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Katyanochek1 [597]
3 years ago
15

Flow around curved height contours requires the incorporation of the centrifugal force. What is the general term to describe the

winds that flow along a curved trajectory above the level where friction plays a role?
Physics
1 answer:
lilavasa [31]3 years ago
4 0

Answer: Gradient Wind

Explanation:

Gradient wind, is the wind that accounts for air flow along a curved trajectory. It is an extension of the concept of geostrophic wind; for example the wind assumed to move along straight and parallel isobars (lines of equal pressure). The gradient wind represents the actual wind better than the geostrophic wind, especially when both wind speed and trajectory curvature are large, because they are in hurricanes and jet streams.

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4.5 x 10kg) - (2.3 x 10 kg)
Degger [83]

Answer:

22 kg

Explanation:

I hope this helps!

6 0
3 years ago
Read 2 more answers
An 8.00 kg mass moving east at 15.4 m/s on a frictionless horizontal surface collides with a 10.0 kg object that is initially at
andrew-mc [135]

Answer:

9.3m/s

Explanation:

Based on the law of conservation of momentum

Sum of momentum before collision = sum of momentum after collision

m1u1 +m2u2 = m1v1+m2v2

m1 = 8kg

u1 = 15.4m/s

m2 = 10kg

u2 = 0m/s(at rest)

v1 = 3.9m/s

Required

v2.

Substitute

8(15.4)+10(0) = 8(3.9)+10v2

123.2=31.2+10v2

123.2-31.2 = 10v2

92 = 10v2

v2 = 92/10

v2 = 9.2m/s

Hence the velocity of the 10.0 kg object after the collision is 9.2m/s

6 0
3 years ago
Help me and you will meet your favorite singer at your door.
Goshia [24]
Sunlight, because it provides a source of energy is the answer because plants also provide a food source (please put as brainliest answer)
4 0
3 years ago
Read 2 more answers
1. A man walks round a park, first walking north for 80m, then turning right and walking
drek231 [11]

Answer:

Total distance travelled = 210m

Explanation:

Distance travelled = 80m + 50m + 10m + 70m

                               = 210m

8 0
3 years ago
An airplane flies eastward and always accelerates at a constant rate. At one position along its path, it has a velocity of 34.5
Inessa05 [86]

Answer:

the acceleration of the airplane is 5.06 x 10⁻³ m/s²

Explanation:

Given;

initial velocity of the airplane. u = 34.5 m/s

distance traveled by the airplane, s = 46,100 m

final velocity of the airplane, v = 40.7 m/s

The acceleration of the airplane is calculated from the following kinematic equation;

v² = u² + 2as

2as= v^2 - u^2\\\\a = \frac{v^2 - u^2}{2s} \\\\a = \frac{(40.7)^2 -(34.5)^2}{2 \times 46,100} \\\\a = 5.06 \ \times \ 10^{-3} \ m/s^2

Therefore, the acceleration of the airplane is 5.06 x 10⁻³ m/s²

5 0
3 years ago
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