A
Explanation:
The earth is spherical. So it's middle part is bulgjng outside. So more sunlight will be incident on the latitude near the equator. This will heat the air and it will rise up. This will cause high pressure difference and polar disturbances.
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Answer to A spring<span> is </span>stretched<span> to a </span>displacement<span> of </span>3.4 m<span> from </span>equilibrium<span>. </span>Then<span> the </span>spring<span> is</span>released<span> and ... </span>Then<span> the </span>spring<span> is </span>released<span> and </span>allowed<span> to </span>recoil<span> to a </span>displacement<span> of </span>1.9 m<span> from</span>equilibrium<span>. The </span>spring constant<span> is </span>11 N/m<span>. What </span>best describes<span> the </span>work involved<span> as the </span>spring recoils<span>? A)87 J of </span>work<span> is performed ...</span>
Answer:

Explanation:
Force can be found by multiplying the mass by the acceleration.

The mass of the roller coaster is 2000 kilograms and the acceleration is 2 meters per second squared.

Substitute the values into the formula.

Multiply.

- 1 kg*m/s² is equal to 1 N
- Therefore our answer of 4000 kg*m/s² is equal to 4000 Newtons

The net force acting on the roller coaster is <u>4000 Newtons.</u>
Answer:
(c) no different than on a low-pressure day.
Explanation:
The force acting on the ship when it floats in water is the buoyant force. According to the Archimedes' principle: The magnitude of buoyant force acting on the body of the object is equal to the volume displaced by the object.
Thus, Buoyant forces are a volume phenomenon and is determined by the volume of the fluid displaced.
<u>Whether it is a high pressure day or a low pressure day, the level of the floating ship is unaffected because the increased or decreased pressure at the all the points of the water and the ship and there will be no change in the volume of the water displaced by the ship.</u>
Answer:
the pressure at B is 527psf
Explanation:
Angular velocity, ω = v / r
ω = 20 /1.5
= 13.333 rad/s
Flow equation from point A to B
![P_A+rz_A-\frac{1}{2} Pr_A^2w^2=P_B+rz_B-\frac{1}{2} pr^2_Bw^2\\\\P_B = P_A + r(z_A-z_B)+\frac{1}{2} pw^2[(r_B^2)-(r_A)^2]\\\\P_B = [25 +(0.8+62.4)(0-1)+\frac{1}{2}(0.8\times1.94)\times(13.333)^2[2.5^2-1.5^2] ]\\\\P_B = 25 - 49.92+551.79\\\\P_B = 526.87psf\\\approx527psf](https://tex.z-dn.net/?f=P_A%2Brz_A-%5Cfrac%7B1%7D%7B2%7D%20Pr_A%5E2w%5E2%3DP_B%2Brz_B-%5Cfrac%7B1%7D%7B2%7D%20pr%5E2_Bw%5E2%5C%5C%5C%5CP_B%20%3D%20P_A%20%2B%20r%28z_A-z_B%29%2B%5Cfrac%7B1%7D%7B2%7D%20pw%5E2%5B%28r_B%5E2%29-%28r_A%29%5E2%5D%5C%5C%5C%5CP_B%20%3D%20%5B25%20%2B%280.8%2B62.4%29%280-1%29%2B%5Cfrac%7B1%7D%7B2%7D%280.8%5Ctimes1.94%29%5Ctimes%2813.333%29%5E2%5B2.5%5E2-1.5%5E2%5D%20%20%5D%5C%5C%5C%5CP_B%20%3D%2025%20-%2049.92%2B551.79%5C%5C%5C%5CP_B%20%3D%20526.87psf%5C%5C%5Capprox527psf)
the pressure at B is 527psf