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kirza4 [7]
3 years ago
14

What are the thee major wind systems

Physics
1 answer:
astra-53 [7]3 years ago
8 0
Trade winds, prevailing westerlies, polar easterlies
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A briefcase sits stationary in an elevator. The mass of the briefcase if 4.5 kg. The elevator then begins accelerating upwards a
Korvikt [17]

Answer:

D. 48.985 N

Explanation:

Newton's second law states that:

\sum F = ma

which means that the net force acting on an object is equal to the product between the object's mass and its acceleration.

The equation of the forces for the briefcase in the elevator therefore is given by:

N-mg=ma

where

N is the normal reaction exerted on the briefcase

(mg) is the weight of the briefcase, with

m = 4.5 kg being its mass

g = 9.8 m/s^2 is the acceleration of gravity

a = 1.10 m/s^2 is the acceleration

Here we chose upward as positive direction.

Solving for N, we find the normal force:

N=mg+ma=m(g+a)=(4.5)(9.81+1.10)=49.095 N

So the closest answer is

D. 48.985 N

3 0
3 years ago
What currents are responsible for powering the movement of tectonic plates ? A. Platonic B. Tectonic C. Convection D. Plutonium
Ivan
The currents responsible for powering the movement of tectonic plates is convection currents which occur in the mantle. As the hotter, less dense liquid rises it displaces the cooler more dense liquid which moves the tectonic plates out of allignment
3 0
3 years ago
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An airplane is flying at an elevation of 5150 ft, directly above a straight highway. two motorists are driving cars on the highw
docker41 [41]
Using trigonometric ratios we can get the distance;
For the first car; The distance from the point on the highway below the plane 
tan = opp/adj
tan(36°) = 5150/x
0.727 = 5150/x
0.727x = 5150
x = 7088.37
For the second car we also use tangent; the distance from the point on the highway below the plane will be; 
tan(56°) = 5150/y
1.483 = 5150/y
1.483y = 5150
y= 3473.72
The we can add the two distances to get how far apart the cars are;
7088.37 + 3473.72 = 10562.09 feet. 
= 10562.09 ft
7 0
3 years ago
A hot-air balloonist, rising vertically with a constant velocity of magnitude v = 5.00 m/s , releases a sandbag at an instant wh
lutik1710 [3]
<span>Data:

Initial velocity upward: Vo = 5.00 m/s ,
Initial position: h = 40.0 m above the ground

Type of motion: free fall.

A) Compute the position of the sandbag at a time 1.05 s after its release.

Equation: y = h + Vo*t - g*(t^2) / 2

y = 40.0 m + 5.00 m/s * 1.05s - (9.8 m/s^2) * (1.05 s)^2 / 2 = 39.8 m

B)Compute the velocity of the sandbag at a time 1.05 s after its release.

Equation: Vf = Vo - g*t

=> Vf = 5.00 m/s - (9.8m/s^2) * (1.05 s) = - 5.29 m/s


Negative sign means that the sandbag is going down.

c) How many seconds after its release will the bag strike the ground?

Equation:

y = yo + Vo*t - g*(t^2) / 2

0 = 40.0 + 5.00t - 4.9 t^2

=> 4.9 t^2 - 5t - 40 = 0

Use the quadratic formula and you get: t = 3.41 s
</span>
7 0
3 years ago
Read 2 more answers
How do you select the appropriate tools for caregiving rated task effectively?​
Gwar [14]

Answer:

Care giving tools like medical devices or any medical supplies which aid the effectiveness of care for the patient will help the quality of work of care giving. ... All of the care giving tool should also be efficient and effective

4 0
3 years ago
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