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Vinvika [58]
3 years ago
15

15 PTS! HELP! BRAINLIEST!

Physics
1 answer:
vodka [1.7K]3 years ago
3 0
I cant do all but I can do number 4
      When force moves ur leg u start skatin and when that happenes ur using a lot of force to move
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The answers to these please
Mrrafil [7]

Answer:

11. Depression

Explanation:

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2 years ago
Instruments in an airplane which is in level flight indicate that the velocity relative to the air (airspeed) is 180.00 km/h and
snow_lady [41]

Answer:

Explanation:

From the given information:

The coordinate axis is situated in the east and north direction.

So, the north will be the  y-axis and the east will be the x-axis

Similarly, the velocity of the plane in regard to the air in the coordinate system will be v_{P/A} = v( cos \theta \ i + sin \theta \ j)

where:

v_{P/A} = velocity of the plane in regard to the air

v = velocity

θ =  angle of inclination of the plane with respect to the horizontal

replacing v = 180 km/ and θ = 20° in above equation, then:

The velocity of the airplane in the coordinate system as:

v_{P} = v_o( cos \phi \ i + sin \phi \ j)

where;

v_p = velocity of the airplane

v_o = velocity

∅ = angle of inclination with regard to the base axis;

Then; replacing  v_o  = 150 km/h and ∅ = 30°

Therefore, the velocity of the plane in the system is :

v_p = v_A + v_{P/A}

v_A=  v_P  -v_{P/A}   --- (1)

v_A= ( 150 cos 30° - 180 cos 20°)i + ( 150 sin 30° - 180sin 20°)j

v_A= (-39.24 km/h)i + (13.44 km/h) j

The magnitude is:

v_A= (-39.24 km/h)i + (13.44 km/h) j

|v_A|^2 = \sqrt{ (-39.24 km/h)^2+ (13.44 km/h)^2}

v_A = 41.48 km/h

The airplane is moving at an angle of the inverse tangent to the abscissa and ordinate.

The angle of motion is:

tan θ = 39.24/13.44

tan θ = 2.9

θ  = tan ^{-1} (2.9)

θ  =  70.97°

The angle of motion is  70.97° from west of north with a velocity of 41.48 km/h.

5 0
3 years ago
Water is a very unique substance because it can exist in all three phases of matter (solid, liquid, gas) within the normal tempe
ahrayia [7]
 the answer is qualitative
5 0
3 years ago
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The earth has a vertical electric field at the surface, pointing down, that averages 100 N/C. This field is maintained by variou
zlopas [31]

Answer:

q = 3.6 10⁵  C

Explanation:

To solve this exercise, let's use one of the consequences of Gauss's law, that all the charge on a body can be considered at its center, therefore we calculate the electric field on the surface of a sphere with the radius of the Earth

          r = 6 , 37 106 m

          E = k q / r²

          q = E r² / k

          q = \frac{100 \ (6.37 \ 10^6)^2}{9 \ 10^9}

          q = 4.5 10⁵ C

Now let's calculate the charge on the planet with E = 222 N / c and radius

           r = 0.6 r_ Earth

           r = 0.6 6.37 10⁶ = 3.822 10⁶ m

           E = k q / r²

            q = E r² / k

            q = \frac{222 (3.822 \ 10^6)^2}{ 9 \ 10^9}

            q = 3.6 10⁵  C

4 0
3 years ago
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