
Andre45 [30]2 years ago
5 0

Answer:

Length, l = 33.4 m

Explanation:

Given that,

Electrical field, E=3\times 10^6\ V/m

Let the electrical potential is, V=10^8\ V

We need to find the length of a thundercloud lightning bolt. The relation between electric field and the electric potential is given by :

V=E\times d\\\\d=\dfrac{V}{E}\\\\d=\dfrac{10^8}{3\times 10^6}\\\\d=33.4\ m

So, the length of a thundercloud lightning bolt is 33.4 meters. Hence, this is the required solution.

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A series of waves with decreasing wavelength labeled radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, gamm
kotegsom [21]

Answer:

label A= radio waves, label C= infrared, Label D= visible Light, Label G= gamma rays.

Explanation:

hope it helped??

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6 0
2 years ago
Read 2 more answers
Among the largest passenger ships currently in use, the Norway has been in service the longest. The Norway is more than 300 m lo
LenaWriter [7]

Answer:

6.33\times 10^8\ kg\cdot m/s

Explanation:

Mass of the ship (m) = 6.9 × 10⁷ kg

Speed of the ship (v) = 33 km/h

First, let us convert the speed from km/h to m/s using the conversion factor.

We know that, 1 km/h = 5/18 m/s

So, 33 km/h = 33\times \frac{5}{18}=9.17\ m/s

Now, we know, the momentum of an object only depends on its mass and speed. Momentum is independent of the length of the object.

So, here, length of the ship doesn't play any role in the determination of the momentum.

Magnitude of momentum of the ship = Mass × Speed

                                                             = (6.9\times 10^7\ kg)(9.17\ m/s)

                                                             = 6.33\times 10^8\ kg\cdot m/s

Therefore, the magnitude of ship's momentum is 6.33\times 10^8\ kg\cdot m/s.

6 0
2 years ago
A child's toy consists of a m = 36 g monkey suspended from a spring of negligible mass and spring constant k. When the toy monke
kolezko [41]

Answer:

Part A - 3N/m

Part B - see attachment

Part C - 4.9 × 10-³J

Part D - E = 1/2kd² + 1/2mv² + mgh

Explanation:

This problem requires the knowledge of simple harmonic motion for cimplete solution. To find the spring constant in part A the expression relating the force applied to a spring and the resulting stretching of the spring (hooke's law) is required which is F = kx.

The free body diagram can be found in the attachment. Fp(force of pull), Ft(Force of tension) and W(weight).

The energy stored in the pring as a result of the stretching of d = 5.7cm is 1/2kd².

Part D

Three forces act on the spring-monkey system and they do work in different forms: kinetic energy 1/2mv² , elastic potential

energy due to the restoring force in the spring or the tension force 1/2kd², and the gravitational potential energy mgh of the position of the system. So the total energy of the system E = 1/2kd² + 1/2mv² + mgh.

8 0
2 years ago
A charge Q is placed on the x axis at x = +4.0 m. A second charge q is located at the origin. If Q = +75 nC and q = −8.0 nC, wha
Stells [14]

Answer:

23.1 N/C

Explanation:

OP = 3 m , OQ = 4 m

PQ = \sqrt{4^{2}+3^{2}}=5 m

q = - 8 nC, Q = 75 nC

Electric field at P due to the charge Q is

E_{1}=\frac{KQ}{PQ^{2}}=\frac{9\times 10^{9}\times 75\times 10^{-9}}{25}=27 N/C

Electric field at P due to the charge q is

E_{2}=\frac{Kq}{PO^{2}}=\frac{9\times 10^{9}\times 8\times 10^{-9}}{9}=8 N/C

According to the diagram, tanθ = 3/4

Resolve the components of E1 along x axis and along y axis.

So, Electric field along X axis, Ex = - E1 Cos θ

Ex = - 27 x 4 / 5 = - 21.6 N/C

Electric field along y axis, Ey = E1 Sinθ - E2

Ey = 27 x 3 /5 - 8 = 8.2 N/C

The resultant electric field at P is given by

E=\sqrt{E_{x}^{2}+E_{y}^{2}}=\sqrt{(-21.6)^{2}+(8.2)^{2}}=23.1 N/C

3 0
2 years ago
In a common but dangerous prank, a chair is pulled away as a person is moving downward to sit on it, causing the victim to land
sattari [20]

Answer:

a) Impulse |J|= 219.4 kgm/s

b) Force F = 2672 N

Explanation:

Given

Height of fall h = 0.50 m

Mass M = 70 kg

Period of collision t = 0.082 s

Solution

The final velocity of the person v is zero since the person will come to rest.

The initial velocity of the person can be calculated by using the "law of conservation of energy".

Initial Kinetic energy = Final potential energy

\frac{1}{2} mu^2=mgh\\\\u = \sqrt{2gh} \\\\u = \sqrt{2 \times 9.81 \times 0.50} \\\\u = 3.13 m/s

a) Impulse

J = final momentum - initial momentum

J = mv -mu\\\\J = 0 - (70 \times 3.13)\\\\J = -219.2 kgm/s

Magnitude of impulse

|J| = 219.1 kgm/s

b) Force

F = \frac{J}{t} \\\\F = \frac{219.1}{0.082} \\\\F = 2672 N

4 0
2 years ago
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