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34kurt
3 years ago
11

In a thunderstorm, the air must be ionized by a high voltage before a conducting path for a lightning bolt can be created. an el

ectric field of about 1.0 × 106 v/m is required to ionize dry air. what would the break- down voltage in air be if a thundercloud were 1.60 km above ground? assume that the electric field between the cloud and the ground is uniform.
Physics
1 answer:
nydimaria [60]3 years ago
4 0
We know that the electric field is equal to 1 E 6 V/m

The distance between the thundercloud and the ground is 1.6km = 1600m

Electric field = Voltage/distance

This means that the breakdown voltage must be equal to

V = 1 E 6 V/m * 1600m = 1.6 E 9 V = 1.6 GV
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In the vertical jump, an athlete starts from crouch and jumps upward to reach as high as possible. Even the best athletes spend
Mumz [18]

Answer:

The ratio of the time he is above ymax /2 to the time it takes him to go from the floor to that height = 0.707

Explanation:

The mathematical derivation and steps is as shown in the attached file.

3 0
3 years ago
What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000V/s?
konstantin123 [22]

Answer:

I=2.71\times 10^{-5}\ A

Explanation:

A 6.0-cm-diameter parallel-plate capacitor has a 0.46 mm gap.  

What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000V/s?

Let given is,

The diameter of a parallel plate capacitor is 6 cm or 0.06 m

Separation between plates, d = 0.046 mm

The potential difference across the capacitor is increasing at 500,000 V/s

We need to find the displacement current in the capacitor. Capacitance for parallel plate capacitor is given by :

C=\dfrac{A\epsilon_o}{d}\\\\C=\dfrac{\pi r^2\epsilon_o}{d}, r is radius

Let I is the displacement current. It is given by :

I=C\dfrac{dV}{dt}

Here, \dfrac{dV}{dt} is rate of increasing potential difference

So

I=\dfrac{\pi r^2\epsilon_o}{d}\times \dfrac{dV}{dt}\\\\I=\dfrac{\pi (0.03)^2\times 8.85\times 10^{-12}}{0.46\times 10^{-3}}\times 500000\\\\I=2.71\times 10^{-5}\ A

So, the value of displacement current is 2.71\times 10^{-5}\ A.

4 0
3 years ago
Given a force of 100 N and a acceleration of 5 m/s, what is the mass
SIZIF [17.4K]

Answer:

force = mass \times acceleration \\ 100 = m \times 5 \\ m =  \frac{100}{5}  \\ m = 20 \:  kg

8 0
2 years ago
Read 2 more answers
A light bulb can be all of the following except
Dennis_Churaev [7]

Answer: A light bulb can be all of the following except option C (a consumer product if it is used to light the office of the board of directors.)

Explanation:

Products are classified as being BUSINESS or CONSUMER products according to the buyer's intended use of the product.

-Consumer products: these are sold goods that are used for personal, family, or household use. The intention of the buyer is for the products to satisfy his personal needs and desires. Example of some of the consumer products include: toothpaste, eatables and clothes.

Business products: products that are not for personal use but for the manufacturing of other goods are called business products.

Therefore a bulb is not serving as a personal use when used to light the office of the board of directors rather it's serving as a business product .

3 0
2 years ago
Use the following half-life graph to answer the following question:
Temka [501]

Answer:

A 1.0 min

Explanation:

The half-life of a radioisotope is defined as the time it takes for the mass of the isotope to halve compared to the initial value.

From the graph in the problem, we see that the initial mass of the isotope at time t=0 is

m_0 = 50.0 g

The half-life of the isotope is the time it takes for half the mass of the sample to decay, so it is the time t at which the mass will be halved:

m'=\frac{50.0 g}{2}=25.0 g

We see that this occurs at t = 1.0 min, so the half-life of the isotope is exactly 1.0 min.

3 0
3 years ago
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