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3 years ago

Probably the deadliest aspect of a thunderstorm is _____. rain thunder lightning hail

2 answers:

max2010maxim [7]3 years ago

8 0

The answer to the given question is lightning.
Lightning is most probably the deadliest aspect of a thunderstorm. It is a sudden electrostatic discharge during an electrical storm. It happens when there is an electrostatic discharge between electrically charged regions of a cloud, which is referred as the intra-cloud lightning or IC, between that cloud and another cloud or the CC lightning, or between a cloud and the ground (CG lightning).

kipiarov [429]3 years ago

3 0

Probably the deadliest aspect of a thunderstorm is lightning .

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Plz help, and show work

pentagon [3]

Answer:

1. 2.67 s

2. 0.1 m/s²

Explanation:

1. Determination of the time taken for the penguin to fall.

Height (h) of cliff = 35 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

35 = ½ × 9.8 × t²

35 = 4.9 × t²

Divide both side by 4.9

t² = 35 / 4.9

Take the square root of both side

t = √(35 / 4.9)

t = 2.67 s

Thus, it will take 2.67 s for the penguin to fall onto the head of a napping polar bear.

2. Determination of the acceleration of the penguin.

Initial velocity (u) = 0 m/s.

Final velocity (v) = 2 m/s.

Time (t) = 20 s

Acceleration (a) =?

a = (v – u)/t

a = (2 – 0)/ 20

a = 2 / 20

a = 0.1 m/s²

Thus, the acceleration of the penguin is 0.1 m/s²

8 0

2 years ago

An average sleeping person metabolizes at a rate of about 80 W by digesting food or burning fat. Typically, 20% of this energy g

Leto [7]

Answer:

-4.7 x 10^-3 J/K-s

Explanation:

The Power generated by metabolizing food = 80 W

The watt W is equivalent to the Joules per sec J/s

therefor power = 80 J/s

20% of this energy is not used for heating, amount available for heating is

==> H = 80% of 80 = 0.8 x 80 = 64 J/s

The inner body temperature = 37 °C = 273 + 37 = 310 K

The entropy of this inner body ΔS = ΔH/T

ΔS = 64/310 = 0.2065 J/K-s

The skin temperature is cooler than the inner body by 7 °C

Temperature of the skin = 37 - 7 = 30 °C = 273 + 30 = 303 K

The entropy of the skin = ΔS = ΔH/T

ΔS = 64/303 = 0.2112 J/K-s

change in entropy of the person's body = (entropy of hot region: inner body) - (entropy of cooler region: skin)

==> 0.2065 - 0.2112 = -4.7 x 10^-3 J/K-s

8 0

3 years ago

You kick a soccer ball with a mass of 2 kg. The ball leaves your foot with a speed of 30 m/s. How much kinetic energy does the b

Lyrx [107]

Answer:

KE= 900 (I think)

Explanation:

KE=½mv²

KE= ½(2)(30)²

KE=½(60)²

KE=30²

KE=900

Hope this helps!

7 0

2 years ago

Read 2 more answers

Write equations for both the electric and magnetic fields for an electromagnetic wave in the red part of the visible spectrum th

Strike441 [17]

Answer:

$E=3.5(8.98*10^{6}x-2.69*10^{15}t)$

$B=1.17*10^{-8}(8.98*10^{6}x-2.69*10^{15}t)$

Explanation:

The electric field equation of a electromagnetic wave is given by:

$E=E_{max}(kx-\omega t)$ (1)

E(max) is the maximun value of E, it means the amplitude of the wave.
k is the wave number
ω is the angular frequency

We know that the wave length is λ = 700 nm and the peak electric field magnitude of 3.5 V/m, this value is correspond a E(max).

By definition:

$k=\frac{2\pi}{\lambda}$

$k=8.98*10^{6} [rad/m]$

And the relation between λ and f is:

$c=\lambda f$

$f=\frac{c}{\lambda}$

$f=\frac{3*10^{8}}{700*10^{-9}}$

$f=4.28*10^{14}$

The angular frequency equation is:

$\omega=2\pi f$

$\omega=2\pi*4.28*10^{14}$

$\omega=2.69*10^{15} [rad/s]$

Therefore, the E equation, suing (1), will be:

$E=3.5(8.98*10^{6}x-2.69*10^{15}t)$ (2)

For the magnetic field we have the next equation:

$B=B_{max}(kx-\omega t)$ (3)

It is the same as E. Here we just need to find B(max).

We can use this equation:

$E_{max}=cB_{max}$

$B_{max}=\frac{E_{max}}{c}=\frac{3.5}{3*10^{8}}$

$B_{max}=1.17*10^{-8}T$

Putting this in (3), finally we will have:

$B=1.17*10^{-8}(8.98*10^{6}x-2.69*10^{15}t)$ (4)

I hope it helps you!

8 0

3 years ago

In a Rutherford scattering experiment a target nucleus has a diameter of 1.34×10-14 m. The incoming α particle has a mass of 6.6

Rasek [7]

Answer:

E = 2.5 x 10⁻¹⁴ J

Explanation:

given,

diameter = 1.33 x 10⁻¹⁴ m

mass = 6.64 x 10⁻²⁷ kg

wavelength is equal to diameter

de broglie wavelength equal to diameter

$\lambda = \dfrac{h}{mv}$

$1.33 \times 10^{-14}= \dfrac{6.626 \times 10^{-34}}{6.64 \times 10^{-27}\times v}$

$v= \dfrac{6.626 \times 10^{-34}}{6.64 \times 10^{-27}\times 1.33 \times 10^{-14}}$

v = 7.5 x 10⁶ m/s

Kinetic energy is equal to

$E = \dfrac{1}{2}mv^2$

$E = \dfrac{1}{2}\times 6.64 \times 10^{-27}\times (7.5\times 10^6)^2$

E = 2.5 x 10⁻¹⁴ J

8 0

3 years ago