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Goshia [24]
2 years ago
15

How do air mass conditions ahead of the squall line support the development of new cell?

Physics
1 answer:
IRISSAK [1]2 years ago
7 0
<span>Storm cells in a squall line typically move from the southwest to the northeast, and as the mature cells in the northeast begin to die off, new ones are formed at the opposite end to advance the line. The air in the southwest corner has strong vertical updrafts that allow new cells to grow and develop into thunderstorms.</span>
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In the sport of parasailing, a person is attached to a rope being pulled by a boat while hanging from a parachute-like sail. A r
scoray [572]

Answer:

W = 506.75 N

Explanation:

tension = 2300 N

Rider is towed at a constant speed means there no net force acting on the rider.

hence taking all the horizontal force and vertical force in consideration.

net horizontal  force:

F cos 30° - T cos 19° = 0

F cos 30° = 2300 × cos 19°

F = 2511.12 N

net vertical force:

F sin 30° - T sin 19°- W = 0

W = F sin 30° - T sin 19°

W =  2511.12 sin 30° - 2300 sin 19°

W = 506.75 N

8 0
2 years ago
A thin beam of light enters a thick plastic sheet from air at an angle of 32.0° with the normal and continues in the sheet at an
Cloud [144]

Answer:

1.36

Explanation:

n_{air} = Index of refraction of air = 1

n_{plastic} = Index of refraction of plastic = ?

i = angle of incidence in air = 32.0° deg

r = angle of refraction in plastic = 23.0° deg

Using Snell's law

n_{air} Sini = n_{plastic} Sinr

(1) SIn32.0 = n_{plastic} Sin23.0

n_{plastic} = 1.36

5 0
2 years ago
The graph represents the heating of water in a pot. At 150 seconds, the water has just reached a boil. If the heat is left on, w
a_sh-v [17]
The temperature will remain constant, at around 100 C, and the volume of water in the pot will decrease, as it turns into steam and floats away from the pot.
4 0
1 year ago
A helium ion of mass 4m and charge 2e is accelerated from rest through a potential difference V in vacuum. Its final speed will
Pavel [41]

Answer:

Final Velocity = √(eV/m)

Explanation:

The Workdone, W, in accelerating a charge, 2e, through a potential difference, V is given as a product of the charge and the potential difference

W = (2e) × V = 2eV

And this work is equal to change in kinetic energy

W = Δ(kinetic energy) = ΔK.E

But since the charge starts from rest, initial velocity = 0 and initial kinetic energy = 0

ΔK.E = ½ × (mass) × (final velocity)²

(Velocity)² = (2×ΔK.E)/(mass)

Velocity = √[(2×ΔK.E)/(mass)]

ΔK.E = W = 2eV

mass = 4m

Final Velocity = √[(2×W)/(4m)]

Final Velocity = √[(2×2eV)/4m]

Final Velocity = √(4eV/4m)

Final Velocity = √(eV/m)

Hope this Helps!!!

8 0
1 year ago
A skydiver is using wind to land on a target that is 50 m away horizontally. The skydiver starts from a height of 70 m and is fa
elena55 [62]

Answer:

Answer:

15.67 seconds

Explanation:

Using first equation of Motion

Final Velocity= Initial Velocity + (Acceleration * Time)  

v= u + at

v=3

u=50

a= - 4 (negative acceleration or deceleration)  

3= 50 +( -4 * t)

-47/-4 = t

Time = 15.67 seconds

6 0
1 year ago
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