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4 years ago

Thunderstorms are violent _________ storms accompanied by thunder and lightning.

Physics

1 answer:

Phoenix [80]4 years ago

7 0

Thunderstorms are violent convective storms accompanied by thunder and lightning.

<span>Thunderstorms are convective storms because it is an effect when there is an intense heating in the Earth’s atmosphere that causes parcels of moist air that rises to the surface of the Earth all throughout the atmosphere. A convective storm also causes tornadoes.</span>

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Robert, a stuntman, projects himself from a height of 50.0 meters in a horizontal direction. He safely lands at a horizontal dis

Olin [163]

To get this we need to know how long he was traveling. To know that we need to solve for the time it took for him to hit the ground. This is given by the vertical height he started at and the equation of motion for falling objects. The position function is
Y= -(9.8/2)t^2 + 50 = 0
t=3.19s
That's how long he's in the air. When he hits the ground he stops moving forward so this is also the amount of time he was moving at 100m/s horizontally. Now use speed distance relationship
X=vt
And solve for v
100m/3.19s = 31.35m/s

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3 years ago

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When the kinetic energy is transformed into another for of energy during the collision or impact

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3 years ago

A train moves at constant velocity of 90km/h. How far will it go in 0.25h.

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3 years ago

An infinitely long line of charge has a linear charge density of 7.50×10^−12 C/m . A proton is at distance 14.5 cm from the line

Nata [24]

Answer:

10.22 cm

Explanation:

linear charge density, λ = 7.5 x 10^-12 C/m

distance from line, r = 14.5 cm = 0.145 m

initial speed, u = 3000 m/s

final speed, v = 0 m/s

charge on proton, q = 1.6 x 10^-19 C

mass of proton, m = 1.67 x 10^-27 kg

Let the closest distance of proton is r'.

The potential due t a line charge at a distance r' is given by

$V=-2K\lambda ln\left (\frac{r'}{r} \right )$

where, K = 9 x 10^9 Nm^2/C^2

W = q V

By use of work energy theorem

Work = change in kinetic energy

$qV = 0.5m(u^{2}-v^{2})$

By substituting the values, we get

$V=\frac{mu^{2}}{2q}$

$-2K\lambda ln\left ( \frac{r'}{r} \right )=\frac{mu^{2}}{2q}$

$- ln\left ( \frac{r'}{r} \right )=\frac{mu^{2}}{4Kq\lambda }$

$- ln\left ( \frac{r'}{r} \right )=\frac{1.67 \times 10^{-27}\times 3000\times 3000}{4\times 9\times 10^{9}\times 1.6\times 10^{-19}\times 7.5\times 10^{-12} }$

$- ln\left ( \frac{r'}{r} \right )=0.35$

$\frac{r'}{r} =e^{-0.35}$

$\frac{r'}{r} =0.7047$

r' = 14.5 x 0.7047 = 10.22 cm

5 0

4 years ago