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Alenkinab [10]
3 years ago
15

Winds that blow from the north and south poles

Physics
1 answer:
Gennadij [26K]3 years ago
8 0
Winds that blow from the north and south poles would be called k<span>atabatic winds. I'm not sure if I spelled that right, but that's the answer I hope.</span>
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The work energy principle states that the change in kinetic energy of an object is equal to the net work done on the object. If
sattari [20]

Answer:

v=\sqrt{2gh}\ m/s

Explanation:

From work energy theorem

Work done by all forces = Change in kinetic energy

Lets take

m= mass of object

h=height from the ground surface

initial velocity of object = 0 m/s

The final velocity of object is v

Work done by gravitational force = m g . h

The final kinetic energy = 1/2 m v²

So

Work done by all forces = Change in kinetic energy

m g h =  1/2 m v² - 0

v² = 2 g h

v=\sqrt{2gh}\ m/s

6 0
3 years ago
A train is accelerating at a rate of 2m/s2. if it's initial velocity is 20 m/s , what is its velocity after 30 seconds
netineya [11]

(v₂-v₁)/t = a

(v₂-20)/30 = 2

v₂-20 = 60

v₂ = 80 m/s

6 0
3 years ago
What statement best describes what it means to maximize your efforts in sports?
Yuliya22 [10]

What statement best describes what it means to maximize your efforts in sports?

D.none of the above

5 0
3 years ago
Two forces are acting on an object. The first force has magnitude F1=33.4 N and is pointing at an angle of θ1=23.8 clockwise fro
marishachu [46]

Answer:

Fe= 28.2 N : Magnitude of the equilibrant (Fe)

β = 18.34° , clockwise from the positive x axis

Explanation:

Concept of the equilibrant

It is called equilibrant  to a force with the same magnitude and direction as the resulting one (in case it is non-zero) but in the opposite direction. Adding vectorially to all the forces (that is to say the resulting one) with the equilibrant you get zero

To solve this problem we decompose the forces given into x-y components to find the resulting force:

Look at the attached graphic

F₁= 33.4 N  , θ₁=23.8° clockwise from the positive y axis (y+)

F₁x= 33.4 *sin23.8° = 13.48 N

F₁y= 33.4 *cos23.8° =30.6 N

F₂=46.1 N ,  θ₂=28.8 counterclockwise from the negative x axis (x-)

F₂x= -46.1 *cos28.8° = -40.4 N

F₂y=  -46.1 *sin28.8° =  -22.2 N

Components of the resultant in x-y R(x,y)

Rx= 13.48 N -40.4 N = - 26.92 N

Ry= 30.6 N  -22.2 N =  + 8.4 N

Components of the equilibrant in x-y Fe(x,y)

Fex= +26.92 N

Fey=  - 8.4 N

Magnitude of the equilibrant (Fe)

F_{e} = \sqrt{(F_{ex})^{2}+{(F_{ey})^{2}  }

F_{e} = \sqrt{(26.92)^{2}+(8.4)^{2}  }

Fe= 28.2 N

Angle the equilibrant makes with the x axis ( β)

\beta = tan^{-1} (\frac{F_{ey} }{F_{ex} } )

\beta = tan^{-1} (\frac-8.4 }{26.92 } )

β = -18.34°                  

β = 18.34° , clockwise from the positive x axis

8 0
3 years ago
Do magnetic field lines ever start or end anywhere?
nevsk [136]

Answer:

No

Explanation:

The lines of the field of a magnet don't begin or stop at anyplace, they generally make shut circles or loops and will proceed inside magnet (however here and there they are not drawn along these lines). We require an approach to show the bearing of the field.  

The field lines of a magnet don't simply end at the magnetic tip. They go directly through it, so that inside the magnet the magnetic field lines indicates from the south to the north pole.

4 0
4 years ago
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