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Fiesta28 [93]
3 years ago
8

During the final stages of descent, asky diver with an

Physics
1 answer:
OLEGan [10]3 years ago
3 0

Answer:

Gravity: downwards

Air drag and air-pressure on the inner surface of the the parachute: Upwards

Explanation:

  • If a sky-diver is in the final stages of his descend with open parachute such that the wind is calm and it does not blows him laterally.
  • In such a condition the air resistance in the form of drag and the pressure force due to the air captured in the parachute are acting in the upward direction which balance the force of gravity on the body. But this situation may occur momentarily and then again the diver must begin to slowly descend.

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A car is travelling at 15 m/s on a horizontal road and stopped after 4 s. The coefficient of kinetic friction between the tires
givi [52]

Answer:

Fr,= umg

umg= ma

a= v/t

umg= mv/t

u= v/gt= 0.38

3 0
3 years ago
At what point is the northern hemisphere pointed farthest away from the sun?
inna [77]
<span>Well, It is the aphelion point, When the Earth is farthest away from the Sun, when the Northern Hemisphere is warm. the Earth is closest to the Sun, or at the perihelion, 2 weeks after the June Solstice, when the Northern Hemisphere is enjoying warm summer months. Well this kind of weather is very nice.</span>
7 0
3 years ago
A steel bar of rectangular cross section (1.5 in. 2 3 .0 in.) carries a tensile load P (see fig- ure). The allowable stresses in
lukranit [14]

Explanation:

Value of the cross-sectional area is as follows.

        A = 1.5 \times 2.30

           = 3.45 in^{2}

The given data is as follows.

          Allowable stress = 14,500 psi

          Shear stress = 7100 psi

Now, we will calculate maximum load from allowable stress as follows.

           P_{max} = \sigma_{a}A

                       = 14500 \times 3.45

                       = 50025 lb

Now, maximum load from shear stress is as follows.

           P_{max} = 2 \times \tau_{a} \times A

                      = 2 \times 7100 \times 3.45

                      = 48990 lb

Hence, P_{max} will be calculated as follows.

       P_{max} = min((P_{max})_{\sigma}, (P_{max})_{\tau})

                  = 48990 lb

Thus, we can conclude that the maximum permissible load P_{max} is 48990 lb.

4 0
3 years ago
A constant force of magnitude F acts on an object of mass 0.04kg initially at rest at a point O. If the speed of the object when
vampirchik [111]

Answer:

F = 100 Newtons

Explanation:

F = ?

m = 0.04kg

u = 0m/s ==> u is just an abbreviation for initial velocity, it is conventional.

s = 50m ==> s is just an abbreviation for distance, it is conventional.

v = 500m/s ==> v is just an abbreviation for final velocity, it is conventional.

v^{2} = u^{2} + 2as\\\\=> a = \frac{v^{2} - u^{2}}{2s}\\a = \frac{500^{2} }{2*50}\\a = 2500ms^{-2}

Then F = ma = 0.04 x 2500 = 100N

5 0
3 years ago
Is it possible for a distance versus-time graph to be a vertical line?
Feliz [49]
If time is the x axis and distance is the y axis then yes, in the case that time is going by but distance remains the same.
4 0
3 years ago
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